Titan Network
Community => General Discussion => Topic started by: avelworldcreator on January 22, 2016, 09:45:19 PM
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Seeing the latest argument in this thread reminds me of the debate earlier on it about the lottery. I noticed no one distiguished between instantaneous probability (single trial) vs probablity over a number of trials. An example would be using a 4-sided dice (yes, I could have used the more common 6-sider but that forces me to use rounded numbers for decimals). One roll has only a 1 in 4 chance of a three showing up, or a 3/4 chance of it not showing up - 75%. TWO rolls gives only a 9/16 or only 56.25% of it not showing up (3/4)^2. At THREE trials the odds for it NOT appearing drops to 42.1875% (3/4)^3, and so on. So the actual odds of winning the lottery by an individual is actually based on the number of times x people play it not just the computed chance of a specific number combination showing. That's why you actually see winners almost every year or so with major wins. The number of people playing actually improves the odds of winning for everyone (increases the number of trials). But what would I know? I just tutor in multivariate calculus and statistics. :P
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Seeing the latest argument in this thread reminds me of the debate earlier on it about the lottery. I noticed no one distiguished between instantaneous probability (single trial) vs probablity over a number of trials. An example would be using a 4-sided dice (yes, I could have used the more common 6-sider but that forces me to use rounded numbers for decimals). One roll has only a 1 in 4 chance of a three showing up, or a 3/4 chance of it not showing up - 75%. TWO rolls gives only a 9/16 or only 56.25% of it not showing up (3/4)^2. At THREE trials the odds for it NOT appearing drops to 42.1875% (3/4)^3, and so on. So the actual odds of winning the lottery by an individual is actually based on the number of times x people play it not just the computed chance of a specific number combination showing. That's why you actually see winners almost every year or so with major wins. The number of people playing actually improves the odds of winning for everyone (increases the number of trials). But what would I know? I just tutor in multivariate calculus and statistics. :P
I don't specifically recalling a discussion point where this fact is significant. I don't recall (I could be wrong) someone asserting what the odds of someone winning are, which is based on how many entries are put in and what their distribution are. However, this statement:
The number of people playing actually improves the odds of winning for everyone
is false. I hope you didn't teach that to anyone. The odds of *someone* winning at all is based on the number of *different* powerball combinations that have been played. Since there are 292,201,338 possible combinations, the odds of *someone* winning on a particular draw if N *different* combinations have been entered is N/292201338. If every single combination has been put in at least once, then the odds are 1.0, or a certainty.
What you're talking about is something different. Suppose there are N different entries in the Powerball, and the rules stated that the Powerball operators were to continue to draw powerball combinations *until* someone won. In that case, the odds of the first one generating a winner are N/P (where P = 292... ). The odds that the first draw does not generate a winner would be 1-N/P. The odds that D successive draws would not generate a winner are (1-N/P) ^ D, meaning the odds of a series of draws not generating a winner get lower over time; the odds of drawing a winner increase with more draws.
However, that's not how powerball is played. Hypothetically speaking, if you bought tickets using the computer to randomly pick the numbers, then the odds of you having the winning number assuming you never draw a duplicate number series ever are 1 - (1-N/P)^T where T is the number of tickets. In effect, given those criteria each random draw is in effect an attempt to hit the winning target randomly, and the odds of never hitting the target follow the same formula. However, the only way to do that is to randomly select your ticket numbers yourself, rejecting duplicates. The lottery computers will not do that for you.
The more people that play, the greater the chances of *someone* winning. But it doesn't increase the chances for everyone to win. Your individual odds are the same no matter what. However, how many people play does have an effect on your statistical return per ticket, because more players both increases the size of the jackpot if you win and increases the probability you'll have to share that jackpot if you win. But unless you coordinate your efforts with those other people, it doesn't affect your personal odds of actually winning at all.
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I don't specifically recalling a discussion point where this fact is significant. I don't recall (I could be wrong) someone asserting what the odds of someone winning are, which is based on how many entries are put in and what their distribution are. However, this statement:
The number of people playing actually improves the odds of winning for everyone
is false. I hope you didn't teach that to anyone. The odds of *someone* winning at all is based on the number of *different* powerball combinations that have been played. Since there are 292,201,338 possible combinations, the odds of *someone* winning on a particular draw if N *different* combinations have been entered is N/292201338. If every single combination has been put in at least once, then the odds are 1.0, or a certainty.
What you're talking about is something different. Suppose there are N different entries in the Powerball, and the rules stated that the Powerball operators were to continue to draw powerball combinations *until* someone won. In that case, the odds of the first one generating a winner are N/P (where P = 292... ). The odds that the first draw does not generate a winner would be 1-N/P. The odds that D successive draws would not generate a winner are (1-N/P) ^ D, meaning the odds of a series of draws not generating a winner get lower over time; the odds of drawing a winner increase with more draws.
However, that's not how powerball is played. Hypothetically speaking, if you bought tickets using the computer to randomly pick the numbers, then the odds of you having the winning number assuming you never draw a duplicate number series ever are 1 - (1-N/P)^T where T is the number of tickets. In effect, given those criteria each random draw is in effect an attempt to hit the winning target randomly, and the odds of never hitting the target follow the same formula. However, the only way to do that is to randomly select your ticket numbers yourself, rejecting duplicates. The lottery computers will not do that for you.
The more people that play, the greater the chances of *someone* winning. But it doesn't increase the chances for everyone to win. Your individual odds are the same no matter what. However, how many people play does have an effect on your statistical return per ticket, because more players both increases the size of the jackpot if you win and increases the probability you'll have to share that jackpot if you win. But unless you coordinate your efforts with those other people, it doesn't affect your personal odds of actually winning at all.
What I meant is that increasing the number of players increases the probability that *someone* will win because that means the number of draws goes up directly. If we distribute the win chance across the number of players then, yes, the odds of improve for everyone. The increase is, admittedly, tiny with numbers this large but it *is* there. When the overall chances of a win go up the individual players chances do not remain static.
If 292,201,338 different combination are entered on a single trial then the odds are certainly 100%! But in practice, even if you had that number people actually playing the probability of 292,201,338 unique values being selected is not 100%. For that to happen you would have to make sure every player's draw is unique and therefore the trials would not be independent.
I simplified the binomial distribution formula. It's what is used when you have multiple, independent trials with two possible outcomes (which the lottery is).
Wikipedia: Binomial Distribution (https://en.wikipedia.org/wiki/Binomial_distribution).
N is the total number of trials in question. If your sample size is equal to the number of trials then N=n and the formula reduces to p^k (obviously the 1-p term and the binomial coefficient reduce to 1 when the sample size and the population are the same value - that was my simplification). Then the next question becomes determining the correct value of p. Should it be chosen to be the instantaneous chance of loss or win? Are we looking for the chance of someone winning every time or losing? In practice p is selected to be the larger value. In this specific case it would be 1-1/292201338 or 292201337/292201338. So our formula is (292201337/292201338)^k. Now the next question is over how many trials does this number drop to 50% or less? When do the chances of winning become equal to, or greater than the chances of losing? We are looking for the solution to the formula (292201337/2992201338)^k<=.5 and that can be calculated by taking the log base p of .5 which is (applying the ceiling function to round) 202,538,534. Notice that is awfully close to the number of possible combinations.
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oh crap did some one get into a Math fight with Arcana ? TAKE COVER !
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oh crap did some one get into a Math fight with Arcana ? TAKE COVER !
He has a reputation? Huh. It took me a week to learn calculus (including differential equations) before I was 19. Been at it about 30 years. Never quit my studies. It's even one of my hobbies.
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He has a reputation? Huh. It took me a week to learn calculus (including differential equations) before I was 19. Been at it about 30 years. Never quit my studies. It's even one of my hobbies.
Arcana is a she, and yes. It's something of a joke 'round City parts.
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Arcana is a she, and yes. It's something of a joke 'round City parts.
Thank you for that correction. Any insult is the unintentional result of my ignorance.
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He has a reputation? Huh. It took me a week to learn calculus (including differential equations) before I was 19. Been at it about 30 years. Never quit my studies. It's even one of my hobbies.
Indeed, She has a reputation, she worked on City of Heroes, and as far as I can tell her position was dealing with the statistical side of things.
Seriously game statistics are no joke, by definition they are calculus, where dynamic changable values are compared and processed against other dynamic changable values. The work of a Statistical engineer is rough, because you have to plan all the numbers carefully, especially Res as it's a percent and you don't want it to be too easily capped, and when res is coming from multiple sources including your powers, others powers and items, you need to carefully consider how much /this/ power should give at max (with and without all enhancements), and then reconsider it based on level and find the lowest base res for the power compared to all the powers and the amounts of damage that will hit on average at the lowest level then compile the lowest res the power can give based on this, however not all res powers are equal so you need to define a high and low range for both the lowest and highest level situations then from there it's a decision as to the distribution based on what kind of powerset it is and what it's hallmarks should be.
and that's if the power isn't a dual effect res/def, and that's if you already have a well defined stat system, if you are starting from scratch and trying to include a whole bunch of stat customization then you have a serious headache.
I hardly think that is the limits of Arcana's ability or expertise either.
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Given I'm one of the math people for CoT I'm familiar with the issues (the math team consists of three people, one guy who also does advanced statistical analysis of map data for a corporation who's clients include national governments, a professor of theoretical physics, and myself. I'm considered the second most skilled person on the subject). In fact I developed a clamping function for certain game stats. I based it on a formula from relativity I've spent a few decades analyzing (I'll try to attach the image here). I'm also internationally published in statistical population modeling with game applications. She tried to rebut one of my statements (and I admit I made my statement a little loosely - I'm not offended) and I corrected her. It just happened to be a specific area of statistical analysis I've spent a couple of decades on. That's not a weakness on her part, in fact she saved me some effort on calculating some values on my return argument. I'd love to have her on our team but we don't have a budget for her (we are all working as volunteers).
Statistics isn't calculus but it is an area where calculus is applied. I suspect that's what you meant.
Oh, and the formula:
(https://images.weserv.nl/?url=gallery.thebasketcasesoftware.com%2Fpiwigo%2Fupload%2F2016%2F01%2F22%2F20160122234009-33be0b0c.png)
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CoT sounds like it's only going to be for smart people.Guess I'm out.
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CoT sounds like it's only going to be for smart people.Guess I'm out.
Nah. It's for people who want to have fun. Of course there are going to be a few "eggheads" around. We're doing the hard work so everyone else can have a good time. In our UI we have both "simple" and "advanced" modes. We are aiming to work with Able Gamers and people with learning disabilities as much as we can without disadvantaging the quality of the experience for everyone else. We are group of volunteers who loved CoH and have spent a lot of our personal time and cash to get it to where it is now and hope to see it be enjoyed and for it last for years to come. It's part of a strategy to get back somehow what we lost with CoH's closing.
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He has a reputation? Huh. It took me a week to learn calculus (including differential equations) before I was 19. Been at it about 30 years. Never quit my studies. It's even one of my hobbies.
Guess I'm a little surprised of someone working on a spiritual heir to City of Heroes project not knowing one of the prominent CoH community members. If you don't mind my asking, do you have any experience with CoH? It would worry me if CoT was trying to recapture the flavor of something the developers hadn't experienced themselves.
Another worry, has nobody from CoT already approached Arcana for at least an extra set of eyeballs looking at design decisions? Pick her brain as much as she is willing to let you!
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CoT sounds like it's only going to be for smart people.Guess I'm out.
Looks like that math degree I didn't get--I got tired of going to school and realized that despite having more hours in upper division math than I had in all of computer science, I was (a) one semester away from the Comp Sci degree (b) three semesters* away from the math degree due to the haphazard way I had taken my classes and (c) three semesters from the dual degree--may get some use.
Never could understand programmers who had an aversion to math...its just way too handy.
*Three semester because the I was three courses short, one of which was a pre-requisite for the other and not offered until the following semester--the school had a horrible habit of offering certain classes only every other year in a single semester.
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Due to some difficult personal circumstances I was away from these forums for an extended period (theft of my equipment,etc. when I'm self-employed and it was a major part of my livelihood - it interrupted my work with CoT as well, but it didn't take much to get back on track when I did). Did I have any experience with CoH? I had transferred my level 50 character that I had created in 2005 to the beta based on a rumor that it would be the last server to close down. My last memories was seeing it kissing the pavement as I saw the the server connection being lost. Remember seeing the "construction" projects for the player arenas. I remember using the map hack to badge hunt and did so in all three "sides". I remember jacking the difficulty level for this character to max (a MA regen scrapper) when I was first learning the game and forgetting I had done it for months. Oddly I still did well doing mostly solo stuff despite having so many of my targets in missions conning red and purple and UV. I even remember buying extra RAM to play the game better and actually taking a personal loan to do it. I think we have only about two people with us that have never played the game. If you look at my posting history on this forum you will find I wrote our original IP agreement trying to protect the contributions of our volunteers. I also took care of doing the actual initial set up of Missing Worlds Media as a corporation, establishing our domain name, and creating and managing our original forums. And of course I had multiple characters on several servers at the end. I miss them all. How much experience in game are you expecting or consider "adequate"? I started playing in early 2005 both on the regular and beta servers.
I'd love to have Arcana on the team! We are quite stretched at the moment and every extra hand is welcome. We need artists, writers, and good coders. Currently I've been working on: The costume creator UI, the web page redesign, the chat system, the game loader and user authentication system, trying my hand at mission writing, trying to figure out how to best do combat simulations to tweak settings, creating UI related art assets, assisting our PR person, and looking into corporate level set up issues such as communications services.
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The number of people playing actually improves the odds of winning for everyone
What I meant is that increasing the number of players increases the probability that *someone* will win because that means the number of draws goes up directly. If we distribute the win chance across the number of players then, yes, the odds of improve for everyone.
K I'm not a math but why would we distribute the added win chance?
The argument struck me as less about math than about language - i.e. the increase in the odds of somone winning vs the increase in "everyone's" odds, i.e. each person's odds.
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K I'm not a math but why would we distribute the added win chance?
The argument struck me as less about math than about language - i.e. the increase in the odds of somone winning vs the increase in "everyone's" odds, i.e. each person's odds.
"Distribute evenly" is just a fancy way of saying I'm dividing by the number of people and giving them an equal "share". Think about it. If you had a cake and the winning chance is the portion that has chocolate icing, and the losing part is the part without, if you spread the frosting more (increase the chances of a win) the chance of a random slice getting frosting gets better. Everyone gets their own cake and each one only gets one random slice of the cake. Not only does the chance of frosting on their slice go up, but chances of more than one person getting frosting goes up too.
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"Distribute evenly" is just a fancy way of saying I'm dividing by the number of people and giving them an equal "share". Think about it. If you had a cake and the winning chance is the portion that has chocolate icing, and the losing part is the part without, if you spread the frosting more (increase the chances of a win) the chance of a random slice getting frosting gets better. Everyone gets their own cake and each one only gets one random slice of the cake. Not only does the chance of frosting on their slice go up, but chances of more than one person getting frosting goes up too.
That's a false analogy. Spreading more icing increases my odds because it reduces the number of losing pieces. That would be more akin to the increased odds from having multiple tickets with different number combinations. [Edit: No it wouldn't, it'd be like if they decided once 100 million tickets were sold they'd draw one less number.]
More people playing definitely increases the odds that someone will win, even multiple people. But it does nothing for the odds of each individual number combination being the winning combination. That'd still be one in 600whatever million. Just as if I roll a die the odds of my getting a six aren't affected by other people, even several hundred million of them, rolling dice as well.
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That's a false analogy. Spreading more icing increases my odds because it reduces the number of losing pieces. That would be more akin to the increased odds from having multiple tickets with different number combinations.
More people playing definitely increases the odds that someone will win, even multiple people. But it does nothing for the odds of each individual number combination being the winning combination. That'd still be one in 600whatever million. Just as if I roll a die the odds of my getting a six aren't affected by other people, even several hundred million of them, rolling dice as well.
I admit it wasn't the best overall image. Quite the hack in fact. But it's not a false analogy. The model was to demonstrate the chances of an individual having the winning number if it got picked; not to demonstrate the increase in odds that a particular number would be the one selected.
Try dice tossing for example using a standard 6-sided die. If I only toss a single die the chances of a 1 showing up is 1 in 6 but the odds of it not showing up is 5 in 6. If I double the number of dice that odds of not showing becomes 25 in 36 which means that if you slice the cake into 36 pieces 25 of them lack frosting. At three dice it goes to 125 in 216 meaning our cake now has 125 pieces without frosting and 91 with. At 4 it is 625 in 1296 (yeah, I'm really doing these numbers in my head at 4am!) which comes to 671 WITH frosting and 625 without. Only 4 trials to more than break even. Is a lottery a fair system though? Nope. It has a lot of similarity to a Ponzi scheme but without the lie you will actually be guaranteed to win if you play. If you got people to play, let the money pool for a while, then split the wealth equally among the actual players, then the system would be fair. Of course this would be less the vig (the public use portion of the monies). Can you attract people to invest in such a system? Yes. Let them win an extra portion (say 10%) lottery style. A definite return with a slight chance of even greater return after a period of investment? With community improvements? Would you play? I know I would.
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Yeah we all agree that more people playing increases the chance someone will win. It was the part about increasing everyone's odds where the quibble came in. Which is why I said it's more of a language disagreement than a math one . You were using 'everyone' in the aggregate sense rather than to mean everyone who played had a better chance of being the winner. So really not even a disagreement at all, just a different use of terms.
The cake example doesn't work for a lottery though since the lottery has a static number of losing number combinations no matter how many people play. You can add all the icing you want but it always goes on the same piece of cake. The more people who play the more icing there is but there's always only the one piece of cake with icing (setting aside of course non-jackpot wins. But I think those payouts are static and that territory was covered in the original conversation.) And of course the more players the higher the likelihood that the iced piece will need to be divided.
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How much experience in game are you expecting or consider "adequate"?
I've gotta say, that whole paragraph comes across a little defensive over an honest question. I'd think just about anyone else participating in this thread would find a dedicated CoH fan not having heard of Arcana almost as odd as, say, not knowing what IOs are. Its just part of the cultural background noise.
If there was one fault I'd assign to the CoT team its that their customer facing has been a bit rough. I'm sure they are an excellent team of developers, programmers, and graphics designers but the public relations could use some attention as well.
I'd love to have Arcana on the team!
Good to hear it! Arcana, anyone from the CoT team approach you if for nothing else than a rough consulting session?
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I've gotta say, that whole paragraph comes across a little defensive over an honest question. I'd think just about anyone else participating in this thread would find a dedicated CoH fan not having heard of Arcana almost as odd as, say, not knowing what IOs are. Its just part of the cultural background noise.
If there was one fault I'd assign to the CoT team its that their customer facing has been a bit rough. I'm sure they are an excellent team of developers, programmers, and graphics designers but the public relations could use some attention as well.
Good to hear it! Arcana, anyone from the CoT team approach you if for nothing else than a rough consulting session?
they will collaborate(or not) depending on desire and or time constraints. He may very well not have heard of arcana if he did not frequent the forums. I knew of her because she kept on putting out thos posts with inconvenient(but mathematically sound) facts that I didnt want to hear. too bad for me. If I had not had time on my hands at work I would not have spent near as much time in the forums.
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I'd think just about anyone else participating in this thread would find a dedicated CoH fan not having heard of Arcana almost as odd as, say, not knowing what IOs are. Its just part of the cultural background noise.
I don't find it odd at all. Granted if you hung out in the CoH forums it was hard to miss Arcanaville. But the vast majority of the playerbase rarely set foot in the forums. In my circle of a dozen friends there's one other who knows for certain who Arcanaville is and another other frequent forum poster who might. The rest of them have no clue who she is; this despite the fact that they were actively managing supergroups, building bases, and hanging with the RP community.
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I've gotta say, that whole paragraph comes across a little defensive over an honest question. I'd think just about anyone else participating in this thread would find a dedicated CoH fan not having heard of Arcana almost as odd as, say, not knowing what IOs are. Its just part of the cultural background noise.
If there was one fault I'd assign to the CoT team its that their customer facing has been a bit rough. I'm sure they are an excellent team of developers, programmers, and graphics designers but the public relations could use some attention as well.
Good to hear it! Arcana, anyone from the CoT team approach you if for nothing else than a rough consulting session?
It wasn't an "honest" question. It was an accusation and it shifted the burden of supporting or denying the claim to the accused. It assumed specific knowledge based on circumstance without establishing the particular situation substantially contributed to the likely existence of such knowledge. If the activity in question (playing the City of Heroes game or any of its connected products) required knowledge of certain persons, or if these peoples' identities were placed in a prominent position for an extended period of time then there would be substance to that question but clearly there wasn't. The game credits that appeared during game start only reflected the studio name or the name of the parent company (Paragon, NC Soft). The team might have been reflected in an "About" menu or on an online site in a relevant section but during normal play there would be no reason to examine these items outside of personal curiosity. Then it didn't just stop there. The accusation was then extended to an entire team of people who had invested a great deal of personal time and money into the project. And then to compound the mistake it was compounded further by trying to mask the inappropriate nature of the accusation by trying to minimize it as an "honest question". In law this is known as a "leading question" in the areas of logic or philosophy it also known as a "complex question", "compound question", or similar. Saying "that whole paragraph comes across as a little defensive for an honest question" is not only victim-blaming but it's frankly, a lie. Then you make an overdrawn appeal to the majority by saying "I'd think just about anyone else participating in this thread would find a dedicated CoH fan of not having heard of Arcana almost as, say, not knowing what IOs are." This is also an example of the "No true Scotsman" argument. My paragraph wasn't defensive - it was angry. You seriously insulted me and people I'm close to.
Our customer facing is rough? Have you checked such gaming publications as Massively? We have one (1) guy handling all our social media, press releases, and forum announcement follow ups. and he's been doing quite well. If you check our Facebook page you will see that we have been rated as "Very responsive" to the public. This is also said by the people who regularly frequent our forum.
And, yes, I'm still a bit angry.
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CoT sounds like it's only going to be for smart people.Guess I'm out.
COH and all good games have their math in the background. I appreciate it, but don't care. I just want a fun game.
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I don't find it odd at all. Granted if you hung out in the CoH forums it was hard to miss Arcanaville. But the vast majority of the playerbase rarely set foot in the forums. In my circle of a dozen friends there's one other who knows for certain who Arcanaville is and another other frequent forum poster who might. The rest of them have no clue who she is; this despite the fact that they were actively managing supergroups, building bases, and hanging with the RP community.
I regret not learning the names of other players and the developers until the last two years of the game. I played from beta to close. I only saw what was on the COH home page and facebook posts. I thought the forums was more for technical support and just general game announcements.
I miss not just the games but the official forums for accessible people and helpful players in general.
....I went to paragonwiki FAR more than the official forums for many of my how-to's and questions...
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It wasn't an "honest" question. It was an accusation and it shifted the burden of supporting or denying the claim to the accused. It assumed specific knowledge based on circumstance without establishing the particular situation substantially contributed to the likely existence of such knowledge. If the activity in question (playing the City of Heroes game or any of its connected products) required knowledge of certain persons, or if these peoples' identities were placed in a prominent position for an extended period of time then there would be substance to that question but clearly there wasn't. The game credits that appeared during game start only reflected the studio name or the name of the parent company (Paragon, NC Soft). The team might have been reflected in an "About" menu or on an online site in a relevant section but during normal play there would be no reason to examine these items outside of personal curiosity. Then it didn't just stop there. The accusation was then extended to an entire team of people who had invested a great deal of personal time and money into the project. And then to compound the mistake it was compounded further by trying to mask the inappropriate nature of the accusation by trying to minimize it as an "honest question". In law this is known as a "leading question" in the areas of logic or philosophy it also known as a "complex question", "compound question", or similar. Saying "that whole paragraph comes across as a little defensive for an honest question" is not only victim-blaming but it's frankly, a lie. Then you make an overdrawn appeal to the majority by saying "I'd think just about anyone else participating in this thread would find a dedicated CoH fan of not having heard of Arcana almost as, say, not knowing what IOs are." This is also an example of the "No true Scotsman" argument. My paragraph wasn't defensive - it was angry. You seriously insulted me and people I'm close to.
Our customer facing is rough? Have you checked such gaming publications as Massively? We have one (1) guy handling all our social media, press releases, and forum announcement follow ups. and he's been doing quite well. If you check our Facebook page you will see that we have been rated as "Very responsive" to the public. This is also said by the people who regularly frequent our forum.
And, yes, I'm still a bit angry.
I am a kickstarter contributer. I have been very happy with MWM communications. admittedly, I dont even look at the lore posts, I prefer to get that in game, but thats just me. I have learned a lot about the nuts and bolts of game design from reading the tech posts, and people like irish girl and VO. I cant wait to see a character creator, but that will come sooner or later. I am curious as to when you will go full time, but thats a big step. VO has become valium online lately, but that is from the outside looking in. they are very busy doing many important tasks to go to alpha. Its easy to sit in my recliner and think that someone should be quicker getting a game going, not so easy to do it.
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From what I remember, City of Titans is going by a "everyone has even stats and everyone follows the same rules" formula. So math will not be as huge of a deal like it was in city of heroes. By following consistency of rules, they'll be able to ensure that powers behave as you always expect them to rather than powers say, changing for pvp. Thats how games become challenging rather than punishing, and challenging games rarely ever need any math. They are made so your challenged but your always learning, which is a game I can get behind.
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From what I remember, City of Titans is going by a "everyone has even stats and everyone follows the same rules" formula. So math will not be as huge of a deal like it was in city of heroes. By following consistency of rules, they'll be able to ensure that powers behave as you always expect them to rather than powers say, changing for pvp. Thats how games become challenging rather than punishing, and challenging games rarely ever need any math. They are made so your challenged but your always learning, which is a game I can get behind.
We use a system called MEDIC for combat and something called "Lensing" for level differences. The math is simple and uniform involving basic operations. We try to keep things in terms of relative percentages for uniformity.
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I am a kickstarter contributer. I have been very happy with MWM communications. admittedly, I dont even look at the lore posts, I prefer to get that in game, but thats just me. I have learned a lot about the nuts and bolts of game design from reading the tech posts, and people like irish girl and VO. I cant wait to see a character creator, but that will come sooner or later. I am curious as to when you will go full time, but thats a big step. VO has become valium online lately, but that is from the outside looking in. they are very busy doing many important tasks to go to alpha. Its easy to sit in my recliner and think that someone should be quicker getting a game going, not so easy to do it.
If you saw the posts about that color picker mockup that was my work (and it's been since improved and had some design issues fixed). And that was part of the character creator design. At the moment we are waiting on an outside contractor to finalize our character base and bill us to complete some stuff. I wish I could say/show more. While waiting I've been working on the login/authentication system (which includes the sign on screen), the game lobby, and maybe the opening cinematics. We are pretty darn close to releasing the character build system as a standalone product but how close is difficult to say when you are waiting on an outside party.
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The number of people playing actually improves the odds of winning for everyone
What I meant is that increasing the number of players increases the probability that *someone* will win because that means the number of draws goes up directly. If we distribute the win chance across the number of players then, yes, the odds of improve for everyone. The increase is, admittedly, tiny with numbers this large but it *is* there. When the overall chances of a win go up the individual players chances do not remain static.
Your calculations are correct but your math is wrong. If I put in a single ticket, the odds of my winning on that draw are one in 292,201,338. Those odds remain constant no matter how many other people enter. The correct way to think out this is to consider that when they draw the balls is just a procedural issue for the lottery: the power ball lottery is mathematically identical to a lottery in which the winning numbers were drawn ahead of time and kept secret, and every ticket entered by a participant is a single attempt to match that draw. When I buy a ticket, the odds of that one ticket matching the winning combination is one in 292,201,338. Those odds are unaffected by any event that happens afterward. If no one else buys a ticket, the odds of that ticket matching the winning numbers is one in 292,201,338. If someone else buys a ticket afterward, the odds remain one in 292,201,338.
There's no such thing as "distributing the win chance among the players." In this context, that is a statistically meaningless statement. The chance of a set of numbers matching the winning powerball combination, either in the real powerball lottery or the modified one above, are independent of all other draws. The only probability that *is* altered by the number of people entering is the chance of having *any* winner. Those odds increase with more entrants. But the odds of any particular *ticket* winning is fixed at one in 292,201,338.
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Try dice tossing for example using a standard 6-sided die.
This is a good analogy to build on. The lottery numbers are big which make them intuitively unwieldy. But conceptually speaking, we can reduce the size of the lottery to a single number: one six sided die. Lets imagine a lottery in which I roll a single six sided die and people are allowed to enter that lottery by entering tickets with a single number, 1 through 6. The odds for a single ticket to win are one in six, I believe everyone will agree. Now suppose A enters the lottery and enters a single ticket, with the number three. A's odds of being a winner are one in six. Question: if B enters the lottery, does that change the odds of A winning? Suppose ten, a hundred, even a thousand more people enter. At this point its almost a certainty that at least one person will win, in fact many people will win and split the pot in that case. However, what are the odds that A is one of the winners?
In all cases the odds of A being one of the winners is one in six. It remains one in six no matter how many other people enter, no matter how those tickets are distributed, and no matter how many winners there ultimately are. It has to be. In fact, consider that powerball winners are allowed to pick their own numbers, *or* they are allowed to let the computer pick them. In that case, the odds of a single ticket winning are based solely on the odds of a random number generator picking the same six numbers as another random number generator in a single run. Once that ticket is printed, that chance is fixed. Nothing can change the odds of that one ticket randomly hitting the correct numbers.
Some people get stuck on conceptualizing statistical problems, the infamous Monte Haul paradox demonstrates that. In that case, nothing short of an explicit experiment will convince them, if that. I would suggest to program one, using smaller numbers to make the runs reasonable. Imagine a lottery in which you have to correctly guess a number from one to a hundred. Write a program that randomly selects a ticket, then randomly selects the winning number. Run a million times, see how many times that ticket wins. Going to guess that will be about ten thousand. Now run a second program identical to the first one, but on top of the initial lottery draw, also generate a thousand more random ticket entries. Now see how often the first one wins. Its going to still be about ten thousand. Using your analysis of the math, you should be able to predict what the first entry's odds should be, if your "win sharing percentage" conjecture is correct. Its going to be wrong. I suspect, however, that before you finish writing the code for this experiment, the error in your thinking will become evident. Because algorithmically, you'll start thinking about the ticket entries as histogram buckets, and then voila.
If that *still* doesn't convince you, then I would love to have you play my lottery. I will engineer it so you have a one in a hundred chance of winning, but because you'll know your win percentage will increase with more entrants, I will computer generate a thousand more entrants. Your chances will now improve to the point where the odds of you winning are significantly higher than the computed odds, and you'll be able to make a lot of money through repeated plays. With luck, I could retire before you determine your error.
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Good to hear it! Arcana, anyone from the CoT team approach you if for nothing else than a rough consulting session?
I've fielded a few questions, but nothing recently. I don't really have the time to be a consistent participant, and so I suspect I would be more of a disruption than an aid: projects like this need reliability of contribution more than anything else in my opinion.
In either case, I don't think it is a reasonable thing to judge any of the development projects on whether or not they include me. I am always open to discussions of any topic, time permitting. However, development teams have to have their own vision of what is right for their games, and the will to stick to it without getting distracted. I have a lot of ideas, having thought about this for a long time and actually put many of those ideas into practice in one form or another, so I'm a really big distraction in that sense.
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I don't find it odd at all. Granted if you hung out in the CoH forums it was hard to miss Arcanaville. But the vast majority of the playerbase rarely set foot in the forums. In my circle of a dozen friends there's one other who knows for certain who Arcanaville is and another other frequent forum poster who might. The rest of them have no clue who she is; this despite the fact that they were actively managing supergroups, building bases, and hanging with the RP community.
I had no problem playing "anonymously" even on my home server. It seems the average City of Heroes player not only had never heard of me, but hadn't really heard of anyone: not me, not Pilcrow, not EvilGecko or Stupid_Fanboy, not Troy Hickman or Mercedes Lackey, not Castle or Geko, Positron was the contact that handed out the really long task force, and Statesman was the guy on the box cover. I'd get a "hey are you..." maybe two or three times a month on global, and of course the other "names" on the server generally recognized me, but if I didn't carry the Paragon Express card, almost no one else recognized my name.
Which, to be honest, was fine with me. The fact that the vast overwhelming majority of players could play the game just fine without knowing any of us, knowing any of what we did, being totally unaware of the meta-discussions about the game going on, and having practically none of the knowledge most forum readers took for granted was something I considered a measure of success for the casual friendly nature of the game. You didn't really need to know *anything* to enjoy playing it on some level. Five year olds did it without any knowledge of the fact that resistance resists resistance debuffs or the difference between accuracy and tohit buffs or what the predicted markov terminus was for SR scaling resistances. You didn't even need to know what enhancements were to play the game.
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Your calculations are correct but your math is wrong. If I put in a single ticket, the odds of my winning on that draw are one in 292,201,338. Those odds remain constant no matter how many other people enter. The correct way to think out this is to consider that when they draw the balls is just a procedural issue for the lottery: the power ball lottery is mathematically identical to a lottery in which the winning numbers were drawn ahead of time and kept secret, and every ticket entered by a participant is a single attempt to match that draw. When I buy a ticket, the odds of that one ticket matching the winning combination is one in 292,201,338. Those odds are unaffected by any event that happens afterward. If no one else buys a ticket, the odds of that ticket matching the winning numbers is one in 292,201,338. If someone else buys a ticket afterward, the odds remain one in 292,201,338.
There's no such thing as "distributing the win chance among the players." In this context, that is a statistically meaningless statement. The chance of a set of numbers matching the winning powerball combination, either in the real powerball lottery or the modified one above, are independent of all other draws. The only probability that *is* altered by the number of people entering is the chance of having *any* winner. Those odds increase with more entrants. But the odds of any particular *ticket* winning is fixed at one in 292,201,338.
The odds of a specific number sequence being generated is 1 in 292,201,338 by any particular randomizer - this is true and this remains constant. That's not the same as the odds of a person having picked that particular number sequence. You actually have multiple independent randomizers here. The lottery mechanism and the players. You are conflating them as one and the same. Each player is an indpendent number generator. Say I have one standard 6-sided red dice and 100 standard 6-sided blue dice and I roll them at one time. What are the odds that any of the blue dice has the same value of the red dice? if it was 200 blue dice? Those odds DO change with the number of independent trials and they improve with increased numbers. This is the same kind of "lottery" that nature plays with the fertilization of eggs. ;D Works pretty well despite some really crappy odds!
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I had no problem playing "anonymously" even on my home server. It seems the average City of Heroes player not only had never heard of me, but hadn't really heard of anyone: not me, not Pilcrow, not EvilGecko or Stupid_Fanboy, not Troy Hickman or Mercedes Lackey, not Castle or Geko, Positron was the contact that handed out the really long task force, and Statesman was the guy on the box cover. I'd get a "hey are you..." maybe two or three times a month on global, and of course the other "names" on the server generally recognized me, but if I didn't carry the Paragon Express card, almost no one else recognized my name.
Those were PEOPLE? You just making up names now aren't you? ;)
I admit I heard of Posi, Castle, and I think I heard of Arcana while the game was up, but never heard of the others. And I played since just before IOs till the end.
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Seriously game statistics are no joke, by definition they are calculus, where dynamic changable values are compared and processed against other dynamic changable values.
Actually, the calculus most people think of when they think of "calculus" - differential and integral calculus - isn't often used to analyze MMOs. In a practical sense, I only really used it once, to perform an approximate analysis of the SR scaling resistances. And to be honest that analysis turned out to be somewhat rubbish which is why I followed that one up with a more discrete precise analysis.
To be frank, most of the math you need to analyze most MMOs out there is simple high school algebra, because to be frank that's all the math most MMO designers are comfortable with. For complex analysis of dynamic situations, you're looking at statistical analysis as the approximation and Markov matrices and monte carlo analyses (aka simulations) as the less-approximation.
The biggest problem isn't the math, or the calculation part of the math. Its the conceptual translation and interpretation of the math as it applies to the game. There were a lot of forum analysts with reasonable quantitative skills that nevertheless could not apply them correctly to the game. Their calculations were correct but those calculations didn't apply correctly to any part of the game. Or sometimes they got tripped up on the precision of their calculations without asking whether those calculations were useful.
Here's my favorite version of math-correct, but mathematician-wrong. Consider the infamous case of stacking resistances or stacking defense. Question, when we stack equal numeric value resistances, does the benefit improve with stacking? Here's the case that it does increase in benefit with stacking.
Suppose we have two powers granting 25% resistance each. If we go from having no resistance to 25% resistance we lower incoming damage to 75% of original. That means it takes 1/.75 = 1.33x the damage to kill us, relative to having no resistances (ignoring regeneration for simplicity sake). But when we go from 25% resistance to 50% resistance, we go from 1/.75=1.33x damage to 0.75/0.50=1.5x damage to kill us, going from 25% resistance to 50% resistance. And in fact if we were to stack one more, we'd go from 0.75/.5=1.5 to 0.5/0.25=2.0x damage to kill us - twice as much with the third stack as without. So clearly, even though each of those powers is numerically identical, stacking them on top of each other improves their net value. Each one gives more benefit than the previous one.
However, here's the case that it does *not* increase the benefit.
We start with an entity with no resistance. We shoot 1000 points of damage at it, and it takes 1000 points of damage (ignoring tohit chance, which just reduces the numbers by a percentage). We now give it a 25% resistance power. It now takes 750 points of damage. The resistance power in effect blocked 250 points of damage. Now we give it another 25% resistance power. It now takes 500 points of damage. That's 250 points less than with just one power. So that second resistance power did exactly the same thing, it blocked 250 points of damage. A third such power would do the same thing: block 250 points of damage. Ergo, when players say stacking resistances gets stronger, they are wrong. Each is just as strong and does just as much as the previous one.
Who's right? The first one. But why? Where's the math error in the second analysis? There isn't one. The problem isn't in the math, the problem is in the head of the second mathematician. Specifically, it is in the question of what is the value of a resistance power? The second analyst looks at resistance powers literally. They block damage. So the analysis focuses on how much damage they block, and since each one blocks the same amount of damage, the second analyst concludes each stack of resistance must have the same value to the player. If two things do the same thing, they must have the same value.
But that's not how an MMO player values resistance. In fact there's no way for that player to directly perceive how much damage is *blocked*. Instead, MMO players value resistance powers the way they do all damage mitigation powers: do they keep me alive? The standard of value is *not* how much they block, but in a sense how much they admit. I called this the admittance effect on the City of Heroes forums. We don't notice the damage we avoid, only the damage we take. And we don't directly perceive the damage, we directly perceive *time* - how long we stay alive, how long before we need a heal, how long we can tank this AV. We directly perceive, in a colloquial sense, "how long does this mitigation work before I die?" And because that's how players judge mitigation, and that's what goes into their build decisions and that's how they will value a power, that's the standard we have to judge them on mathematically if we want to make a useful statement to players.
As mathematical analysts, we have to be useful to the non-mathematicians. So we have to figure out what players want, and tune our analysis to match. In this case, the mathematically correct analysis looks at increasing survivability. And that survivability varies inversely proportional to damage admittance - the inverse of damage mitigation. When we go from 1/.75 to 0.75/0.5 to 0.5/0.25 we go from 1.33 to 1.5 to 2.0. That is escalating return on investment. The other analysis where we go from 250 to 250 to 250 is quantitatively correct, but its not talking about the same thing players talk about when they talk about "value." It takes the word "value" and perverts it into something that is mathematically correct but inconsistent with what players themselves mean, and that's deception at best and intellectual dishonesty at worst.
This is the hard part. Not the calculations. Have Wolfram Alpha do the calculations for you. The hard part is translating the world, and people's perceptions of the world, into math in the first place. You don't need calculus, you need comprehension.
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Those were PEOPLE? You just making up names now aren't you? ;)
I admit I heard of Posi, Castle, and I think I heard of Arcana while the game was up, but never heard of the others. And I played since just before IOs till the end.
Pretty sure Aaron Williams played too (he lives near me and Mercedes Lackey is within about 2-300 miles). He's the creator of Nodwick and he had his characters from that series in strips as CoH characters. That's the only reason I even knew he played. I suspected the developers played and for all I know I might have been on missions with them. I didn't know for sure and didn't care. If they had ever been revealed to me and even given me the time of day I'd at most used them as an information source about the game i.e. strategy guide. At most it would have been the idle chatter I had with anyone else or shop talk. Would I have held them as inspirational? Yes. It's why I'm doing what I'm doing now. They showed what is possible.
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The odds of a specific number sequence being generated is 1 in 292,201,338 by any particular randomizer - this is true and this remains constant. That's not the same as the odds of a person having picked that particular number sequence. You actually have multiple independent randomizers here. The lottery mechanism and the players. You are conflating them as one and the same. Each player is an indpendent number generator. Say I have one standard 6-sided red dice and 100 standard 6-sided blue dice and I roll them at one time. What are the odds that any of the blue dice has the same value of the red dice? if it was 200 blue dice? Those odds DO change with the number of independent trials and they improve with increased numbers. This is the same kind of "lottery" that nature plays with the fertilization of eggs. ;D Works pretty well despite some really crappy odds!
Yes, the odds of *any* matching change with more trials, because the *any* changes with more trials. But that's congruent to saying that the more entries there are, the greater the chance that there is a winner, that among all of them, there is at least one winner. But that is not what you said. You said the odds increase for a single player, that the odds of a specific ticket winning increase with more entries. That is not the same thing you are saying here.
Because your wording is inconsistent and ambiguous, you should specify what you mean with precision. I am a lottery player. I buy a single lottery ticket with a single set of (potential) winning numbers. One ticket, one entry. Question: do the odds of me specifically winning change with the number of people who enter the lottery. True or false.
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And, Arcana? You got to know I just shared that posting on game play with in our internal gameplay channel don't you? ;D
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Those were PEOPLE? You just making up names now aren't you? ;)
I admit I heard of Posi, Castle, and I think I heard of Arcana while the game was up, but never heard of the others. And I played since just before IOs till the end.
Pilcrow was the winner of the first and only "most valuable forum poster" contest. He edged me out by some number of votes the community people were too merciful to inform me about. He's also the person who is credited with suggesting a system similar to enhancement diversification before that system was created and released by the devs, among other things.
EvilGeko was a relatively infamous personality on the forums, noted primarily for promoting the regeneration powerset and making controversial (but generally interesting) statements. He was the generally less curmudgeonly version of Venture, perhaps.
Stupid_Fanboy was less well known generally, but a frequent poster on the forums in the Scrapper forums in particular, and most famous for catching the "discount error" in Claws, which caused the devs to work on a complete revamp of the powerset.
Troy Hickman is a comic book writer who wrote the second City of Heroes comic book series, and more relevant to players of the game his "Smoke and Mirrors" story from the comic books was translated into Twilight Son's task force (http://paragonwiki.com/wiki/Twilight%27s_Son_Task_Force).
You know Mercedes Lackey as Victoria Victrix here. She was involved in some Hail Mary activities here, and there are rumors she occasionally writes stuff (http://www.amazon.com/s/ref=nb_sb_noss_2?url=search-alias%3Daps&field-keywords=mercedes+lackey).
Geko was the forum handle for the original "powers guy" in charge of basically everything you'd associate Castle with, but before Castle.
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And, Arcana? You got to know I just shared that posting on game play with in our internal gameplay channel don't you? ;D
Joshex will be happy to know he's been immortalized in a game dev chat somewhere.
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Yes, the odds of *any* matching change with more trials, because the *any* changes with more trials. But that's congruent to saying that the more entries there are, the greater the chance that there is a winner, that among all of them, there is at least one winner. But that is not what you said. You said the odds increase for a single player, that the odds of a specific ticket winning increase with more entries. That is not the same thing you are saying here.
Because your wording is inconsistent and ambiguous, you should specify what you mean with precision. I am a lottery player. I buy a single lottery ticket with a single set of (potential) winning numbers. One ticket, one entry. Question: do the odds of me specifically winning change with the number of people who enter the lottery. True or false.
It changes with the number of times you play as a certainty. But the situation here is if the chance of there being a winner increases with the size of the group does that affect the individual chance? For a group of a given size N in which there is a winner the chance of me being that winner is w=1/N. And if the group chance is P=p(N) then my chance is v=Pw or v=p(N)/N. So, yes, the number of players does affect my chance of winning because my situation is no longer independent of the other trials; it's part of an aggregate.
Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.
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Those were PEOPLE? You just making up names now aren't you? ;)
I admit I heard of Posi, Castle, and I think I heard of Arcana while the game was up, but never heard of the others. And I played since just before IOs till the end.
Not sure if kidding. If not, well, the posters (Pilcrow, EvilGeko, Stupid_Fanboy) referenced all had well over 10k posts, some specializing in frequenting some forums more than others. It's quite easy to not know about them if you did not frequent the forums, but on the other hand if you did regularly visit the forums and did not notice them (or Arcanaville, and a few others), them most likely you were only frequenting a few specific forums such as the role-playing forum or a server specific forum. Even then there was a good chance to notice some highly active posters.
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It changes with the number of times you play as a certainty. But the situation here is if the chance of there being a winner increases with the size of the group does that affect the individual chance? For a group of a given size N in which there is a winner the chance of me being that winner is w=1/N. And if the group chance is P=p(N) then my chance is v=Pw or v=p(N)/N. So, yes, the number of players does affect my chance of winning because my situation is no longer independent of the other trials; it's part of an aggregate.
Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.
My chance of winning is still one in 600whatever million, as is my chance of being one of the winners. More people buying tickets doesn't give me more number combinations or make them select more winning combinations or make them select fewer numbers, any of which would actually increase my chance of winning. In a lottery the number of winning and losing combinations is static and independent of the number of tickets sold and the number of tickets which have each combination.
Again, there's a language problem here. 'Increased odds of there being a winner' is just not equivalent to 'everyone's odds of winning go up.'
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Those were PEOPLE? You just making up names now aren't you? ;)
Whether internet people count as people is of course very debatable :P
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Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.
I think your math is correct, right up to the part where you try to express it in English. It seems you agree with my the odds of the ticket winning do not change, no matter how many other people enter. That's correct, and there's no disagreement there. But then you leave math, and extrapolate colloquially, and make a statement that is blatantly false. Since my chance of winning never changes, my chance of being one of the winners also does not change. Those two are mathematically distinct things, but you've calculated one and then simply asserted the two are the same. Try calculating directly. What are the odds not of winning, but of being one of the winners, if I am the only ticket entered. Now calculate directly, not indirectly, what are the odds of being one of the winners, if a million other people enter.
The odds of a winner existing at all increases with more entries, but the odds I am one of them do not. I'm beginning to wonder if you're making the same error I made earlier in the thread. When I first calculated the odd of winning, I made the incorrect assumption that if there was no winner, all previous entries were still alive: that you could enter once, and if there was no winner you still had a chance to win until there was a winner. That's not true for powerball. If there is no winner, all previous entries lose and in effect there is a brand new drawing held *only* with the tickets purchased since the last drawing, with the prize rolling over into the new lottery. That's the only explanation I have for why you would think the number of entries affects the odds of winning for a particular person. If I'm the *only* entry and tickets actually rolled over with the prize, then my odds of winning are 100% - because there has to be a winner eventually and I'm the only entry. It'll take about 40,000 years for me to win, but I will eventually win. For more entries to *increase* the chances for me to be one of the winners requires a different set of lottery rules that don't match Powerball.
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Whether internet people count as people is of course very debatable :P
Whether internet people count as people is debatable because you can count on internet people debating internet people about everything internet people count on.
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I had no problem playing "anonymously" even on my home server.
Honestly I didn't know who you are either because I never went to the official forums after the death of i12 pvp. I only remembered you because I remember someone in the HUB channel with your name always trying to solo a Pylon quicker than the last time, put 2 and 2 together with your forum title then thought, "oh I remember you"
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My chance of winning is still one in 600whatever million, as is my chance of being one of the winners. More people buying tickets doesn't give me more number combinations or make them select more winning combinations or make them select fewer numbers, any of which would actually increase my chance of winning. In a lottery the number of winning and losing combinations is static and independent of the number of tickets sold and the number of tickets which have each combination.
Again, there's a language problem here. 'Increased odds of there being a winner' is just not equivalent to 'everyone's odds of winning go up.'
Actually the number was barely over two million. :P
Ok. If the number of players increases the chance of any of the players selecting the winning number goes up. I wrote that as P=p(N). This is a degenerate form of P=p(q, N) where q is the number of people of the population that have picked the random number. For my examples I simply treated q=1. p(q,N,r) is just a variation of the binomial distribution nCk*r^k*(1-r)^(n-k).
If P is the chance of a win and someone wins then the chance of being that person is 1/N. You become a member of a group being selected from at random; you are a member of a trial - a random "value". So your odds are P/N of you being selected as the winner. Of course this isn't a linear function. I'm going to use a 4 sided dice here for simplicity... Ok. I got bit by my own math. :)
I just ran it through a spreadsheet.
(I hope this copy/paste doesn't look too crappy).
1 25.00% 25.00%
2 43.75% 21.88%
3 57.81% 19.27%
4 68.36% 17.09%
5 76.27% 15.25%
6 82.20% 13.70%
7 86.65% 12.38%
8 89.99% 11.25%
9 92.49% 10.28%
10 94.37% 9.44%
11 95.78% 8.71%
12 96.83% 8.07%
13 97.62% 7.51%
14 98.22% 7.02%
15 98.66% 6.58%
16 99.00% 6.19%
17 99.25% 5.84%
18 99.44% 5.52%
19 99.58% 5.24%
20 99.68% 4.98%
21 99.76% 4.75%
22 99.82% 4.54%
23 99.87% 4.34%
24 99.90% 4.16%
25 99.92% 4.00%
26 99.94% 3.84%
27 99.96% 3.70%
28 99.97% 3.57%
29 99.98% 3.45%
30 99.98% 3.33%
31 99.99% 3.23%
32 99.99% 3.12%
33 99.99% 3.03%
34 99.99% 2.94%
35 100.00% 2.86%
It looks like the odds of the individual go down as the win chances go up by increased number of players. Of course this is on a single trial. Repeated trials it would go up of course. Still, I stand corrected. For some reason I didn't think dividing by N would make for such a massive drop in odds.
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The last few days of reading this thread have been the most fun I've had in a few years. Keep it up! :D
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The last few days of reading this thread have been the most fun I've had in a few years. Keep it up! :D
Heck I'm laughing at my own gaffe here. ;D
A number between 0 and 1 being divided by increasingly large whole numbers? Of course it gets smaller and smaller.
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I am eating dinner with my team Arcana shirt on :P ;D
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I am eating dinner with my team Arcana shirt on :P ;D
She and I have both made different errors on the subject. I came to mine at about 3am in the morning and was overtired. I was just seeing the value in the numerator going up exponentially and the denominator "only" going up in a linear manner. It's a pattern I've seen often and I've come to expect certain results. I neglected to take into account the range of values for both. I was just taking umbrage at people accusing me of an entirely different error. I didn't err on the win-chance being distributed or that the group's win chance increasing though. Or even that the instantaneous probabilities were being misused.
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Your chances in powerball won't get smaller either. Odds of your number being the winner would still be one in however many possible combinations there are. The odds of there being a winner are variable. Your 'pot odds,' as they say in poker, are variable. Not the odds of your number being the winning number.
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Your chances in powerball won't get smaller either. Odds of your number being the winner would still be one in however many possible combinations there are. The odds of there being a winner are variable. Your 'pot odds,' as they say in poker, are variable. Not the odds of your number being the winning number.
You make the assertion, but can you show how that is true? I've done the math and demonstrated the results. You are still using the odds of a specific value being generated in a single trial. It is part of the overall analysis but it's incomplete.
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You make the assertion, but can you show how that is true? I've done the math and demonstrated the results. You are still using the odds of a specific value being generated in a single trial. It is part of the overall analysis but it's incomplete.
I'm not a math (that doesn't sound nearly as good as "i'm not a science") so I can't demonstrate anything mathematically other than that my checkbook is balanced (and demonstrations are always easier with really small numbers :D). Arcana's shown it numerous times in her posts though.
But to attempt to answer your question anyway - powerball is a specific value being generated in a single trial - i.e. the ping pong ball drawing. So those are the odds we're using.
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I'm not a math (that doesn't sound nearly as good as "i'm not a science") so I can't demonstrate anything mathematically other than that my checkbook is balanced (and demonstrations are always easier with really small numbers :D). Arcana's shown it numerous times in her posts though.
But to attempt to answer your question anyway - powerball is a specific value being generated in a single trial - i.e. the ping pong ball drawing. So those are the odds we're using.
Actually it's one of several values being generated in multiple trials. The lottery organization's randomizer and the random selections of each of the players over multiple games. That's what I was talking about with that "red dice/blue dice" example earlier. The odds of a person being born and surviving to adulthood is even worse odds but we have a population in the billions (I think the number is somewhere in the order of 1 in 10^66 power against).
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Actually it's one of several values being generated in multiple trials. The lottery organization's randomizer and the random selections of each of the players over multiple games. That's what I was talking about with that "red dice/blue dice" example earlier. The odds of a person being born and surviving to adulthood is even worse odds but we have a population in the billions (I think the number is somewhere in the order of 1 in 10^66 power against).
My last attempt, I long since should have given up and left it to smarter people. It's not multiple trials, it's a unique single trial multiple times. Only the money carries over from drawing to drawing. The tickets are valid for only one drawing.
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You make the assertion, but can you show how that is true? I've done the math and demonstrated the results.
The fact that you are doing math implies a problem. The odds of a single ticket matching the results of the powerball draw, given that the ticket contains one possible number sequence out of a set of (presumably) equally likely values is equal to 1/N, where N is the number of possible winning combinations. That's essentially an axiom of probability. No other independent choice can change those odds: that's also axiomatically the definition of "independent" in statistics. Given the first axiom and the second, there should never be a calculation that changes the odds of a single ticket winning.
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If P is the chance of a win and someone wins then the chance of being that person is 1/N. You become a member of a group being selected from at random; you are a member of a trial - a random "value". So your odds are P/N of you being selected as the winner.
Nope. You are multiplying two statistical factors that are not independent factors. Consider this: suppose the lottery asks the entrant to pick a number from 1 to 1000. Suppose exactly one person enters, and guesses 12 as his entry. What are the odds of someone winning this lottery? One in a thousand. What are the odds of you being that winner? The same: one in a thousand. It is not P/N.
If it was possible to know, out of all the people who entered the lottery, how many different possible winning numbers were chosen by at least one person and that number was n, and the total possible winning combinations was N, then the odds of *someone* winning would be n/N. The odds that you would be one of them, if you put in only a single ticket, would be 1/n. The actual odds of you being the winner would then be (1/n) * (n/N). Notice that ends up being 1/N, which is also the odds of winning the lottery computed directly: you put in one entry, and the odds of that one entry matching the winning combination is 1/N.
You math is still correctly calculated but misapplied. Somehow, you are invoking the set n, the total nunber of *different* combinations entered out of all the people who enter P. That quantity has nothing to do with the discussion at hand, but its an improperly defined value in your calculation assumptions.
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You seriously insulted me and people I'm close to.
To recap: you didn't know who Arcana was, as a self-confessed numbers hobbyist, in a CoH-related forum, in a thread were darn near every other post is a numbers post by Arcana. There's nothing wrong with that at all but as I said I was a "little surprised" of that being the case and asked "If you don't mind my asking, do you have any experience with CoH?"
And *that* makes you angry and is considered an insult? As you admitted in the next post there actually are a couple of CoT developers who never played CoH. You honestly do not see how that could have possibly have been a valid question? And I'm at a complete loss of how a question specifically for you somehow become an insult to "people [you're] close to"?
I'll offer this, no insult was intended. But I won't offer up an apology for imagined slights.
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I sure hope those who never played CoH pre-shutdown will be behind the content/powers. Sorry, no offense, but you will not capture what makes CoH, CoH. IP sell and all, with someone behind the wheel who has never touched the game before :/
Technically, Geko had no experience playing CoH before designing the powers system. I'm not sure he had any experience with long division before designing the powers system. Decimal points and percentages also seemed to be not in his wheelhouse. Come to think of it, I'm not convinced his calculator had a number four on it. Not sure what happened to Champions Online during development, except to say you shouldn't get your copy of Excel off the back of a truck in Hong Kong.
I'm not exactly sure why someone who had never played City of Heroes before would volunteer to try to make a replica of that game for little money or recognition, but we should probably cut them some slack until they release actual content to judge. If I was hired by NCSoft to conduct an audit of the development of City of Heroes back in 2003, I would have recommended they set fire to it for the insurance money. City of Heroes at launch was a dumpster fire of an MMO, but it was a warm and soothing flame. It is hard to tell what people will like until they see it. Fairness dictates that we judge product, and not resumes.
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To recap: you didn't know who Arcana was, as a self-confessed numbers hobbyist, in a CoH-related forum, in a thread were darn near every other post is a numbers post by Arcana. There's nothing wrong with that at all but as I said I was a "little surprised" of that being the case and asked "If you don't mind my asking, do you have any experience with CoH?"
And *that* makes you angry and is considered an insult? As you admitted in the next post there actually are a couple of CoT developers who never played CoH. You honestly do not see how that could have possibly have been a valid question? And I'm at a complete loss of how a question specifically for you somehow become an insult to "people [you're] close to"?
I'll offer this, no insult was intended. But I won't offer up an apology for imagined slights.
They haven't even released a beta of the game yet, and already they are replicating the experience of communicating with the devs. That's a good sign?
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They haven't even released a beta of the game yet, and already they are replicating the experience of communicating with the devs. That's a good sign?
Back in the CoH forum days there were a coupla incidents where devs said stuff that they probably should not have said. One in particular comes to mind, an incident where a dev spoke about the PvP player base that was not what one could call an tactful statement (I bet MM3 remembers this one, along with several others here). Now I didn't care for CoH PvP at all, to me it was a waste of resources to pour into a bottomless pit, the game wasn't built with PvP in mind and would have taken a foundational overhaul to make it PvP competent, but even this non-PvPer saw that posting and went "whoa, there is a person who lacks the customer service oriented skills to be making public statements about his product".
I suspect avelworldcreator is this sort of developer. Probably a great guy if you knew him in real life, probably has the best of intentions at heart. But the impression this reader gets reading his posts is that of someone who is perhaps a little too thin-skinned to be posting in a forum with somewhat limited people skills. I'm just not sure of what to make of someone who feels the need to brag of their math skills. Not everyone is cut out for customer-facing duties, nor should everyone be held to that standard. But these people probably need to be kept out of the public eye as much as possible to prevent missteps.
But looking back I can see avelworldcreator might have had a valid point in one area, in my trying to be a little less laser-focus damning in my critique I ended up painting with way too broad of a brush. I should not have said that the CoT team could use some PR work when precisely what I meant to say was that avelworldcreator, a CoT team member, could use some PR work. To all members of the CoT team I will indeed apologize for my careless statement, keep up the good work! As for avelworldcreator specifically, just from your postings in the past couple of days I do not believe you are serving as a good public face for the CoT team. If you want to be angry and insulted over *that*, now that I understand. My impression stands. But hey, I never cared for Posi's forum postings either so you're in good company. If you're ever in the Virginia area (after the SNOWPOCALYPSE passes) drop me a message and I'll buy you a beer, see if we don't grind each other's gears as much in person.
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Is it wrong that my inner baby statistician has had to squee a little during these last few pages? ;)
EDIT: I too did not know of Arcana before I came to this forum, though I think my wife did mention 'Arcanaville' when we were working on her guide to the Master of Statesman Task Force. I wasn't much of a forums guy.
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Nope. You are multiplying two statistical factors that are not independent factors. Consider this: suppose the lottery asks the entrant to pick a number from 1 to 1000. Suppose exactly one person enters, and guesses 12 as his entry. What are the odds of someone winning this lottery? One in a thousand. What are the odds of you being that winner? The same: one in a thousand. It is not P/N.
If it was possible to know, out of all the people who entered the lottery, how many different possible winning numbers were chosen by at least one person and that number was n, and the total possible winning combinations was N, then the odds of *someone* winning would be n/N. The odds that you would be one of them, if you put in only a single ticket, would be 1/n. The actual odds of you being the winner would then be (1/n) * (n/N). Notice that ends up being 1/N, which is also the odds of winning the lottery computed directly: you put in one entry, and the odds of that one entry matching the winning combination is 1/N.
You math is still correctly calculated but misapplied. Somehow, you are invoking the set n, the total nunber of *different* combinations entered out of all the people who enter P. That quantity has nothing to do with the discussion at hand, but its an improperly defined value in your calculation assumptions.
Arcana, I already spotted my own error. Of course I multiplied the factors. The percent chance OF a winning draw is applied to the chance OF a given person being the one to draw it. The word "of" indicates multiplication.
There is a number games played, G.
In each game the players act to create a pool of possible winning numbers of size V each game. For the sake of simplification I will treat this as a constant and it will be the same value each game.
In each pool there will be a certain percentage of duplicates, D. Also being treated as a constant for the same reasons as above.
The pool of unique numbers, U is equal to V*(1-D).
p is the percent chance of a specific number sequence being generated in the game drawing.
The chance of a player having picked a specific entry in the pool is 1/V for each game.
The chance of one of the numbers in the pool being drawn is 1-(1-p)^(U*G)
I'm not going to compute the odds of one of the duplicate values being drawn. This is complicated enough as is.
Let's recap. The chance of a given pool producing a wining entry is random with a binomial distribution, meaning the chances of a winning entry being produced increases with each game.
The chance of a given player having that winning entry is random with a flat distribution.
Both random selections are independent of each other.
The odds of two random selections occurring together is the product of their individual probabilities. R=(1-(1-p)^(U*G))/V
The odds of a random individual winning the lottery is 1-((1-(1-p)^(U*G))/V)^G if the same people play each game. The odds of an individual winning changes from game to game within these constraints.
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The odds of an individual winning changes from game to game within these constraints.
The odds of an individual ticket winning is 1/V, no matter what. No matter how many games are played, and no matter how many tickets are sold. That number combination has one chance to come up in a set of V numbers. It always has one chance to come up. Every draw, it has one chance to come up. One time.
If it's a drawing out of two tickets, A and B, you'd win statistically half the time if you pick either letter. No matter how many tickets are sold. No matter how many games are played.
** If 600 million people played once, you'd split the winnings approximately down the center, 300 million ish people would win.
** If two people played 600 million times, you'd split the winnings approximately down the center, both individuals winning approximately 300 million times.
This chance of one individual = one ticket = one chance does not change with the number of tickets sold or the potential number pool.
It only changes if one individual purchases more than one ticket. Then their chances go up. Statistically insignificantly after about 4 tickets, but up nonetheless.
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YIKES giant walls of text and math!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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To recap: you didn't know who Arcana was, as a self-confessed numbers hobbyist, in a CoH-related forum, in a thread were darn near every other post is a numbers post by Arcana. There's nothing wrong with that at all but as I said I was a "little surprised" of that being the case and asked "If you don't mind my asking, do you have any experience with CoH?"
And *that* makes you angry and is considered an insult? As you admitted in the next post there actually are a couple of CoT developers who never played CoH. You honestly do not see how that could have possibly have been a valid question? And I'm at a complete loss of how a question specifically for you somehow become an insult to "people [you're] close to"?
I'll offer this, no insult was intended. But I won't offer up an apology for imagined slights.
To recap: I just started back on these forums after a long hiatus and that was AFTER following a link from elsewhere to this topic and I did so because the topic was potentially relevant to my own work. I read the beginning entry. Skimmed past a number of other entries and focused on the last few pages of more than a thousand such that happened within the last few weeks. I paid attention more to the conversations than the posters even then. This is not enough to evaluate the statistical frequency of a given individual poster or to become familiar with that person or her self-descriptions. Arcana's level of posting in these forums does not relate in any manner or form to the any other person's experience with the City of Heroes game. Your "surprise" was based on overdrawn and egocentric assumptions. You said and asked more in that post than "If you don't mind me asking...". You related my experience, or the potential lack thereof, to the possibility that I and the people involved with the project would have no insight or understanding of the game we are trying to capture the spirit of. You made an accusation BEFORE you presented that question. Your question was to shift the burden of proof from yourself to me. THAT is what was the basis of insult. If the question had been presented by itself I would not likely have been insulted. But that was not the case and your attempt to compartmentalize your question from the rest of your statements is disingenuous. You accused me of possibly being deceitful or incompetent and they you have the gall to say my insult is imagined? Take that question and convert it to a statement of what you believed about me and my circumstances at the time. Consider what action were you trying to justify in relation to me or my associates based on my reply. That's right, you were demanding >>I<< take responsibility for your future conduct.
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We're going to nip this in the bud... feel free to take this to PMs, but y'all getting dangerously close to personal at this point, so I suggest not even continuing this line of conversation at all. Just FYI I'll delete posts that further it, no matter who's posting.
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YIKES giant walls of text and math!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
If you would keep posting builds this would happen less. :P :P
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The odds of an individual ticket winning is 1/V, no matter what. No matter how many games are played, and no matter how many tickets are sold. That number combination has one chance to come up in a set of V numbers. It always has one chance to come up. Every draw, it has one chance to come up. One time.
If it's a drawing out of two tickets, A and B, you'd win statistically half the time if you pick either letter. No matter how many tickets are sold. No matter how many games are played.
** If 600 million people played once, you'd split the winnings approximately down the center, 300 million ish people would win.
** If two people played 600 million times, you'd split the winnings approximately down the center, both individuals winning approximately 300 million times.
This chance of one individual = one ticket = one chance does not change with the number of tickets sold or the potential number pool.
It only changes if one individual purchases more than one ticket. Then their chances go up. Statistically insignificantly after about 4 tickets, but up nonetheless.
Again, I ask that you prove your assertion. You have stated my argument is in error by simply dismissing it as untrue. My analysis is directly based on an understanding of binomial distributions. Please show how a lottery is not a event that has two possible outcomes for a given trial and that each game is not instance of such a trial in a sequence of such.
Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number. The phrase "no matter what" means independent of outside influences but that is not the case. The outside influence in this case is the actual lottery drawing. You ARE correct that the odds of a person having one of the numbers in V is 1/V every time. But there is a chance that two or more people may choose the same set of numbers as well.
Where you get 600mil vs 300mil when the number of combinations of lottery numbers is barely over 200 mil is unexplained. But here is a shocker there is an actual chance of all 600million people all picking a losing number and there is also a chance (much lower but it still exists) that all 600 million people can pick the winning number. Crazy things like this happen I watched one joke about those very kind of odds with dice, throw something like 20 of them, and they ALL came up sixes! Trying figuring the odds of something like that. Not that they all came up with the same number but that it happened right after he did it to prove how unlikely it was to happen. It's something called the "Law of Large Numbers".
I just connected the 1/V chance to the chance that any number in V would be drawn BECAUSE that first chance was dependent on the second. The math of that second selection mechanism is a bit more messy. And it gets equally messy when you play more than once. "The more times you play the more likely you are to win" is a truism. It isn't limited to just four chances to be statistically significant.
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They haven't even released a beta of the game yet, and already they are replicating the experience of communicating with the devs. That's a good sign?
I hope so. ;D
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Again, I ask that you prove your assertion.
Where you get 600mil vs 300mil when the number of combinations of lottery numbers is barely over 200 mil is unexplained.
ok now you're getting agge confused with me. The 600 i was saying yesterday was based on my faulty memory from the original conversation. I should have known never to not use google as my memory. my memory now tells me it's 1 in 292 million and change.
the 600 million agge just used was as far as i can tell just a large number out of a hat.
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ok now you're getting agge confused with me. The 600 i was saying yesterday was based on my faulty memory from the original conversation. I should have known never to not use google as my memory. my memory now tells me it's 1 in 292 million and change.
the 600 million agge just used was as far as i can tell just a large number out of a hat.
Oh, didn't know that number had been used by you before. I'm actually surprised this conversation got so deep. I'm just a little confused by people saying "you are wrong because of this" just after I found my actual mistake and pointed it out. :o I found my goof actually quite hillarious but I guess you have to be more of a math geek to catch it.
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Again,
That was the first time I've posted on this subject.
And this is the first time anyone's ever mistaken me for Arcanaville. lol
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Except that the mistake we've all been talking about isn't a math mistake but a 'this is how this particular lottery works' mistake. Your catching your own math mistake is the only problem I recall anyone having with your math thus far. I refer you to Arcana's
The fact that you are doing math implies a problem.
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The error here is that the chance any member of member of V must first be a winning number.
The whole point of a lottery is at the end, you draw a number. So yes, a member of V must be a winner. ;) Whether or not it is chosen by a player is completely irrelevant. A number is always chosen.
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Except that the mistake we've all been talking about isn't a math mistake but a 'this is how this particular lottery works' mistake. Your catching your own math mistake is the only problem I recall anyone having with your math thus far. I refer you to Arcana's
I found her statement illogical. Statistical analysis without math? That seems akin to doing cooking without ingredients.
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The whole point of a lottery is at the end, you draw a number. So yes, a member of V must be a winner. ;)
At least we hope so! :P
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the 600 million agge just used was as far as i can tell just a large number out of a hat.
Correct. :)
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But here is a shocker there is an actual chance of all 600million people all picking a losing number and there is also a chance (much lower but it still exists) that all 600 million people can pick the winning number.
If you read carefully, I very specifically said statistically. Statistically, out of 600 million tickets, the choices between A & B are typically going to be very close to even unless someone is (or a bunch of someones are) gaming the system (and most people who try to game the system typically lose, unless you game it so hard you break probability (http://www.nytimes.com/1992/02/25/us/group-invests-5-million-to-hedge-bets-in-lottery.html?pagewanted=all)).
Also: the statistical likelihood of EVERY player picking the winning versus the losing number in an A/B choice is equal. ;)
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That was the first time I've posted on this subject.
And this is the first time anyone's ever mistaken me for Arcanaville. lol
The "again" referred to what someone else said. Not Arcana though. If you have just gone through a long explanation with cites and and detailed analysis you might get a little prickly when some just says "you are wrong" without further explanation. It's like "Seriously?"
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I found her statement illogical. Statistical analysis without math? That seems akin to doing cooking without ingredients.
Her statement wasn't illogical. Her point was that there's no need to do a statistical analysis when the odds are already a given.
Oh and agge, hate to burst your bubble but i think you were confused with me rather than with arcana. the shame :gonk:
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Your chances in powerball won't get smaller either. Odds of your number being the winner would still be one in however many possible combinations there are. The odds of there being a winner are variable. Your 'pot odds,' as they say in poker, are variable. Not the odds of your number being the winning number.
is hardly saying "you're wrong" without further explanation.
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Her statement wasn't illogical. Her point was that there's no need to do a statistical analysis when the odds are already a given.
Oh and agge, hate to burst your bubble but i think you were confused with me rather than with arcana. the shame :gonk:
Well, this is the first time I've been mistaken for you, too!
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The "again" referred to what someone else said. Not Arcana though. If you have just gone through a long explanation with cites and and detailed analysis you might get a little prickly when some just says "you are wrong" without further explanation. It's like "Seriously?"
I think it's entirely possible you are already prickly and need to take a day or so away ;)
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So much math. :O It makes this double-English-major whimper in fear and sadness.
Meanwhile, can we make all the snow around my house go away? I think I need a fire blaster out here, perhaps.
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If you read carefully, I very specifically said statistically. Statistically, out of 600 million tickets, the choices between A & B are typically going to be very close to even unless someone is (or a bunch of someones are) gaming the system (and most people who try to game the system typically lose, unless you game it so hard you break probability (http://www.nytimes.com/1992/02/25/us/group-invests-5-million-to-hedge-bets-in-lottery.html?pagewanted=all)).
Also: the statistical likelihood of EVERY player picking the winning versus the losing number in an A/B choice is equal. ;)
Yep, with an A/B decision with equal odds for each then the odds for all win/lose are indeed equal. But when the odds AREN'T equal? Things get a bit more messy. Here the question is what are the odds of any V being picked is actually affected by the size of V.
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Her statement wasn't illogical. Her point was that there's no need to do a statistical analysis when the odds are already a given.
Oh and agge, hate to burst your bubble but i think you were confused with me rather than with arcana. the shame :gonk:
But my argument was that the odds were NOT completely a given. What is given was just the number of combinations of lottery picks that could be chosen from. That does form part of the odds but that's like baking a cake with only flour. You need it, but there are other things that go with it to make the final product.
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Well, this is the first time I've been mistaken for you, too!
Certainly wouldn't want to make a habit of it :D
I'm mistaking myself for someone else too I think. I ran out of ways to say the same thing pages ago and said I was deferring to smarter people to take over. You and Arcana appeared to do just that and yet here I am posting again.
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is hardly saying "you're wrong" without further explanation.
So what part is the explanation why the numbers don't get smaller after I just showed the results of direct calculations where they did?
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Certainly wouldn't want to make a habit of it :D
I'm mistaking myself for someone else too I think. I ran out of ways to say the same thing pages ago and said I was deferring to smarter people to take over. You and Arcana appeared to do just that and yet here I am posting again.
I keep hoping someone mistakes me for someone like Bill Gates and pays me accordingly. That plan hasn't worked out so far. :gonk:
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So what part is the explanation why the numbers don't get smaller after I just showed the results of direct calculations where they did?
Again, you're asking for a math rebuttal when the point we've all been trying to make is that you don't need to do any more math.
And wow, I can't go 15 minutes without posting. Ok really this time :-X
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Again, you're asking for a math rebuttal when the point we've all been trying to make is that you don't need to do any more math.
And wow, I can't go 15 minutes without posting. Ok really this time :-X
Umm. I made a claim. Went through the trouble and math to try to prove the claim. Found out I made a mistake and even proved the opposite of what I claimed. Figured that was the end of it. Laughed about my mistake even. Suddenly I am being told I am wrong about something else without explanation. I know the areas where my analysis can be challenged and error shown if it is possible to do so, but no one has approached those topics. What was my conclusion? That the odds of playing the lottery really suck if you only play one time and those odds get worse if you are competing with other people for the chance to win. The odds improve if you keep playing but the chances of you even breaking even financially make it a doubtful exercise if you are really trying to get the big money. I have better luck playing blackjack in a casino (I really do! I make money usually playing it).
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So much math. :O It makes this double-English-major whimper in fear and sadness.
You majored in English twice?
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Umm. I made a claim. Went through the trouble and math to try to prove the claim. Found out I made a mistake and even proved the opposite of what I claimed. Figured that was the end of it.
The problem is that when the fact is that a particular value should be X, proving it is first lower than X, then reversing and saying it is higher than X, would be just as wrong. However:
Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number. The phrase "no matter what" means independent of outside influences but that is not the case. The outside influence in this case is the actual lottery drawing. You ARE correct that the odds of a person having one of the numbers in V is 1/V every time. But there is a chance that two or more people may choose the same set of numbers as well.
This is an error in parsing. The assertion is: for any single ticket entry, the odds of that ticket winning is 1/V, where V is the total *number* of possibilities. V is not a *set* and therefore there is no such thing as being a member of V. V is not the set of possible entries, V is a scalar quantity. This is important because as I said earlier, you are making ambiguous statements which confuse the issue. This sentence: "the error here is that the chance any member of member of V must first be a winning number" parses to gibberish. Ignoring the phrase doubling, this still makes no sense. A ticket has a single lottery entry, comprised of a set of numbers but is fundamentally a single selection out of 292,201,338. When the lottery draw occurs, a set of numbers will be drawn that will fundamentally be a single selection out of a selection space of 292,201,338 possibilities. The odds of winning is one in 292,201,338. That is true no matter how many other people enter. If no one else enters, your odds of winning are one in 292,201,338. If a billion more people enter, regardless of what numbers they pick, your odds of matching the winning powerball draw is still one in 292,201,338. You may need to share the winnings with someone else - likely in this case. But your odds of actually matching the winning draw are still one in 292,201,338. If no one wins, your odds of winning were still one in 292,201,338, and you just happened to lose.
There is no "set" of winning numbers in this context. There is a "set" of winning numbers in the sense of the way the balls are picked out of the machine, but a *single* lottery draw is composed of five balls drawn from a set of 69 and one ball drawn from a set of 26. That set of balls comprises a single lottery draw. You have that one chance to have the winning sequence. Every powerball lottery draw gives every *entry* one chance to either win or lose. You could have multiple winning tickets. You could have none. But every ticket gets one chance to match, or not match, the powerball draw. Each ticket has a separate chance of doing so.
In any case, it is irrelevant whether someone else also picks the same numbers or not and whether you need to share the winnings or not. Your odds of winning are one in 292,201,338. You may need to share the winnings if someone else happens to pick the same ticket numbers and that sequence wins. That is irrelevant to whether you will win or not.
I have a feeling this is going to be one of those times when I am no longer trying to convince someone they are very wrong, but rather just trying to make sure everyone else knows that they are wrong, and more importantly why. The quantitatively minded are already convinced, I'm sure. But for those that aren't mathematically inclined, here's an intuitive way to explain why it cannot possibly be true that your odds of winning the powerball with a single ticket entry cannot change in any way based on how many other people happen to enter that lottery. If that were true, it would be mathematically true no matter how large the lottery is. So lets look at the most simple (what mathematicians would call the most degenerate) case. Lets say we're asking contestants to call a coin toss. Basically, instead of pulling 5 white balls with the numbers one through sixty nine on them and another ball with the numbers one through twenty six on them, we're just going to flip a coin. On one side of the coin is the number 1, and on the other side is the number 2. To enter, you write down either a 1 or a 2 on your ticket. This is the most simple lottery imaginable.
What are your odds of winning? Obviously, 50/50 I hope even the least mathematically inclined of you are certain of. Okay. Now lets say I also enter the lottery. What are the odds that your ticket matches the coin toss? Still 50/50? Why? Is it because no matter what I do, if your ticket has a 1 then the coin toss will either be a 1 or a 2, and you have an even chance to win? Even if a hundred other people enter? Sure, isn't your odds of calling a coin toss the same whether you do it by yourself or a stadium of other people do it at the same time?
Congratulations, you all get an A in Arcanaville statistics 101. When you try to guess a number, the odds you'll get it right is not affected by whether you're the only person guessing or any number of other people try to guess at the same time. A trillion aliens throughout the galaxy could try to guess that same number at the same time you do, and it can't hurt or help you out. If you are trying to guess a coin toss, your odds are 50/50 no matter what. If you are trying to guess the roll of a six sided die, your odds are one in six no matter what. If you are trying to guess the six numbers that will pop out of the powerball lottery machine, your odds are one in 292,201,338 no matter what. No matter how many other people try to do the same thing, no matter how many of them succeed or fail, your odds remain one in 292,201,338. Because your odds of winning are determined at the moment you make the selection: at that moment you have a 50/50 chance of calling that coin toss, a one in six chance of calling that die roll, and a one in 292,201,338 chance of calling the powerball. The only thing that matters is how many possibilities there are, and how many guesses you get. If I give you one guess, then the odds are one in X, where X is the number of possibilities.
This is a fundamental principle of statistics. It is, in fact, *the* fundamental principle of statistics. Get this wrong, and you can't get anything else right, except by, heh, random chance. This is not a debatable thing, nor is it an esoteric thing. You cannot claim to understand statistics and get this wrong. Getting this wrong in statistics is like claiming the Earth is larger than the Sun but claiming to have a Ph.D in astronomy.
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Congratulations, you all get an A in Arcanaville statistics 101. When you try to guess a number, the odds you'll get it right is not affected by whether you're the only person guessing or any number of other people try to guess at the same time. A trillion aliens throughout the galaxy could try to guess that same number at the same time you do, and it can't hurt or help you out. If you are trying to guess a coin toss, your odds are 50/50 no matter what. If you are trying to guess the roll of a six sided die, your odds are one in six no matter what. If you are trying to guess the six numbers that will pop out of the powerball lottery machine, your odds are one in 292,201,338 no matter what. No matter how many other people try to do the same thing, no matter how many of them succeed or fail, your odds remain one in 292,201,338. Because your odds of winning are determined at the moment you make the selection: at that moment you have a 50/50 chance of calling that coin toss, a one in six chance of calling that die roll, and a one in 292,201,338 chance of calling the powerball. The only thing that matters is how many possibilities there are, and how many guesses you get. If I give you one guess, then the odds are one in X, where X is the number of possibilities.
This is a fundamental principle of statistics. It is, in fact, *the* fundamental principle of statistics. Get this wrong, and you can't get anything else right, except by, heh, random chance. This is not a debatable thing, nor is it an esoteric thing. You cannot claim to understand statistics and get this wrong. Getting this wrong in statistics is like claiming the Earth is larger than the Sun but claiming to have a Ph.D in astronomy.
Unless you can figure out how to program a "luck" stat which increases likelihood of things going in your favor and make a luck power which scales up based on the number of enemies/players in range. as a developer such things are open to our hands to put into a game world, but not the real world.
would be funny though, you max enhance the luck of the luck power and walk into the golden giza and as you merely pass slot machines they pour out coins and the people at the tables shiver in sight of you.
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You majored in English twice?
Bachelor of Arts, 2001 (Major: English; Minor: Professional Education, grades 9-12). Master of Arts, 2003 (Major: English).
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The problem is that when the fact is that a particular value should be X, proving it is first lower than X, then reversing and saying it is higher than X, would be just as wrong. However:
This would be true if that was the case but I have yet to see direct argument of this nature, just tautologies.
This is an error in parsing. The assertion is: for any single ticket entry, the odds of that ticket winning is 1/V, where V is the total *number* of possibilities. V is not a *set* and therefore there is no such thing as being a member of V. V is not the set of possible entries, V is a scalar quantity. This is important because as I said earlier, you are making ambiguous statements which confuse the issue. This sentence: "the error here is that the chance any member of member of V must first be a winning number" parses to gibberish. Ignoring the phrase doubling, this still makes no sense. A ticket has a single lottery entry, comprised of a set of numbers but is fundamentally a single selection out of 292,201,338. When the lottery draw occurs, a set of numbers will be drawn that will fundamentally be a single selection out of a selection space of 292,201,338 possibilities. The odds of winning is one in 292,201,338. That is true no matter how many other people enter. If no one else enters, your odds of winning are one in 292,201,338. If a billion more people enter, regardless of what numbers they pick, your odds of matching the winning powerball draw is still one in 292,201,338. You may need to share the winnings with someone else - likely in this case. But your odds of actually matching the winning draw are still one in 292,201,338. If no one wins, your odds of winning were still one in 292,201,338, and you just happened to lose.
There is no "set" of winning numbers in this context. There is a "set" of winning numbers in the sense of the way the balls are picked out of the machine, but a *single* lottery draw is composed of five balls drawn from a set of 69 and one ball drawn from a set of 26. That set of balls comprises a single lottery draw. You have that one chance to have the winning sequence. Every powerball lottery draw gives every *entry* one chance to either win or lose. You could have multiple winning tickets. You could have none. But every ticket gets one chance to match, or not match, the powerball draw. Each ticket has a separate chance of doing so.
In any case, it is irrelevant whether someone else also picks the same numbers or not and whether you need to share the winnings or not. Your odds of winning are one in 292,201,338. You may need to share the winnings if someone else happens to pick the same ticket numbers and that sequence wins. That is irrelevant to whether you will win or not.
I have a feeling this is going to be one of those times when I am no longer trying to convince someone they are very wrong, but rather just trying to make sure everyone else knows that they are wrong, and more importantly why. The quantitatively minded are already convinced, I'm sure. But for those that aren't mathematically inclined, here's an intuitive way to explain why it cannot possibly be true that your odds of winning the powerball with a single ticket entry cannot change in any way based on how many other people happen to enter that lottery. If that were true, it would be mathematically true no matter how large the lottery is. So lets look at the most simple (what mathematicians would call the most degenerate) case. Lets say we're asking contestants to call a coin toss. Basically, instead of pulling 5 white balls with the numbers one through sixty nine on them and another ball with the numbers one through twenty six on them, we're just going to flip a coin. On one side of the coin is the number 1, and on the other side is the number 2. To enter, you write down either a 1 or a 2 on your ticket. This is the most simple lottery imaginable.
What are your odds of winning? Obviously, 50/50 I hope even the least mathematically inclined of you are certain of. Okay. Now lets say I also enter the lottery. What are the odds that your ticket matches the coin toss? Still 50/50? Why? Is it because no matter what I do, if your ticket has a 1 then the coin toss will either be a 1 or a 2, and you have an even chance to win? Even if a hundred other people enter? Sure, isn't your odds of calling a coin toss the same whether you do it by yourself or a stadium of other people do it at the same time?
Congratulations, you all get an A in Arcanaville statistics 101. When you try to guess a number, the odds you'll get it right is not affected by whether you're the only person guessing or any number of other people try to guess at the same time. A trillion aliens throughout the galaxy could try to guess that same number at the same time you do, and it can't hurt or help you out. If you are trying to guess a coin toss, your odds are 50/50 no matter what. If you are trying to guess the roll of a six sided die, your odds are one in six no matter what. If you are trying to guess the six numbers that will pop out of the powerball lottery machine, your odds are one in 292,201,338 no matter what. No matter how many other people try to do the same thing, no matter how many of them succeed or fail, your odds remain one in 292,201,338. Because your odds of winning are determined at the moment you make the selection: at that moment you have a 50/50 chance of calling that coin toss, a one in six chance of calling that die roll, and a one in 292,201,338 chance of calling the powerball. The only thing that matters is how many possibilities there are, and how many guesses you get. If I give you one guess, then the odds are one in X, where X is the number of possibilities.
This is a fundamental principle of statistics. It is, in fact, *the* fundamental principle of statistics. Get this wrong, and you can't get anything else right, except by, heh, random chance. This is not a debatable thing, nor is it an esoteric thing. You cannot claim to understand statistics and get this wrong. Getting this wrong in statistics is like claiming the Earth is larger than the Sun but claiming to have a Ph.D in astronomy.
I never said directly "V" itself was a set. I said " But there is a chance that two or more people may choose the same set of numbers as well. ". The lotter involves each player selecting a set of numbers i.e. {15, 27, 33,34,15}, by an arbitrary method, with each number distinctive (with the possible exception of a "powerball" number), and each selected from a finite range of integral values. How did you conclude that I was equating those independent sets with a variable used to enumerate. I was pointing out the possibility of duplicate selection which does affect the distribution. (I can see I was also getting very tired at close to 2am when I said "any member of member "). How about about the set of selected sets "V" where the "v" is the quantity of chosen sets? Does this word salad satisfy you?. The lottery picks are a set of sets each game. The size of the set containing the other sets is finite and quantifiable.
The lottery device produces 292,201,338 possible set combinations but that is not the set of POTENTIALLY winning combinations. THAT was previously generated by the players. This means the odds of there even being a winning ticket is not independent of the previous actions of the players. If the device does not select one of the previously generated combinations the odds of a person being a winner isn't 1/v - it's ZERO. It's the win chance of zero divided by the number of unique ticket combinations - which is still ZERO. Tell me again how the odds of winning are always either 1/v or 1/292201338? Tell me again how the odds of winning are independent of the previously selected set of combinations? Let me use that word again to drive my point home: ZERO. Those are your base odds. ZERO or 100%. It's a coin toss. Tell me why it's not such. This should be good. But wait! That's only if the odds of a winning are even. This is a WEIGHED coin toss. Did you get that part? "Weighed"? How is it weighed? By the number of unique sets chosen previously by the players and the weighing is calculated by the binomial distribution formula. Again, I want it explained why this is not the case. So the actual base odds are x% lose and (1-x)% win. It also means the base odds are NOT solely dependent on the possible lottery combinations. 1/v? Those are the odds of a player having a picked a specific combination. It's NOT the odds of the player having picked a WINNING combination. Again I want a direct explanation why that is NOT the case. First you have to calculate the odds of there even being a winning combination in the set, and then you have to calculate the odds of a given member of the set being that combination.
What's going on? Because the lottery is NOT exclusively a process of independent selection. It combines both independent selection with dependent selection. Dependencies alter the final odds.
Your analogy is false. The lottery is not a game where the players are trying to guess a winning value selected previously. The winning number is not generated in advance with all possible values being a possible winning choice. The lottery system system is actually the one doing the "guessing" from the choices offered by the players. It's doing a random drawing from a "bag" where there is a chance it draws nothing at all.
[Edit: Just trying to catch typos].
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Unless you can figure out how to program a "luck" stat which increases likelihood of things going in your favor and make a luck power which scales up based on the number of enemies/players in range. as a developer such things are open to our hands to put into a game world, but not the real world.
would be funny though, you max enhance the luck of the luck power and walk into the golden giza and as you merely pass slot machines they pour out coins and the people at the tables shiver in sight of you.
Huh! I'll pass that on. I think I might have seen something of that nature in the works but I'll ask. Thanks for that idea. May not be used and it may not be unique, but still worth pondering.
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I never said directly "V" itself was a set. I said " But there is a chance that two or more people may choose the same set of numbers as well. ".
No. I quote:
Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number.
There are no members of V in the context of the post you quoted. V was a number. You yourself stated as such when you acknowledged that the probability was being asserted as "1/V" so I know you read it right. Then you asserted there were members of V.
The lotter involves each player selecting a set of numbers i.e. {15, 27, 33,34,15}, by an arbitrary method, with each number distinctive (with the possible exception of a "powerball" number), and each selected from a finite range of integral values. How did you conclude that I was equating those independent sets with a variable used to enumerate. I was pointing out the possibility of duplicate selection which does affect the distribution. (I can see I was also getting very tired at close to 2am when I said "any member of member "). How about about the set of selected sets "V" where the "v" is the quantity of chosen sets? Does this word salad satisfy you?. The lottery picks are a set of sets each game. The size of the set containing the other sets is finite and quantifiable.
That's just obfuscation. The powerball lottery is mathematically equivalent to this situation, which is actually a textbook-class homework assignment:
An integer between one and 292,201,338 is selected at random. A player is asked to guess the number, given the range of possibilities.
Question one: What are the odds of the player guessing the correct number, given one guess.
Question two: Do these odds change if other people make guesses, if no one is allowed to know whether any of the guesses are correct until all guesses are made?
The answer to the first question is: "one in 292,201,338." The answer to the second question is: "no." Period the end.
The lottery device produces 292,201,338 possible set combinations but that is not the set of POTENTIALLY winning combinations. THAT was previously generated by the players. This means the odds of there even being a winning ticket is not independent of the previous actions of the players.
That's correct. The odds of there being a winning combination is n/N, where n is the number of different combinations generated by the players, and N is 292,201,338. No one has said otherwise.
If the device does not select one of the previously generated combinations the odds of a person being a winner isn't 1/v - it's ZERO. It's he win chance of zero divided by the number of unique ticket combinations - which is still ZERO. Tell me again how the odds of winning are always either 1/v or 1/292201338? Tell me again how the odds of winning are independent of the previously selected set of combinations?
That is also true. IF the device does not select a number chosen by one of the players, then the odds of any particular person winning are zero.
Let me use that word again to drive my point home: ZERO. Those are your base odds. ZERO or 100%. It's a coin toss. Tell me why it's not such. This should be good. But wait! That's only if the odds of a winning are even. This is a WEIGHED coin toss. Did you get that part? "Weighed"? How is it weighed? By the number of unique sets chosen previously by the players and the weighing is calculated by the binomial distribution formula. Again, I want it explained why this is not the case. So the actual base odds are x% lose and (1-x)% win. It also means the base odds are NOT solely dependent on the possible lottery combinations. 1/v? Those are the odds of a player having a picked a specific combination. It's NOT the odds of the player having picked a WINNING combination. Again I want a direct explanation why that is NOT the case.
Sure. The odds of *at least one player* picking the winning combination, IF the device has actually picked one of the combinations actually entered, is 100%. That's because the premise is: the lottery picked one of the entries actually chosen by at least one player. However NO ONE IS TALKING ABOUT THAT PROBABILITY, INCLUDING YOU. The statement being argued, and you've repeated yourself, is that the odds of A SPECIFIC PERSON winning change based on the number of other players entering. THAT IS STILL FALSE.
The odds of a specific ticket winning are still 1/N. Why? The odds of a specific ticket winning IF THE LOTTERY MACHINE ACTUALLY PICKS A WINNING NUMBER are 1/n, where n is the number of different combinations actually entered into the lottery. Since one of them was actually picked GIVEN THE PREMISE THAT THE LOTTERY MACHINE PICKED A WINNING NUMBER, the odds of that ticket winning are 1/n, and n depends on the number of people who entered.
HOWEVER, THAT ASSUMES THE MACHINE PICKED A NUMBER ACTUALLY ENTERED. The odds of THAT happening itself are not 100%. Therefore, you cannot say "the odds of a ticket winning depend on how many other people enter." You can only say "the odds of a ticket BEING THE WINNING TICKET DURING A DRAWING IN WHICH THERE IS A DECLARED WINNER depends on the number of people entered. That's obvious: if you are the only entrant at all, and the lottery declares that there is a winner, it has to be you. But the odds of a ticket winning are, according to basic probability, the odds of there actually being a winner, multiplied by the odds of you being that winner if there is a winner. The second part, 1/n, is the part you are claiming are the odds of a ticket being a winner. THAT IS FALSE. The true odds of a ticket being a winner are the odds of there actually being a winner - the odds of your premise being true - multiplied by the odds of that winner being that ticket. And the odds of your premise being true - that the lottery machine actually picked a winner - is easy to determine. Given that n was the number of different combinations entered, the odds of one of them being picked are n/N. So the odds of the ticket being the winner are n/N times 1/n. In other words, 1/N. And that's invariant to the number of entries.
The direct explanation for your error is that you performed calculations presuming there was a winner, and calculated what the odds were of a single ticket being that winner. But the odds of there being a winner at all are not 100%. Sometimes there is no winner at all. And obviously, when there is no winner, no ticket wins, and that affects the overall odds of a specific ticket actually winning. You are not calculating the odds of a ticket winning over all possible lottery outcomes. You are calculating the odds of a ticket winning over a subset of them. And if you do that, then you have to state what calculation you are doing correctly.
What you are trying to say, incorrectly, is that in all powerball lottery draws in which there is a declaration of a winning ticket, the odds that your ticket is the winning ticket are affected by how many people entered.. And everyone would agree with that, because that's obvious. IF there is a winner at all, the odds that it is your ticket rise if there are less entrants, and fall if there are more of them. In the degenerate case where you are the only entrant, the odds you are the winner are 100% - it has to be you. In the opposite degenerate case where there are an extremely large number of entries, the odds that you will be one of them fall to a low of 1 in 292,201,338. Obviously, the odds can't get any lower than that. If the lottery says there was a winner, the odds of your ticket being the winner are higher if less people entered. HOWEVER, before the drawing is held and before you know if there actually was a winner or not, the odds of a particular ticket being the winning ticket are still 1 in 292,201,338. Your odds of winning don't change as more people enter.
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For some reason after every post from Arcana the response I keep hearing in my head is Vizzini from The Princess Bride saying "INCONCEIVABLE".
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Okay, so..let me get this straight..avel comes in first speaking about how limited and confounded the old engine for CoH/The topic of 'just how dedicated are you to this "legacy" CoH and die hard fans(with the added mix of passive-aggressive subversion of the CoH fans here to support/play COT(mind you, I will support it, but not above CoH i23. I feel others will most likely act in the same manner.) Then, someone speaks about something math-based, Arcana is mentioned, someone brings up the lottery/chances of winning, and then avel spirals into this argument with Arcana. Someone then accuses avel of not playing/being active in CoH because they didn't know arcana, and they get offended. :-\(I know I'm missing abit here and there but I think I got the jest of things).
P.S: Avel, I know you mean well and like was posted before, I'm sure your an absolutely swell person! :D, but, coming into a CoH revival thread, talking about how old and decreped CoH's engine was(very true on that one)/proceeding to question, if not directly, why stay with CoH legacy when COT will offer a much more modern engine, then(and, this is just how I took your post on CoH, I'm sure this is absolutely incorrect in founding) using subtle passive-aggressive undertones and diction to describe our dedication to a game that is a bit dated to divert our attention over to COT.(The deal as it stands, if I am correct, is that the legacy won't even be on that old engine(disk image), but instead will be on a new engine used by a third-company that is yet to be decided(at the time.) As well as APR making use of UE4 and the IP(along with y'all's very own character creator as well :D))Forgive me for this assertion, but, you must understand how this makes you look to me, and, presumably from the post that followed after your initial post, others in this forum as well.
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Arcanaville Statistics 201: how to calculate probabilities for people who don't like math.
The basic idea behind calculating probabilities comes down to two basic principles and very simple math. The first one is what I call the counting rule, and it is this:
If there is a set of equally likely occurrences, and you are interested in a subset of them, the odds of what you're interested happening are just p/N, where p is the number of different things you are interested in, and N is the total number of possible occurrences.
That's simple enough: it is just counting. What are the odds of rolling a two on a six sided die? One in six. Six ways for the die to roll, only one of them is a two, so the odds are 1:6, or about 16.7%. What are the odds of rolling an even number? Three in six, or 3/6, or 1/2, or 50%. Easy. Just count.
But what if it is difficult to count? What if the different things that can happen are not all equally likely? The second rule is what I call the divide and conquer rule:
If the odds of a set of things happening is p, and the odds of any one of those things in the set is q, then the odds of any one of those things happening is p * q
You attack something with an attack that has a 10% chance to crit. What are the odds of getting a critical hit, if your tohit chance is 85%? It is just 0.85 * 0.1 = 0.085, or 8.5%. First you have to get a hit at all, then you have to actually trigger the critical.
Most of statistical analysis is just about applying these rules correctly, and sometimes it can be non-obvious. Statistics is actually less about math in terms of calculation, and more about applying the rules correctly. And it is an open secret in the world of statistics that the rules are often not applied correctly. The fact that a large percentage of math professors get the Monte Haul paradox wrong is a testament to the fact that even though they know the rules, they often let their intuition override proper mathematical reasoning. Even though that paradox has a trivial mathematical resolution every math professor should know.
How do we analyze something like the lottery. We apply the rules, without letting intuition or word games get in the way. What are the odds of a single ticket matching the powerball draw? Well, first lets figure out how many possibilities there are. The powerball lottery draws five numbers from a set of 69 balls labelled one through sixty-nine, and a special "powerball" from a separate set of balls labeled one through twenty-six. How do we determine how many ways there are to do that?
Lets start with the first five. Obviously there are 69 possible ways to draw the first ball, because there are 69 of them. Once that first ball is draw, how many possibilities are there for the second draw? 68, because in every case there are only 68 balls left. Then 67, then 66, then 65. Using the two rule above, you might guess there are 69*68*67*66*65 ways to draw those balls, and you'd be correct.
However the Powerball lottery does not require you to pick all of the balls in the correct order. It only requires you to get all the numbers right. In other words, according to the rule of powerball, the draw "1,2,3,4,5" is considered identical to the draw "5,4,3,2,1." So actually, the number we calculated above is larger than the actual number of possibilities, because we've double counted many of them (more than double, actually). Now what? Well, actually we can still figure this out, because we can use our rules in reverse. The first rule, the counting rule, says that if you want to know what the odds of something happening are, you just count up the possibilities. In this case, we've counted too high. What we need to figure out is, for every "real" possibility, how many times have we repeatedly counted the same thing, and reduce our number by the overage.
Consider a simple case where we pick two numbers. We could pick 1,2, or 2,1. Both are the same "powerball" draw of a one and a two. We've counted the same possibility twice. So we'd divide by two to get the true number of possibilities. We'd normally calculate the total possibilities as 69* 68 = 4692, but because we counted everything twice the real number is 69*68/2 = 2346. There are two thousand three hundred forty six ways to draw two powerball balls. Math people would say there are 4692 permutations of two balls, but only 2346 combinations. Permutations are when order matters, combinations are when order doesn't matter.
So what happens when we pick five balls. Well, we can play a neat trick here. We want to know how many ways we overcounted those five ball sequences. And we know how to do that already. In effect, it is like we have five balls in a bag, and we want to know how many possible ways - permutations in this case - there are to pull those balls out. And that's just 5 * 4 * 3 * 2 * 1. So there are 120 ways you can order five balls. So for every possible combination of powerball, we've overcounted by 120 times. So we take 69*68*67*66*65 and divide by 120, and we get 11,238,513. That's the number of possible ways to draw the first five balls. The actual "powerball" is a separate ball picked from a separate set of 26 balls. There are exactly 26 ways to do that. So the total number of possibilities is 11,238,513 * 26 = 292,201,338. Only *one* of those possibilities is selected at drawing time. So there's only one winning sequence out of 292,201,338 possible sequences they could draw. So if you enter a ticket and that ticket has a specific sequence of numbers on it, the odds that that specific sequence will match the drawn sequence is one in 292,201,338. That's where those numbers come from.
This is all following the rules for probability. Notice that number doesn't rely on anything besides counting possibilities. In effect, we've mostly just used rule one. But what if you want to look at more complicated things, things that can confuse people. What if you wanted to answer the question: if the powerball authority says there's a winning ticket out there somewhere, what are the odds that it is my ticket?
That's more complicated, but the rules still work. It is just that now, you'll likely need to do more math because there are too many possibilities to count easily. You could, in theory, but the numbers are too big. Basically, if you know there is a winner somewhere, then the total number of possibilities for what came out of the powerball machine aren't 292,201,338 anymore, at least not necessarily. The new piece of information we have - that there was a winner - means the machine could have only picked a sequence that someone has entered. If there are a billion entries into the powerball, the odds are pretty good that every single sequence was played by at least one person, but if only a million entries went in then it is impossible that all 292 million were played, because there aren't enough entries to do that. If we knew how many tickets were entered, and if we made the assumption that all ticket sequences were selected randomly (they aren't in reality, but it is not an unreasonable approximation most of the time) we could then calculate how many sequences are *likely* to have been played, out of all of them. Unfortunately, that's Arcanaville statistics 401. Let's just assume we can. If big N is 292,201,338, we call little n the number of sequences that were actually played by players, and we do what mathematicians do when they don't know something: use the letter as if we knew what it was.
Since the total possibilities for the draw are no longer big N but little n, the odds of one specific ticket being the winner is no longer 1/N, but 1/n. Less possibilities, so the odds change. But does this mean that the odds of a ticket winning depend on how many tickets are entered? No, and that's because we started by assuming there was a winner. What do the rules say about what the odds of winning are overall?
Well, they are 1/N, because the rules say so. But what if we want to analyze this the hard way? Well then why are you reading this post: go read a stats text book. Okay, fine. The rules say that to figure out what the odds of a specific ticket winning, we can try to count up all possibilities. First, there is the case where there is actually a winner. We know there are n possibilities there. Of them, one out of n is a specific ticket sequence. The odds of a specific ticket winning *if there is a winner* is one out of n. But that's not the set of all possibilities. What about the case where there is no winner? Well, if n of the draws picks a winner, then the rest don't. That's N-n. In all of them, that specific ticket doesn't win, obviously. So the odds of the ticket winning, using the counting rule, is that it is one out of n + (N-n) = N. There are n possibilities when there is a winner, N-n where there is no winner, and the total is the sum of the two. That's obviously N, but that's how that works.
Notice that no matter how we choose to do it, the easy way or the long way, the odds are the same. That's because they have to be. The point of view shouldn't change the odds. When two different mathematical methods of calculating something generate different answers, the problem is generally not with the math, but with the mathematician. In fact, it can be a valuable way to double check yourself. If you know two ways to do something and you do them both and they agree, that would tend to suggest you haven't made an error. If they don't, pretty sure you did.
Man, this got longer than I thought it would. But it is still shorter than a stats textbook. If there any City of Heroes players out there still awake, consider this final thing. We used to calculate the odds of hitting a target all the time. Well, I did anyway. We needed to know the tohit chance, which is kind of like asking how many possible tohit rolls are hits. We knew how many possibilities there were in total, because we roll 0-100 (there's a teeny tiny complication here that isn't important to this discussion, lets leave that alone for now). We then calculate the odds of hitting as all the ways you can hit, divided by all the ways there are to roll tohit total. "75% chance to hit" was just another way of saying 75 rolls hit out of 100 rolls. Or, because the tohit system seemed to round off to the nearest .0001, you could say 7500 rolls hit out of 10000. Did we ever ask whether anything else was rolling tohit rolls at the same time? Of course not. Because that doesn't matter. All that matters in this case is how many possibilities hit, and how many total. Divide, done.
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You majored in English twice?
Maybe it was English and American?
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But what if we want to analyze this the hard way? Well then why are you reading this post: go read a stats text book.
... Have you actually encountered a stats text in the wild written remotely as comprehensibly as your writeups have been over the years? Because textbooks generally suck pretty harshly, and statistics hasn't in my experience been so much an exception as an exemplar.
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... Have you actually encountered a stats text in the wild written remotely as comprehensibly as your writeups have been over the years? Because textbooks generally suck pretty harshly, and statistics hasn't in my experience been so much an exception as an exemplar.
Strongly agreed. I have read exactly one good book on statistics (Lady Luck, by Warren Weaver), and everything else has been utterly atrocious.
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No. I quote:
"Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number."
There are no members of V in the context of the post you quoted. V was a number. You yourself stated as such when you acknowledged that the probability was being asserted as "1/V" so I know you read it right. Then you asserted there were members of V.
I also pointed out in the same paragraph that I was tired and apparently hadn't spotted that error in my proofreading. This is a strawman and you know it.
"The lotter involves each player selecting a set of numbers i.e. {15, 27, 33,34,15}, by an arbitrary method, with each number distinctive (with the possible exception of a "powerball" number), and each selected from a finite range of integral values. How did you conclude that I was equating those independent sets with a variable used to enumerate. I was pointing out the possibility of duplicate selection which does affect the distribution. (I can see I was also getting very tired at close to 2am when I said "any member of member "). How about about the set of selected sets "V" where the "v" is the quantity of chosen sets? Does this word salad satisfy you?. The lottery picks are a set of sets each game. The size of the set containing the other sets is finite and quantifiable."
That's just obfuscation.
What am I supposed to be hiding by just describing mechanical structure? Because that is what "obfuscation" is. It's a deliberate act to hide something. You accused me of being deceitful. You made the accusation now support it or withdraw the claim.
The powerball lottery is mathematically equivalent to this situation, which is actually a textbook-class homework assignment:
An integer between one and 292,201,338 is selected at random. A player is asked to guess the number, given the range of possibilities.
First of all at NO point is it a single integer in the range from 1 to 292,201,338. It's a set of much smaller integers in combination. Worse. The number you are quoting? I just checked it:
How many Powerball combinations are there? (http://www.ask.com/hobbies-games/many-powerball-combinations-5d156463b6b22620)
There are 175,223,510 combinations of Powerball numbers. Five numbers are drawn from a pool of 59 white balls, and the final Powerball is drawn from a separate pool of 35 red balls
I suspect that value is 35*59*58*57*56*55 but I haven't checked it. (Update: checked it: 21,026,821,200 is the result of my calculations. 59C5 * 35 to be exact.)
I don't know which lottery you are describing but it's not one I have ever heard of!
That is NOT how the lottery is done. The number is NOT selected in advance and hidden from those choosing. What happens is the players create a "bag" of of numbers sets each being one of 175,223,510 combinations. The "pick" is then generated from the range of combinations in a public, televised, drawing and if it is "in the bag" then, and only then, is there a winner. What does that mean? It means there are v+1 possible outcomes NOT 175,223,510 each drawing. Either one of v or none is the outcome.
Don't give me that "textbook model or assignment" garbage. Not only is that an appeal to authority, but you haven't even bothered to check your authorities for accuracy. Textbooks often simplify their examples or assignments for clarity or for the expected capacity of the student to complete the assignment in a reasonable amount of time - and they have been found to have gotten them wrong before. I've actually found a solution presented in a TEACHER'S edition of a math text to be in error and I've had my math teachers also discover such over the years.
Question one: What are the odds of the player guessing the correct number, given one guess.
Question two: Do these odds change if other people make guesses, if no one is allowed to know whether any of the guesses are correct until all guesses are made?
The answer to the first question is: "one in 292,201,338." The answer to the second question is: "no." Period the end.
This is correct only within the applicable frame of reference but you have not shown the frame of reference you are using is the one that applies. The evidence is to the contrary.
That's correct. The odds of there being a winning combination is n/N, where n is the number of different combinations generated by the players, and N is 292,201,338. No one has said otherwise.
That is also true. IF the device does not select a number chosen by one of the players, then the odds of any particular person winning are zero.
Sure. The odds of *at least one player* picking the winning combination, IF the device has actually picked one of the combinations actually entered, is 100%. That's because the premise is: the lottery picked one of the entries actually chosen by at least one player. However NO ONE IS TALKING ABOUT THAT PROBABILITY, INCLUDING YOU. The statement being argued, and you've repeated yourself, is that the odds of A SPECIFIC PERSON winning change based on the number of other players entering. THAT IS STILL FALSE.
Support that claim with evidence. Basic classroom instruction - show your work. Show that the odds of a given, arbitrary person winning of the possible candidates is restricted to a single factor. You haven't done that yet. You've only made blanket assertions. It's not false because you claim it to be false, but only because you have taken the steps to do that. That's not math; it's the practice of being accountable.
The odds of a specific ticket winning are still 1/N. Why? The odds of a specific ticket winning IF THE LOTTERY MACHINE ACTUALLY PICKS A WINNING NUMBER are 1/n, where n is the number of different combinations actually entered into the lottery. Since one of them was actually picked GIVEN THE PREMISE THAT THE LOTTERY MACHINE PICKED A WINNING NUMBER, the odds of that ticket winning are 1/n, and n depends on the number of people who entered.
In a single paragraph you have both asserted the outcome is both independent of the lottery machine and also dependent. Which will it be? The odds of a specific ticket winning is not 1/N. That's only the odds of it being independently selected from a set of possibilities of size N. It only becomes a chance of being a winning ticket if, and only if, the set in question actually contains a winning entry. This is a dependency created by separate source. You are forgetting an important detail. Percentages aren't merely numbers. They represent fractions of a range. What is the range here? The set of 175,223,510 base combinations. That range is modified by arbitrary points marked by the players. The lottery machine picks a member from that set at random. The base odds of a winner is the chance of the marked combination being selected. It's the same mechanics as a roulette wheel.
HOWEVER, THAT ASSUMES THE MACHINE PICKED A NUMBER ACTUALLY ENTERED. The odds of THAT happening itself are not 100%.
Didn't I already say that?
Therefore, you cannot say "the odds of a ticket winning depend on how many other people enter."
Why not?
You can only say "the odds of a ticket BEING THE WINNING TICKET DURING A DRAWING IN WHICH THERE IS A DECLARED WINNER depends on the number of people entered.
Uh, no. That's not a complete statement. The odds of there even being a declared winner is directly affected by the number of unique entries. See "roulette wheel". These odds are distributed binomially.
That's obvious: if you are the only entrant at all, and the lottery declares that there is a winner, it has to be you.
Yep, and the odds of that happening in a Powerball drawing are one in 175,223,510.
But the odds of a ticket winning are, according to basic probability, the odds of there actually being a winner, multiplied by the odds of you being that winner if there is a winner.
Which is exactly what I said. ;D
The second part, 1/n, is the part you are claiming are the odds of a ticket being a winner.
You might want to recheck that against what I actually said. I notice this time you didn't directly quote me. Why not?
THAT IS FALSE.
And it would be false - if that's what I actually claimed. I gave as my derived formula of individual win chance p(N)/N. I even ran it through a spreadsheet and published the results and pointed out an error I had made. Funny thing about a public forum - claims about what was said or not said can be directly checked.
The true odds of a ticket being a winner are the odds of there actually being a winner - the odds of your premise being true - multiplied by the odds of that winner being that ticket. And the odds of your premise being true - that the lottery machine actually picked a winner - is easy to determine. Given that n was the number of different combinations entered, the odds of one of them being picked are n/N. So the odds of the ticket being the winner are n/N times 1/n. In other words, 1/N. And that's invariant to the number of entries.
Good! You FINALLY address the issue of whether my use of the binomial is flawed. When I first brought it in to play I was considering not a single lottery drawing but multiple. In a single drawing p(N) = n/N is correct and your conclusions are true (That assumes that n is limited to unique entries). In multiple drawings it is not and the binomial formula applies instead.
The direct explanation for your error is that you performed calculations presuming there was a winner, and calculated what the odds were of a single ticket being that winner.
Where did you get that idea? I did my calculations presuming there was a selection process between the existence of any winner and there being none as the first step.
But the odds of there being a winner at all are not 100%. Sometimes there is no winner at all. And obviously, when there is no winner, no ticket wins, and that affects the overall odds of a specific ticket actually winning. You are not calculating the odds of a ticket winning over all possible lottery outcomes. You are calculating the odds of a ticket winning over a subset of them. And if you do that, then you have to state what calculation you are doing correctly.
That's odd. Show me exactly where I did that. I rather distinctly remember a heavy use of the word "zero". I even made a point of emphasizing it. I also spoke the base odds were the result of a weighed "coin-toss". In short I first considered there being NO winner, and then I considered the chance of there being a winner. Even when I was describing my binomial calculations I considering in terms of the chance of failure over a number of trials.
What you are trying to say, incorrectly, is that in all powerball lottery draws in which there is a declaration of a winning ticket, the odds that your ticket is the winning ticket are affected by how many people entered..
Again, that is NOT what I said. And no point did I make such a qualified statement. That would be in direct contradiction to the formula I actually used of p(N)/N. In fact my original formula was only p(N). I introduced 1/N as a coefficient later.
And everyone would agree with that, because that's obvious. IF there is a winner at all, the odds that it is your ticket rise if there are less entrants, and fall if there are more of them. In the degenerate case where you are the only entrant, the odds you are the winner are 100% - it has to be you. In the opposite degenerate case where there are an extremely large number of entries, the odds that you will be one of them fall to a low of 1 in 292,201,338. Obviously, the odds can't get any lower than that.
Actually they can. The number of entries can exceed the number of possible lottery combinations. There is nothing to cap that number except player availability. That's a factor entirely independent of the lottery.
If the lottery says there was a winner, the odds of your ticket being the winner are higher if less people entered. HOWEVER, before the drawing is held and before you know if there actually was a winner or not, the odds of a particular ticket being the winning ticket are still 1 in 292,201,338. Your odds of winning don't change as more people enter.
The odds of there even being a winning ticket is not independent of the number of players. And the odds of a ticket being a winning ticket depends on the size of the pool of potentially winning tickets 1/n (or 1/v or whatever you want to use to reflect the size of that pool). When one, and only one person, enters the drawing the odds are at their worst at 1 in whatever the base number of combinations are. The odds change with additional unique entries. If I am rolling a standard dice and there is only one possible matching value the odds are 1/6. If there are 2 unique values it's 1/3, and so on.
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It would help if you looked at the right year's powerball format before going off on Arcana's numbers.
Back to :-X Really good things seem to happen when I don't talk.
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Maybe still tired?
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Just to change the subject :). I am a bit confused with the compatibility with windows. Being in networking I am unsure if we covered it and I just forgot, but why does some old games seem to have an issue running on newer windows? Such as I was able to play Star Wars: Rebellion on windows 98 and no issue, but as soon as I put in on 8 or 8.1 it cannot run. This is with toying around with the compatibility settings.
I have run into this problem with several games when I upgraded from Win98 to XP. The main cause is, I believe, because Win98 was on a Windows platform, and XP (and everything after it) is on an NT platform (basically the internal structure, as was states in a previous post by someone else). And, DOSBox is the way to go to run most of those games pre-XP. Although, a few still have issues running. I still can't seem to get Battlespire to run, even with DOSBox.
Um, AvelWolrdCreator.
The chances of choosing the same numbers that the person pulling little white balls out of a cage gets won't change depending on whether that person pulling the little white balls out of the cage before or after everyone else picks a number. If you were to take all the possible combinations of the little numbered balls (lets just use 292,201,338 as an example) and ask 500,000 people to pick a single number between 1 and 292,201,338. The chances that any one individual picking the same number that the Guy in Charge reads from the random number generator on his computer will be, I'm sure you would have guessed that I would say, 1 in 292,201,338. That will not change for any of the 500,000 people that were asked to pick a number. It won't change for any individual if the number of people asked to pick was increased by a factor of 100, or decreased by a factor of 1000. The fact that the number of possible outcomes is based on the combinations of a number of little white balls is irrelevant. One out of the total number of combinations is any one individual's chance of wining the lottery.
Personally, I think you and Arcana are discussing two almost entirely different points of almost similar issues. You, I believe are talking about the chances of having a winning ticket (with the assumption that there is a winning ticket). Arcana is talking about the chance of a specific ticket having the same number sequences as the person pulling balls out of a cage.
You're comparing Golden Delicious apples to Granny Smith apples.
As always, correct me if I'm wrong.
Dangit people. Let me post!!! >:(
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It would help if you looked at the right year's powerball format before going off on Arcana's numbers.
Back to :-X Really good things seem to happen when I don't talk.
It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate. ;D
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I have run into this problem with several games when I upgraded from Win98 to XP. The main cause is, I believe, because Win98 was on a Windows platform, and XP (and everything after it) is on an NT platform (basically the internal structure, as was states in a previous post by someone else). And, DOSBox is the way to go to run most of those games pre-XP. Although, a few still have issues running. I still can't seem to get Battlespire to run, even with DOSBox.
Um, AvelWolrdCreator.
The chances of choosing the same numbers that the person pulling little white balls out of a cage gets won't change depending on whether that person pulling the little white balls out of the cage before or after everyone else picks a number.
Well, that depends on if you know what's being drawn or not. :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner (I'm only considering the case of all the numbers being generated being used for selection - the game includes lesser prize categories too).
As the number of possible tickets increase the chance of any of them being a winner increases directly.
If you ever go to a casino with a lottery game you can actually watch the whole process in action (I've actually done this).
The lottery is closer to Bingo or a roulette wheel than just a number guessing game.
Never said that. If you were to take all the possible combinations of the little numbered balls (lets just use 292,201,338 as an example) and ask 500,000 people to pick a single number between 1 and 292,201,338. The chances that any one individual picking the same number that the Guy in Charge reads from the random number generator on his computer will be, I'm sure you would have guessed that I would say, 1 in 292,201,338. That will not change for any of the 500,000 people that were asked to pick a number. It won't change for any individual if the number of people asked to pick was increased by a factor of 100, or decreased by a factor of 1000. The fact that the number of possible outcomes is based on the combinations of a number of little white balls is irrelevant. One out of the total number of combinations is any one individual's chance of wining the lottery.
This would be true if was just a guessing game. Roulette is based on guessing. Bingo is just being lucky enough to have the right card with the right numbers on it (which makes it even closer than roulette to being like the lottery). With lottery you get to "choose your card" or just let the computer pick one out for you. I think this is even allowed in some versions of Bingo.
Personally, I think you and Arcana are discussing two almost entirely different points of almost similar issues. You, I believe are talking about the chances of having a winning ticket (with the assumption that there is a winning ticket). Arcana is talking about the chance of a specific ticket having the same number sequences as the person pulling balls out of a cage.
You got it! Almost. I'm working first with the odds of their being a winning ticket. That's that "p(N)" part. I'm not actually assuming there is even a winning ticket at all, just the possibility. Then I'm applying that possibility to the chances of any person having that ticket if it exists - that's the "1/N" part. Two parts - one problem. And I'm not limiting things to a single lottery drawing but a number of such drawings over time. It's really just the Law of Large Numbers but I'm actually doing the math.
You're comparing Golden Delicious apples to Granny Smith apples.
As always, correct me if I'm wrong.
Dangit people. Let me post!!! >:(
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First of all at NO point is it a single integer in the range from 1 to 292,201,338. It's a set of much smaller integers in combination.
It's a pretty basic problem to show that they are mathematically equivalent, which is what proofs are based on. Arcana detailed the steps to showing them as equivalent earlier in the thread.
That range is modified by arbitrary points marked by the players. The lottery machine picks a member from that set at random. The base odds of a winner is the chance of the marked combination being selected. It's the same mechanics as a roulette wheel.
The whole "set of entries defined by the players" is a massive unnecessary overcomplication of the problem. The lottery machine picks a number (or a set of numbers, they're equivalent), whether anyone bought any tickets or not.
Arcana attempted to show that the complex overcomplication does cancel itself out (N ends up in both the numerator and the denominator), but you rejected the logical progression out of hand.
When I first brought it in to play I was considering not a single lottery drawing but multiple.
Here's the root of the problem. There is only one drawing. Why are you considering multiple drawings? A ticket is only valid for a single drawing. It almost seems like you have a fundamental misunderstanding of how lotteries work.
That's the reason that it can also be shown as equivalent if the winning numbers are determined before or after everyone buys tickets.
The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.
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Well, that depends on if you know what's being drawn or not. :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner
No, that would be a raffle. It's been stated multiple times in the thread that there doesn't have to be a winner, even if everyone didn't already know that from the increasing jackpot that made everyone start talking about powerball in the first place.
It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate. ;D
funny because when i type 'powerball odds' into google I get
"On October 4, 2015, the Powerball format changed again; the white-ball pool increased from 59 to 69 while the Powerball pool decreased from 35 to 26. While this improved the chance of winning any prize to 1 in 24, it also lengthened the jackpot odds to 1 in 292,201,338."
in a little box at the top of the page, prior to any of the links to sites. And it did that the first time I ever googled the odds yesterday so it's not just a tailored result caused by me never clearing my history.
and as long as I'm talking again, damn me, Arcana saying something is a textbook example, which is a pretty common idiomatic phrase, then proceeding to state it in the format of an actual textbook example, is not an appeal to authority.
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The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.
and also to keep from going to jail :D
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Well, that depends on if you know what's being drawn or not. :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner
That's not how Powerball works. Powerball draws a sequence of numbers completely at random, and then checks to see if any of the entries matches. If no ticket matches, the the lottery is declared to have no winner. All tickets are then discarded (if they win secondary prizes they are paid off, but we're only considering winning the jackpot) and the lottery starts over completely from scratch, except the money allocated to the jackpot prize is "rolled over" to the next drawing. That's how Powerball can end up with a billion dollar prize. It starts small, and every time there is no winner the pot rolls over to the next lottery. But you have to buy new tickets to enter that contest. As Vee mentions above, you are talking about a raffle where the winner is chosen from a set of entries. Powerball doesn't work that way. Powerball takes a set of 69 balls each with the numbers one through 69 and draws five of them randomly. That generates a combination of five numbers from 69. Then it draws another ball from a different set of balls numbered one through 26. The powerball number can theoretically be identical in number to one of the other previously drawn balls, although the first five must obviously be different.
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It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate. ;D
I never said that. I said the powerball lottery drawing is mathematically equivalent to choosing a single integer from one to 292,201,338. The process of picking balls and having people choose six numbers rather than one is just a human convenience. Much like IP addresses. This is not an IP address: 10.1.1.1. That's a human convenience for expressing IP addresses aka dotted decimal. The actual IP address that corresponds to what a human being would write as 10.1.1.1 is 167837953 in decimal, or 00001010000000010000000100000001 in binary. In other words, an IP address is an unsigned 32-bit integer. A single number. For anyone who hasn't done this before, I suggest trying this: ping 167837953. See what happens. Your computer knows what I do: an IP address is a single number, and when human beings type it using four numbers it has to be converted into a single number first. When you give it a single number, it says "oh good" and uses that directly without conversion.
The lottery operators know what they are doing. They engineer the number of balls and the way they are chosen to create a very specific chance of winning per ticket entered. And the lottery operators both know what that chance is, and are required by law to publish it. The odds of winning the powerball (given the current drawing rules) is one in 292,201,338. If those odds varied based on the number of people who entered the lottery, they would be required by law to state that. They do not. The reason why they don't require people to pick a number between one and 292,201,338 is gamesmanship, and nothing more. Mathematically the lottery would operate the same if that were the case.
Finding and recognizing mathematically equivalent or mathematically congruent situations is a foundational principle of analytical mathematics.
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The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.
There was a time in the not too distant past where small-time slot machine operators were allowed to tinker with their machines to change their payouts. Failure to understand exactly how statistics worked sometimes caused them to make serious payout mistakes. Once when I was in Reno I came across a small-ish establishment that had a slot machine that took dimes. It wasn't well-played, specifically because it took dimes and most people don't have a lot of those in their pockets. But I happened to stop and take a look at it, and someone had altered the payout table in such a way that while the payout favored the house if you played one dime, and if you played three dimes, it actually favored the platyer if you played two dimes. I had three dimes. I put two into the machine. A half hour later, I had emptied the machine and had to figure out what to do with a giant bucket of dimes.
There are days when I wish anyone who wanted to challenge my understanding of probability would have to create a slot machine expressing their ideas so I could play it. Then I would no longer have to debate the point too much. I could just break the house, fist pump a couple of times, and ask the next challenger to step forward. Technically, there is a way to do that for the case of someone who believes the odds of a lottery ticket winning are dependent on the number of entries posted. Any mismatch between the calculated odds and the true odds provides an exploitable arbitrage opportunity.
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Basically from this what you been saying is what this guys says:
http://www.durangobill.com/PowerballOdds.html
As near as I can tell, that person's calculations and logic are valid, with one exception. At the very end he calls lotteries "zero-sum games." That's kind of true, but usually the term "zero-sum" refers to the notion that anything any participant wins must come from another participant's lossess. He calls entities like the government "participants" but in a games-theoretical sense, and in the general colloquial sense most people mean, the government is not a "participant." The lottery operators are really the house, not a player. And it is theoretically possible to break the house in a lottery like the powerball, although it is astronomically unlikely.
See, the jackpot is funded by the participants, and if there are multiple winners they must all share the same prize. So it is impossible for the jackpot winners to win more than a certain fraction of the total money spent on lottery tickets. However, there is no such limit on the secondary prizes: anyone who wins a secondary prize is entitled to the full value of it, outside of parimutuel laws in some states. So it is at least mathematically possible that a large group of people all play the same numbers, hit the five white numbers and miss the powerball and all become entitled to the $1,000,00 second place prize. There's no limit mathematically on how high the losses for the lottery could go, although I suspect there are terms and conditions to account for this possibility. It is extremely unlikely any group of people is going to break the bank of the powerball lottery, but this sort of thing on a smaller scale has happened before (https://www.lotterypost.com/news/112702).
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It's a pretty basic problem to show that they are mathematically equivalent, which is what proofs are based on. Arcana detailed the steps to showing them as equivalent earlier in the thread.
I know they are equivalent if there is actually only one contest in play. But there are several contests being played and that breaks the equivalence somewhat. Now if she said "Makes a drawing equivalent to an integer between 1 and R" then there is no dispute. She got onto me about being precise in my word usage. Getting onto me when I do the same is hypocritical.
I notice you didn't dispute me questioning the actual value she used though. You might notice I resorted to using R as the number because I have found three possible distinct values for what it could be. Saves typing and avoids pointless dispute.
The whole "set of entries defined by the players" is a massive unnecessary overcomplication of the problem. The lottery machine picks a number (or a set of numbers, they're equivalent), whether anyone bought any tickets or not.
That the machine generates numbers regardless of tickets bought is a given. But the odds of a winner depend directly on the number of unique tickets actually bought - that a significant factor not an "unnecessary overcomplication of the problem". If zero tickets are bought the odds of a winner are 0 in R or just zero. If one ticket is bought it is one in R, and so one. u(N)/R.
Arcana attempted to show that the complex overcomplication does cancel itself out (N ends up in both the numerator and the denominator), but you rejected the logical progression out of hand.
I didn't reject it out of hand. I even allowed it as a possibility (actually that was an error but that comes next).
But the number of unique tickets bought is not n/N or even n/R. If the number of tickets bought is v, then the percentage of the total number bought follows a binomial distribution. It's not a linear result. Even if R tickets are bought the percentage of tickets that are unique is not automatically 100%. If u(v) is the percentage of unique tickets bought then the number of unique tickets, U, is u(v)*R. Since the odds of a winning draw is U/R then u(v) is the percent (notice the value that actually dropped?). u(v) is simply a derivative of the binomial distribution formula. Arcana ignored the issue of unique value - I didn't. Now if lim v→∞ u(v) DOES equal a 100% chance of all possible tickets being bought but I really doubt there is an infinite number of buyers. ;D
We both agreed that the odds of a person having the winning ticket is one in v, and we even agreed that by the rules of statistics the final odds were the product of that and the odds of there being a winning ticket. So the final formula still winds up being u(v)/v.
Here's the root of the problem. There is only one drawing. Why are you considering multiple drawings? A ticket is only valid for a single drawing. It almost seems like you have a fundamental misunderstanding of how lotteries work.
Really? There was, and will only be, only one lottery drawing? It won't happen at least once a week or 365.25/7 weeks a year?. Amazing! I thought it was otherwise! ... Yes I'm being sarcastic. I'm considering multiple drawings because I understand the binomial distribution AND how the lottery works. "The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N." That "drawn with replacement" is your ticket sales between each drawing. And the definition of "success" is the greater value when the there is only a single drawing. I was considering what the win chance of a random player was over a series of lottery drawings - which happen at least 52 times a year.
That's the reason that it can also be shown as equivalent if the winning numbers are determined before or after everyone buys tickets.
Actually they ARE equivalent as long as the no one knows the winning selection before they buy and the players confident the drawing isnt rigged. But they are not equivalent events if the issue is making sure the drawing is public for accountibility reasons. Which kind of nullifies that "not know the winning selection" part.
The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.
Yep. In fact I just looked up Bingo odds. The lottery is just "Bingo Lite". If you ever go to a casino with a lottery board you will see a bunch of squares that people pick in advance (their tickets) and some guy drawing balls. What do you see in a bingo game? The players with their "tickets" (bingo cards) and some guy - drawing balls. The lottery just has fewer slots and more limited game variations.
Bingo - FAQ (http://wizardofodds.com/ask-the-wizard/bingo/)
And pay attention to that very first question. See that formula with exponents? Yeah. That's an application of the binomial formula.
And now go down to the third question. He's computing odds according to the number of players. Notice it varies? Do you notice as the number of players goes up the chances of a winner appear quicker goes up?
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There was a time in the not too distant past where small-time slot machine operators were allowed to tinker with their machines to change their payouts. Failure to understand exactly how statistics worked sometimes caused them to make serious payout mistakes. Once when I was in Reno I came across a small-ish establishment that had a slot machine that took dimes. It wasn't well-played, specifically because it took dimes and most people don't have a lot of those in their pockets. But I happened to stop and take a look at it, and someone had altered the payout table in such a way that while the payout favored the house if you played one dime, and if you played three dimes, it actually favored the platyer if you played two dimes. I had three dimes. I put two into the machine. A half hour later, I had emptied the machine and had to figure out what to do with a giant bucket of dimes.
Lesson learned: Arcana goes to Vegas with me if I ever go. :O
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Bingo - FAQ (http://wizardofodds.com/ask-the-wizard/bingo/)
And pay attention to that very first question. See that formula with exponents? Yeah. That's an application of the binomial formula.
And now go down to the third question. He's computing odds according to the number of players. Notice it varies? Do you notice as the number of players goes up the chances of a winner appear quicker goes up?
Oooh. I get why you're refusing to see logic. You're talking about multiple lottery draws and formulating chances over time. Well, no one else is.
Everyone else - literally everyone else - is talking about a single lottery draw. And has always been since the very beginning of the topic. A single ticket in a single lottery has 1/V chance of being chosen. Period, end of story, finito. (And it never changes. Lotteries are singular events. If no one wins, all tickets are voided. Buy another ticket. That ticket has a 1/V chance of being chosen, too.)
Stop talking about multiple lottery draws. lol
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If you ever go to a casino with a lottery board you will see a bunch of squares that people pick in advance (their tickets) and some guy drawing balls.
You mean Keno? Your ticket doesn't stay valid for multiple draws in it either.
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Arcana -
Ok. Let's keep it simple. Here are the point that you will have to show I'm in error for me to accept that is the case. Now I must point out that on a previous point you stated out that I had made an error (though you did not point to the error I actually made), I checked my work, and found - lo and behold! - I had in fact made and error. I even posted my results with my error for all to see. I was prepared to let the matter drop after that point. I didn't initiate any further conversation other that to point it out as a lesson in checking your work. You started the matter back up, with me as the target of accusation. You have gone so far of accusing of deceit even though I've been quite forthcoming. Now, this is what you need to demonstrate if you are going be able to support your accusations against me.
1. Show that the number on each ticket is not an independent random selection.
2. That if it IS an independent random selection that the probility of selected values does not follow a binomial distribution.
3. That if R is the number of possible lottery numbers, and if the number of tickets sold, v, approaches R, then the number of unique tickets, U, also approaches R, and when v= R then U=R.
4. That the odds of a winning ticket, W, is not U/R.
5. That the odds of a given ticket being a winning ticket, P, is not W/v.
And a bonus:
6. Show the odds of winning does not improve with repeated playing.
7. And that if the odds of that repeated playing does not also have a binomial distribution.
But... you have already "corrected me" by asserting conditions 4 and 5 are already true. That's definitely not a good start. Now why don't you see if you can deal with 1..3. Because that's what you are going to have to demonstrate as untrue if you are to show my formula P=U(v)/v is also untrue. That formula and how it is derived is my entire argument. It's the one I used to generate my spreadsheet results. If I am so mathetically dense and incompetent as you are definitely try to publicly argue (and, yes, you HAVE used words to that effect), then doing that should be trivial for you.
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Oooh. I get why you're refusing to see logic. You're talking about multiple lottery draws and formulating chances over time. Well, no one else is.
Funny thing is that it doesn't matter if you consider multiple draws or not. Read on.
I know they are equivalent if there is actually only one contest in play. But there are several contests being played and that breaks the equivalence somewhat.
It doesn't break it all since each drawing is a completely isolated event. If one drawing is mathematically equivalent to generating a single random number out of 292,201,338, then multiple drawings are equivalent to generating several numbers between 1 and 292,201,338.
I notice you didn't dispute me questioning the actual value she used though.
I try to keep my posts short and concise rather than nitpicking every single little thing. Helps avoid TL;DR syndrome.
But the odds of a winner depend directly on the number of unique tickets actually bought - that a significant factor not an "unnecessary overcomplication of the problem".
It's unnecessary because the odds of there being a winner at all is a completely useless piece of information unless you're the powerball commission trying to drive up the jackpot. If I'm buying a ticket, I only care what the odds of me winning a large amount of money are. Those are clearly printed on the label, easy to verify, and do not depend on how many people also buy tickets.
Really? There was, and will only be, only one lottery drawing? It won't happen at least once a week or 365.25/7 weeks a year?. Amazing! I thought it was otherwise! ... Yes I'm being sarcastic.
Your odds across multiple drawings is just the number of drawings you enter times the odds of winning a single drawing. Since everything resets between drawings except for the jackpot size (which doesn't affect how the game is played), you only need to model a single drawing.
I'm considering multiple drawings because I understand the binomial distribution AND how the lottery works. "The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N."
I'm familiar with biniomal distributions, they're what is responsible for the bell curve when rolling multiple dice, etc. However trying to model the powerball as one doesn't make sense -- p is such a small number that you would need an absurdly large n in order to have more than one success in the sample. I suppose if you were ultra-rich and wanted to find out your odds of winning long term when buying millions of tickets every drawing.
Since tickets are independent of each other, you can't just apply it across everyone at once, either. Well, you can, but then it simply gives you the odds of someone, somewhere winning after N draws, which again doesn't seem like a particularly useful thing to know.
I was considering what the win chance of a random player was over a series of lottery drawings - which happen at least 52 times a year.
That's the problem, you weren't. In order to do that over time, you would have to divide the probability by the number of tickets sold, which fluctuates with each drawing. If all you care about is the odds of a single player winning, it's much simpler to just multiply instantaneous odds by the number of drawings they participate in.
The lottery is just "Bingo Lite". If you ever go to a casino with a lottery board you will see a bunch of squares that people pick in advance (their tickets) and some guy drawing balls. What do you see in a bingo game? The players with their "tickets" (bingo cards) and some guy - drawing balls. The lottery just has fewer slots and more limited game variations.
Powerball and Bingo are mechanically different and do not share the same probabilities. In Bingo, numbers are cumulatively covered on your card and drawing continues until someone wins. State does not reset between every drawing. That's why the site you linked uses exponents to calculate win chances after a certain number of balls drawn.
By contrast, Powerball resets everything between drawings, and the show of drawing balls one by one is meaningless as it's impossible to win before all of the balls are drawn.
binomial formula.
Honest suggestion: You might want to try using tools other than a hammer once in a while.
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Oooh. I get why you're refusing to see logic. You're talking about multiple lottery draws and formulating chances over time. Well, no one else is.
Everyone else - literally everyone else - is talking about a single lottery draw. And has always been since the very beginning of the topic. A single ticket in a single lottery has 1/V chance of being chosen. Period, end of story, finito. (And it never changes. Lotteries are singular events. If no one wins, all tickets are voided. Buy another ticket. That ticket has a 1/V chance of being chosen, too.)
Stop talking about multiple lottery draws. lol
When I brought up this topic I thought it amusing because it was so off-topic to the thread when I first saw it as (I thought) long-dead topic - I think I even pointed that out. I mainly was pointing out that if someone kept playing their chances would improve. I initiated the conversation with multiple lottery draws and showed the formula. For the current argument on the topic I was the OP. And in that positing mutiple drawings was part of the argument.
Ok. The point that Arcana and I are sticking on is related to the odds of a winning ticket being drawn in the first place. We are using two different formulas. The formulas are quite similar but differ on a single item. The number of possible winning tickets being drawn from. Her's is basically (if I have read it right) just "v" where v is the number of ticket sold, and mine is u(v). Both of us agree that the odds of there being a winner depends on the number of tickets sold (if there is no tickets sold nothing drawn by the lottery will produce a winner). We even agree that the chance of a person having a winning ticket is the chance of a winner times the chance of a person having a winning ticket. But the choice of that "v" vs. "u(v)" makes a huge difference in outcome.
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You have gone so far of accusing of deceit even though I've been quite forthcoming.
No, she didn't. She said something you said was obfuscation. You can be obfuscating without being deceitful.
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*Watches all the math/statistics talk*
(https://images.weserv.nl/?url=images.sodahead.com%2Fblogs%2F000348543%2Fdavid_letterman_xlarge.jpeg)
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Ok. The point that Arcana and I are sticking on is related to the odds of a winning ticket being drawn in the first place. We are using two different formulas. The formulas are quite similar but differ on a single item. The number of possible winning tickets being drawn from. Her's is basically (if I have read it right) just "v" where v is the number of ticket sold, and mine is u(v). Both of us agree that the odds of there being a winner depends on the number of tickets sold (if there is no tickets sold nothing drawn by the lottery will produce a winner). We even agree that the chance of a person having a winning ticket is the chance of a winner times the chance of a person having a winning ticket. But the choice of that "v" vs. "u(v)" makes a huge difference in outcome.
Golden Delicious meet Granny Smith.
*Watches all the math/statistics talk*
(https://images.weserv.nl/?url=images.sodahead.com%2Fblogs%2F000348543%2Fdavid_letterman_xlarge.jpeg)
Oh no.
Who's going to write more of my favorite Valdemar books?!? :'(
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Ok. The point that Arcana and I are sticking on is related to the odds of a winning ticket being drawn in the first place. We are using two different formulas. The formulas are quite similar but differ on a single item. The number of possible winning tickets being drawn from. Her's is basically (if I have read it right) just "v" where v is the number of ticket sold, and mine is u(v). Both of us agree that the odds of there being a winner depends on the number of tickets sold (if there is no tickets sold nothing drawn by the lottery will produce a winner). We even agree that the chance of a person having a winning ticket is the chance of a winner times the chance of a person having a winning ticket. But the choice of that "v" vs. "u(v)" makes a huge difference in outcome.
Actually, this is wrong in several respects:
1. My sticking point is not what formulas you use, but rather what I said it was every single time I've posted about the subject. The original statement being discussed was "the odds of a ticket having the winning sequence are dependent on the number of people who enter the powerball." I assert that is false, you have asserted several times that statement is true. However, none of your math, regardless of the accuracy of the calculation, supports that statement.
Calculation can be correct, irrefutable, and nevertheless completely invalid. I assert the Earth has three moons. Proof: 1+1+1=3. Ergo, three moons.
2. I never said the odds of a winning ticket being drawn were equal to "v" where v was the number of entries. Claiming that makes me wonder if you are even reading my posts at all. What I said several times is that if n is the number of distinct sequences entered then the odds of a winning ticket being drawn at all are n/N. n, the numnber of distinct entries, is something I never gave a formula for. Several reasons. First, lottery entries aren't actually random and independent. Some people enter one entry chosen randomly. Some people enter many times. Most of the people who enter many times pick different sequences for each entry, because why enter the same entry multiple times? This means that the function f such that f(e) = n where e is the number of entries is non-trivial. For my purposes, however, it doesn't matter how to calculate it, because it ultimately doesn't matter what the value of n is within the context of the question, which I will remind you is do the odds of a specific ticket winning the powerball lottery change with the number of other entries submitted. The calculations, which I presented several times now, show that ultimately n factors out of that expression, so the function f is ultimately irrelevant.
3. "We even agree that the chance of a person having a winning ticket is the chance of a winner times the chance of a person having a winning ticket. " No, we don't agree on this either. The chance of a person having a winning ticket is the chance of a person having a winning ticket. It is not the chance of having a winning ticket times another factor. What I said was that the chance of a specific ticket having the winning combination is equal to the odds of a winning sequence being drawn at all, times the odds of selecting the winning sequence from the set of all entered sequences. Which I've shown mathematically is equal to the odds of the ticket matching the winning sequence out of all possible sequences.
All three of these things are so obvious in my posts I doubt anyone who has stayed awake long enough to read them would have any difficulty quoting the passages specifically.
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Arcana -
Ok. Let's keep it simple. Here are the point that you will have to show I'm in error for me to accept that is the case. Now I must point out that on a previous point you stated out that I had made an error (though you did not point to the error I actually made), I checked my work, and found - lo and behold! - I had in fact made and error. I even posted my results with my error for all to see. I was prepared to let the matter drop after that point. I didn't initiate any further conversation other that to point it out as a lesson in checking your work. You started the matter back up, with me as the target of accusation. You have gone so far of accusing of deceit even though I've been quite forthcoming. Now, this is what you need to demonstrate if you are going be able to support your accusations against me.
1. Show that the number on each ticket is not an independent random selection.
2. That if it IS an independent random selection that the probility of selected values does not follow a binomial distribution.
3. That if R is the number of possible lottery numbers, and if the number of tickets sold, v, approaches R, then the number of unique tickets, U, also approaches R, and when v= R then U=R.
4. That the odds of a winning ticket, W, is not U/R.
5. That the odds of a given ticket being a winning ticket, P, is not W/v.
And a bonus:
6. Show the odds of winning does not improve with repeated playing.
7. And that if the odds of that repeated playing does not also have a binomial distribution.
But... you have already "corrected me" by asserting conditions 4 and 5 are already true. That's definitely not a good start. Now why don't you see if you can deal with 1..3. Because that's what you are going to have to demonstrate as untrue if you are to show my formula P=U(v)/v is also untrue. That formula and how it is derived is my entire argument. It's the one I used to generate my spreadsheet results. If I am so mathetically dense and incompetent as you are definitely try to publicly argue (and, yes, you HAVE used words to that effect), then doing that should be trivial for you.
I'm sorry, but this is word salad. You are asserting that for me to meet your criteria of demonstrating you are in error, I am required to:
a) Show that the number on each ticket is not an independent random selection.
b) That if it IS an independent random selection that the probility of selected values does not follow a binomial distribution.
Going to stop right there. You want me to prove that the numbers on the tickets are not independent random selections, and if they are the selected values do not follow a binomial distribution. Those are essentially contradictory things. Moreover, none of these things is actually necessary to prove you wrong. None of those statements directly supports the statement "the odds of a ticket having the winning sequence is dependent on the number of people who enter." You keep talking about the probability of a ticket having a number that matches a binomial distribution of other selections. I have no idea why.
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Funny thing is that it doesn't matter if you consider multiple draws or not. Read on.
Oh, I am.
It doesn't break it all since each drawing is a completely isolated event. If one drawing is mathematically equivalent to generating a single random number out of 292,201,338, then multiple drawings are equivalent to generating several numbers between 1 and 292,201,338.
Each drawing represents more than one contest. You can treat each contest as multiple drawings at the same time, but trying to account for the interdepencies between the draws simply makes things needlessly complicated. Saying that you draw 1 of R combinations is no more complicated than saying you draw 1 of R integers and it actually avoids confusion and is more accurate.
I try to keep my posts short and concise rather than nitpicking every single little thing. Helps avoid TL;DR syndrome.
Good practice but you ducked my complaint of hypocrisy.
It's unnecessary because the odds of there being a winner at all is a completely useless piece of information unless you're the powerball commission trying to drive up the jackpot. If I'm buying a ticket, I only care what the odds of me winning a large amount of money are. Those are clearly printed on the label, easy to verify, and do not depend on how many people also buy tickets.
What is printed on the ticket is a simplified number. It is indeed a form of odds. It's the odds of a specific value being drawn. That is all.
The odds of being a winner is useless - if every drawing guaranteed a winner. But the odds of all tickets being a loser exists as a strong possibility and affects the final odds. If you have ever played roulette you know there is a some outcomes you can't put your chips on - the house always wins. If you are calculating odds in that game you can't ignore that possibility. And that game and the lottery are largely equivalent. The number of tickets sold affect the odds of there being a winner and in no small way.
Your odds across multiple drawings is just the number of drawings you enter times the odds of winning a single drawing. Since everything resets between drawings except for the jackpot size (which doesn't affect how the game is played), you only need to model a single drawing.
As I've been onto Arcana about already, show your work. Don't just make the claim, prove it. Show that the chances of continuing to play the lottery over several games does not increase your chances of winning or does not follow a binomial distribution. Show that the formula you just presented is the correct one. What I see is P(G)=G*P(1) as your assertion. So far I've always tried to show my work, even when I found it produced an error I showed my mistakes and explained them. I may not always be perfect in this but I make a good faith effort to try. And if someone points out an omission I try to correct. If something is true you should be able to show why that is the case. It's called "accountability".
I'm familiar with biniomal distributions, they're what is responsible for the bell curve when rolling multiple dice, etc. However trying to model the powerball as one doesn't make sense -- p is such a small number that you would need an absurdly large n in order to have more than one success in the sample. I suppose if you were ultra-rich and wanted to find out your odds of winning long term when buying millions of tickets every drawing.
Really? In practice p is supposed to be chosen to be the larger of two alternates. It's supposed to represent the chance of success, not failure. In this of u, p is at least (1-1/R). p is the chance that a given ticket is unique. The distribution winds up being (1-1/R)^v. When you consider that v is probably larger than a million u(v) starts to develop statistical significance. And that's just the start.
When you consider the chance of a person of a given person winning over a number of trials ("games"), p is the LOSING odds for the player - it's the greater value. You are figuring the odds of someone LOSING each every one of G games. p=1-(u(v)/v) so L=(1-(u(v)/v)^G
Since tickets are independent of each other, you can't just apply it across everyone at once, either. Well, you can, but then it simply gives you the odds of someone, somewhere winning after N draws, which again doesn't seem like a particularly useful thing to know.
Tickets are independent of each other - until you consider the question of uniqueness.
That's the problem, you weren't. In order to do that over time, you would have to divide the probability by the number of tickets sold, which fluctuates with each drawing. If all you care about is the odds of a single player winning, it's much simpler to just multiply instantaneous odds by the number of drawings they participate in.
Oh, I agree that's an issue. I didn't ignore it but I didn't bring it up because I was still trying to find a workable solution. But I don't think your method solves the problem; not saying it doesn't, just that that I get a gut feeling that the suggested solution isn't complete. The problem is v is no longer a constant. we have v=V(G). That's the problem with any model really. A single factor can cause the complexity of any analysis to explode. It explains our desire to simplify things but we run the risk of possibly oversimplifying the problem. But that was the part of the origin of statistics - an effort to simplify and homogenize. First thing you learn is how to derive the mean value. Later you learn that's not the whole story and you learn things like mode and median. Then comes questions of distribution and variance. And you STILL aren't done with learning. You also have to figure out how to plan your models and ask the questions the model brings. You have to deal with errors of type alpha and beta. And getting into the multivariate brings substantially more complexity.
Powerball and Bingo are mechanically different and do not share the same probabilities. In Bingo, numbers are cumulatively covered on your card and drawing continues until someone wins. State does not reset between every drawing. That's why the site you linked uses exponents to calculate win chances after a certain number of balls drawn.
The main differences and the person continuing to draw until there is a winner and the greater of slots on your ticket/card. That is why I first used the model of a roulette wheel to compare with as it is the closest game mechanically to the lottery in several other ways. Unfortunately getting any straightforward information of figuring odds has been a problem.
But...
I carried on with that site. And there is a direct section on the Powerball lottery and odds.
Powerball Lottery (http://wizardofodds.com/games/lottery/powerball/)
The part I love is close to the bottom. The "Return To Player Jackpot Size" table. He relates number of tickets sold to the odds of winning. Looks great... until you distribute the winning odds across the number of tickets sold. Which is both mine and Arcana's argument!
By contrast, Powerball resets everything between drawings, and the show of drawing balls one by one is meaningless as it's impossible to win before all of the balls are drawn.
Quite true. The only reason for that show is to demonstrate each number drawn is the result of chance and has not been weighed towards certain results. The accountability isn't meaningless though. The theatrics? The actors present? Not needed.
Honest suggestion: You might want to try using tools other than a hammer once in a while.
But I LOVE my hammer! I keep finding all these nails to hit! (what do you mean he isn't a nail? He's upright and has a hard head! Oh wait! That's ME!) ;D :P
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You mean Keno? Your ticket doesn't stay valid for multiple draws in it either.
I was going to reply to this one as well, but I am choosing to refrain from commenting on someone professing to have any competency regarding that "lottery board game" in casinos.
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Don't just make the claim, prove it. Show that [the chances of](sic, I assume) continuing to play the lottery over several games does not increase your chances of winning...
Literally no one has claimed that. Obviously if you enter two drawings you have a better chance of winning than if you only entered one. What people have claimed is that your continuing to play the lottery over several games does not alter the odds of any particular drawing. Just as if you flipped a coin twice you'd have better odds of one of them being heads, but the odds on each flip would still be even.
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*Watches all the math/statistics talk*
(https://images.weserv.nl/?url=images.sodahead.com%2Fblogs%2F000348543%2Fdavid_letterman_xlarge.jpeg)
Yeah, I think it has probably crossed over the horizon to "everyone who can be convinced already is, and is now just taking bets on how long before I summon the algebra gods to smite the infidel." I used to have an overabundance of hope in these areas on the official forums. Back then I used to wonder if Dr Rock** had days like this on the Euro forums. I heard people used to just confirm his stuff and move on. It took me almost two years to get my mitigation formulas generally accepted. It was probably a metric system thing.
** Dr Rock was probably the most well known forum poster no one on the (US) forums ever heard of. I was told he was one of the early quants Euro-side and used to post all manner of guides and even made some spreadsheet tools that predate mine by over two years. I exchanged a couple PMs with him, but we never had any lengthy discussions, mores the pity. I have no idea what eventually happened to him, because I think he stopped playing before the US-Euro merge.
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Huh! I'll pass that on. I think I might have seen something of that nature in the works but I'll ask. Thanks for that idea. May not be used and it may not be unique, but still worth pondering.
I got the Idea from Fire Emblem (GBA) luck is determined in several things there, it works as more of a randomizer than anything else.
in an MMO context I regret to say there's not many ways to work with it without it becoming a game breaking "holy thing that all must have"
when it comes to luck, if you didn't already build it into your stat system from the beginning it will definitely break something or mean reworking the entire stat system which could be a nightmare if you already have powers and sets built around a different system.
so yeah, consider it, and more power to you if you can find a way to add it in without breaking things lol.
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I remember several RPGs from when i was a kid (i want to say some of the final fantasy games even) having a luck stat that wasn't explained in the manual (yeah, i'm old) and didn't seem to change with leveling or be able to be enhanced. And of course I'd always have a vague idea of what it was probably supposed to do by virtue of knowing what the word 'luck' meant but it always kind of bothered me anyway. I think at some point I decided it must be a holdover from pen and paper systems and was just an 'under the hood' randomizer of some sort and just quit worrying about it.
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I'm sorry, but this is word salad. You are asserting that for me to meet your criteria of demonstrating you are in error, I am required to:
a) Show that the number on each ticket is not an independent random selection.
b) That if it IS an independent random selection that the probility of selected values does not follow a binomial distribution.
Going to stop right there. You want me to prove that the numbers on the tickets are not independent random selections, and if they are the selected values do not follow a binomial distribution. Those are essentially contradictory things. Moreover, none of these things is actually necessary to prove you wrong. None of those statements directly supports the statement "the odds of a ticket having the winning sequence is dependent on the number of people who enter." You keep talking about the probability of a ticket having a number that matches a binomial distribution of other selections. I have no idea why.
Really? First the dismissive statement "that's just word salad". That's an evasion.
The only statement there that needed better clarification is "2" (I used numbers, not letters for enumeration). That should be the "probability of the selected values BEING UNIQUE does not follow a binomial distribution"
Show me exactly where I talk about A ticket having a number that matches the binomial distribution of other tickets. I speak of the probability of selected values - a plural; a group - when speaking of any distribution .
I gave a list of possible rebuttals. All you had to do demonstrate - not just make an assertion - how ANY of them were untrue. This is a requirement in any branch of mathematics, theoretical or applied. It distinguishes between the amateur and the professional. Objective demonstration of your claims. A professional does not resort to insults, ad hominems, dismissals of the other party's concerns, questions, or viewpoints, or avoids direct requests to substantiate any claim made. The professional accepts admissions of error by another person and doesn't continue to press the subject. The professional aims to be critical without being accusatory. A professional can confident that if he (or she) is called into court to testify on an area they are considered expert that they will be able to present their conclusions and fully support them but are prepared to admit error if that is discovered when those are examined. A professional is prepared also prepared to do the same in the court of their peers. A professional is able to distinguish between what is subjective opinion, assumptions, and emotional appeals or other logical fallacies and what is objective evidence, argument, and logical analysis.
And this point I have no other choice to conclude you are not a professional in any field. You are a highly skilled amateur who has developed a following based on a few past accomplishments, emotional argument, and esoteric language. You sound good on the surface but when pressed, or you feel challenged, the internal rot begins to leak out. I find you arrogant and needlessly argumentative. You avoid taking responsibility for your claims and you depend on browbeating the other party in order to "win". I've had disputes with other people in this forum but a lot of it has been miscommunication or loose statements with no ill-will intended. I've been caught on mistakes and I've readily owned up to them when I recognized I had been sloppy in my work. I don't fear failure. In fact mistakes I can walk away from, even when I've been seriously injured, are things I've laughed about. I came to these forums almost three years ago to find out if there was a way to rescue the game and the community I had come to love and care about. I've spent almost all my free time and other personal resources to this end in the way that seemed the most viable and I began here; the record still exists in the archives of these forums. I tried to limit the number of personal claims since I most recently showed up. I've admitted to only one superlative though I can admit to more. I've gone through direct assessment of my abilities by demonstrably trained and accomplished professionals who had been proven qualified to do so - something you have failed to do - and have done well. That's all that needs to be said on that topic. The only thing you have actually proven to me? That you are a board troll. I do not feed trolls when I discover them. Goodbye.
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Lesson learned: Arcana goes to Vegas with me if I ever go. :O
My win/loss ratio in casinos is quite good too. I play the slots for fun only and not for money - I play blackjack if I'm going for that; my usual take is 10 times what I started with. I only failed once when I let my emotions cloud my judgement. If I find a broken machine like she did I report it; I consider to do otherwise theft and fraud.
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A professional does not resort to insults, ad hominems, dismissals of the other party's concerns, questions, or viewpoints, or avoids direct requests to substantiate any claim made.
Good to know.
If I may be permitted some hyperbole (and why wouldn't I? everyone knows hyperbole is the greatest thing ever), I'd liken this whole conversation to a geometry teacher asking a student for the measure of an interior angle in an equilateral triangle, the student responding with several elegant proofs of the pythagorean theorem, the teacher saying that's wrong because the question was about angles in an equilateral triangle, not the sides in a right triangle, and the student then refusing to accept that they're wrong unless the teacher disproves the pythagorean theorem.
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Good to know.
If I may be permitted some hyperbole (and why wouldn't I? everyone knows hyperbole is the greatest thing ever), I'd liken this whole conversation to a geometry teacher asking a student for the measure of an interior angle in an equilateral triangle, the student responding with several elegant proofs of the pythagorean theorem, the teacher saying that's wrong because the question was about angles in an equilateral triangle, not the sides in a right triangle, and the student then refusing to accept that they're wrong unless the teacher disproves the pythagorean theorem.
I suspect I'm the target of that example but I can take a friendly poke ;D
I'd happily buy you a beer while you probably poke at me some more. 8)
[Edit: fixed the spelling error "by" when I meant "buy". Darn those homonyms! :o )
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I remember several RPGs from when i was a kid (i want to say some of the final fantasy games even) having a luck stat that wasn't explained in the manual (yeah, i'm old) and didn't seem to change with leveling or be able to be enhanced. And of course I'd always have a vague idea of what it was probably supposed to do by virtue of knowing what the word 'luck' meant but it always kind of bothered me anyway. I think at some point I decided it must be a holdover from pen and paper systems and was just an 'under the hood' randomizer of some sort and just quit worrying about it.
It's usually just a positive modifier to ability scores as well as defense and attack. The name might change to something other than "luck" but the mechanic remains the same. The example that comes readily to me is the Paragon template in the D&D 3.0 Epic Rule Book.
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Each drawing represents more than one contest.
How so? Unless you care about the amount won in the jackpot (which as far as I know no one in this thread has given more than a passing mention to), there is only one contest that is relevant to the person who wants to know their odds of winning it. Can you show that the other independent contests somehow affect the chance of the ticket that I bought matching the random result?
Saying that you draw 1 of R combinations is no more complicated than saying you draw 1 of R integers and it actually avoids confusion and is more accurate.
The integer equivalency is useful because the probability of a single die roll being a certain number is a well known (http://alumnus.caltech.edu/~leif/FRP/probability.html) and accepted value (http://mathworld.wolfram.com/Dice.html) that can be taken as an axiom and used as a baseline to get to some common ground.
Assigning a number to each possible combination allows the problem to be reduced to a single die roll. It's well known that it reduces to 1/s and I don't think that's in dispute here.
Good practice but you ducked my complaint of hypocrisy.
I disagree with it, but it's irrelevant to the discussion. Attacking the person making the argument doesn't change the facts or contribute in any meaningful way.
What is printed on the ticket is a simplified number. It is indeed a form of odds. It's the odds of a specific value being drawn. That is all.
It's the number that is most relevant to an individual playing the lottery, and most importantly, it is not affected by how many other people play.
The odds of being a winner is useless - if every drawing guaranteed a winner. But the odds of all tickets being a loser exists as a strong possibility and affects the final odds.
I still don't see how that affects an individual's chances at all. The odds of all tickets losing don't affect anything other than jackpot rollover. The odds of a specific number (set of numbers, whatever) coming up remain fixed at 1/292201338.
If you have ever played roulette you know there is a some outcomes you can't put your chips on - the house always wins. If you are calculating odds in that game you can't ignore that possibility. And that game and the lottery are largely equivalent. The number of tickets sold affect the odds of there being a winner and in no small way.
The probability of matching a specific number on a Roulette wheel is a better analogy than Bingo, so long as you confine the analysis to betting on one number and ignore the payout odds. However, since Roulette has a much smaller number of outcomes, most statistical analyses of it focus on how best to improve your chances of beating the house in a situation where it is feasible to win multiple times during a session. In that case the analysis becomes about how much you win relative to how much you lose, taking payout proportions into account.
As I've been onto Arcana about already, show your work.
I'm not really sure how you expect me to "show my work" for X*Y. On the surface it seems like a rigged game to ask for a proof when there isn't even agreement if the methodology for applying the theorems is sound. Let me gather a little more information about what you're considering as a valid starting point and I'll see what I can do.
Show that the chances of continuing to play the lottery over several games does not increase your chances of winning
I never said it didn't. That it increases your chances should be obvious to anyone.
or does not follow a binomial distribution.
I did mistakenly say earlier that it was a simple n*1/v problem. After giving it a moment's thought, it became obvious that is incorrect -- playing the lottery 292 million separate times does not guarantee a win, as it would if all the tickets were purchased for a single draw. Indeed it follows your (tongue in cheek here) beloved binomial distribution; though only as far along as the number of times you individually play, so it stays exceedingly small.
However, that's orthogonal to the point I was trying to make and to the assertion you made that started this whole thing:
The number of people playing actually improves the odds of winning for everyone
What I meant is that increasing the number of players increases the probability that *someone* will win because that means the number of draws goes up directly. If we distribute the win chance across the number of players then, yes, the odds of improve for everyone. The increase is, admittedly, tiny with numbers this large but it *is* there. When the overall chances of a win go up the individual players chances do not remain static.
Later I think you may have revised that assertion to say it lowered the chances (forgive me, there are a lot of posts to search through), but unless I'm missing something, the last I heard you were still claiming that more people playing somehow altered an individual's chance of matching the jackpot.
So, before I spend any more time on it, is that what you're saying? Below is what I'm saying, do you agree or disagree with them?
* The odds of one individual ticket matching the winning powerball numbers in a single drawing are 1/292,201,338.
* The odds of one person winning the powerball jackpot in a single drawing increase linearly with the number of unique tickets they purchase (n/292,201,338).
* The odds of one person winning the powerball jackpot in their lifetime increase nonlinearly with the number of drawings they participate in during that lifetime. Using binomial terms, this would be a probability mass function of f(1; n, 3.4223x10^-9), where n is the number of drawings they enter.
* The above odds do not change regardless of how many other people play the powerball. Even in the degenerate case of an infinite number of other players, the odds of my ticket matching the draw are still the same, my jackpot winnings as a result would simply approach zero.
Powerball Lottery (http://wizardofodds.com/games/lottery/powerball/)
The part I love is close to the bottom. The "Return To Player Jackpot Size" table. He relates number of tickets sold to the odds of winning. Looks great... until you distribute the winning odds across the number of tickets sold.
Yes, because that table is all about the return, not the odds of winning. Of course the jackpot is split if there is more than one winner. But nobody (except perhaps you?) is talking about the size of the jackpot, since the original statement was just "More people playing improves the odds of winning for everyone".
Finally, it will take a few minutes to find everything, but I'm in the process of splitting this off to a separate thread, as it has veered WAAAAAY off-topic, even moreso than is normal for this thread.
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Good to know.
If I may be permitted some hyperbole (and why wouldn't I? everyone knows hyperbole is the greatest thing ever), I'd liken this whole conversation to a geometry teacher asking a student for the measure of an interior angle in an equilateral triangle, the student responding with several elegant proofs of the pythagorean theorem, the teacher saying that's wrong because the question was about angles in an equilateral triangle, not the sides in a right triangle, and the student then refusing to accept that they're wrong unless the teacher disproves the pythagorean theorem.
Except this started with, and should end with a simple question, repeated beyond the point of reasonableness: do the odds of a powerball ticket having the winning combination change based on the number of additional tickets entered. There is only one correct answer to this question, and it is "no." Someone who says "banana" is talking about something else. Someone who says "yes, because banana" is just plain wrong. Also crazy, but wrong regardless.
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How so? Unless you care about the amount won in the jackpot (which as far as I know no one in this thread has given more than a passing mention to), there is only one contest that is relevant to the person who wants to know their odds of winning it. Can you show that the other independent contests somehow affect the chance of the ticket that I bought matching the random result?
Hypothetically speaking, you can think of the lottery as n separate attempts to win, one for each ticket, except the problem is that those are not independent drawings because of course the powerball winning sequence is the same for every draw. That lack of independence means that *sometimes* your analysis will work, and *sometimes* it will not. For example, if you treat the powerball as n independent attempts to win, your calculations will show that for all n there is always a finite chance of having no winner regardless of the distribution of the n entries. In practice, however, if the set of all entered sequences is covering, then the odds of a winner are 100% and thus the odds of having no winner are zero.
Ironically, the viewpoint that you can treat the powerball as n separate draws *only* works to consider the case of calculating the odds of any single ticket winning. In that case, that viewpoint will calculate those odds as being 1/N, where N is the number of different possible draws. Those odds would be independent of all other draws, which is the correct answer. But when you try to calculate anything else across all entries, you'll generally get the wrong answer.
Still wrong, but in an ironic way.
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I gave a list of possible rebuttals. All you had to do demonstrate - not just make an assertion - how ANY of them were untrue. This is a requirement in any branch of mathematics, theoretical or applied. It distinguishes between the amateur and the professional. Objective demonstration of your claims. A professional does not resort to insults, ad hominems, dismissals of the other party's concerns, questions, or viewpoints, or avoids direct requests to substantiate any claim made. The professional accepts admissions of error by another person and doesn't continue to press the subject. The professional aims to be critical without being accusatory. A professional can confident that if he (or she) is called into court to testify on an area they are considered expert that they will be able to present their conclusions and fully support them but are prepared to admit error if that is discovered when those are examined. A professional is prepared also prepared to do the same in the court of their peers. A professional is able to distinguish between what is subjective opinion, assumptions, and emotional appeals or other logical fallacies and what is objective evidence, argument, and logical analysis.
And this point I have no other choice to conclude you are not a professional in any field. You are a highly skilled amateur who has developed a following based on a few past accomplishments, emotional argument, and esoteric language. You sound good on the surface but when pressed, or you feel challenged, the internal rot begins to leak out. I find you arrogant and needlessly argumentative. You avoid taking responsibility for your claims and you depend on browbeating the other party in order to "win". I've had disputes with other people in this forum but a lot of it has been miscommunication or loose statements with no ill-will intended. I've been caught on mistakes and I've readily owned up to them when I recognized I had been sloppy in my work. I don't fear failure. In fact mistakes I can walk away from, even when I've been seriously injured, are things I've laughed about. I came to these forums almost three years ago to find out if there was a way to rescue the game and the community I had come to love and care about. I've spent almost all my free time and other personal resources to this end in the way that seemed the most viable and I began here; the record still exists in the archives of these forums. I tried to limit the number of personal claims since I most recently showed up. I've admitted to only one superlative though I can admit to more. I've gone through direct assessment of my abilities by demonstrably trained and accomplished professionals who had been proven qualified to do so - something you have failed to do - and have done well. That's all that needs to be said on that topic. The only thing you have actually proven to me? That you are a board troll. I do not feed trolls when I discover them. Goodbye.
It must be sad, thinking you're the only sane person in the world and simply lacking the ability to demonstrate it.
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So, before I spend any more time on it, is that what you're saying? Below is what I'm saying, do you agree or disagree with them?
* The odds of one individual ticket matching the winning powerball numbers in a single drawing are 1/292,201,338.
* The odds of one person winning the powerball jackpot in a single drawing increase linearly with the number of unique tickets they purchase (n/292,201,338).
* The odds of one person winning the powerball jackpot in their lifetime increase nonlinearly with the number of drawings they participate in during that lifetime. Using binomial terms, this would be a probability mass function of f(1; n, 3.4223x10^-9), where n is the number of drawings they enter.
* The above odds do not change regardless of how many other people play the powerball. Even in the degenerate case of an infinite number of other players, the odds of my ticket matching the draw are still the same, my jackpot winnings as a result would simply approach zero.
I'm glad you took a step back and returned to the original claims. I truly hope that Avelworldcreator can cool off and come back to continue these talks because they have lead to some of the most interesting posts in a long time.
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Except this started with, and should end with a simple question, repeated beyond the point of reasonableness: do the odds of a powerball ticket having the winning combination change based on the number of additional tickets entered. There is only one correct answer to this question, and it is "no." Someone who says "banana" is talking about something else. Someone who says "yes, because banana" is just plain wrong. Also crazy, but wrong regardless.
Yeah. Just thought I'd try a quasi parable since neither direct refutation nor analagous examples have worked over 8 pages.
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I'm glad you took a step back and returned to the original claims. I truly hope that Avelworldcreator can cool off and come back to continue these talks because they have lead to some of the most interesting posts in a long time.
Which ones were the interesting ones? I know why I decided to jump into the original powerball discussion: numerical analysis is something I do both professionally and casually, so on the official forums you basically couldn't keep me away from that. But once we determined that no, you probably shouldn't buy powerball tickets unless you have money to burn, the rest drifts off into general probability and statistics. I didn't think even I could make that interesting.
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As near as I can tell (there's a lot to go through), the issue here seems to centre around the question of whether Powerball picks a winning combination from those selected by players. As I understand it, it does not. If it did, however, then obviously the number of players would affect your probability of winning--though not in a trivial-to-calculate manner, since it actually depends on the number of unique combinations selected rather than the number of players. I don't think there's any reasonable way to calculate number of combinations selected as a function of the number of players, except perhaps by statistical analysis if one had access to sufficient data, as there are issues of psychology involved.
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As near as I can tell (there's a lot to go through), the issue here seems to centre around the question of whether Powerball picks a winning combination from those selected by players. As I understand it, it does not. If it did, however, then obviously the number of players would affect your probability of winning--though not in a trivial-to-calculate manner, since it actually depends on the number of unique combinations selected rather than the number of players. I don't think there's any reasonable way to calculate number of combinations selected as a function of the number of players, except perhaps by statistical analysis if one had access to sufficient data, as there are issues of psychology involved.
It is a possibility, but that's contradicted by the fact that when powerball was replaced with a lottery with explicitly defined rules, the problem persisted. That suggests the problem is a more general one.
Also, it has been repeated in the thread that *if* powerball were to select the winning combination from the set of all entries, and not the set of all possible draws, then the odds of winning would be one in n, where n was the total number of different sequences entered. n would be some function of all entries, but the reality of what that value is can't be determined purely from a statistical function because as you say, psychology is involved. It can be *estimated* by a statistical function because many people elect to have the computer randomly select their ticket. Those tickets would be distributed in random fashion which if converted to a histogram and sorted would then have a binomial character. But the binomial calculation isn't really useful directly, because it calculates the likelihood of numbers being picked multiple times. It does not directly calculate n, the number of likely different numbers selected.
Actually, this related problem has an analog in game design. Suppose City of Heroes were to create a set of something, let's say enhancements, and then allowed you to buy them in the Paragon store but only allowed you to buy unopened "packs" of them each containing a random one. If you wanted the complete set, on average how many of those do you have to buy?
The interesting thing is that this number escalates a bit faster than I think most developers (who generally are not math-proficient) realize**. Suppose the set of things we're talking about is an enhancement set with five things in it. The best way I can think of to decompose this problem is to consider step by step what the odds of getting a different one are, and then convert those odds into an expectation of number of purchases, and then sum them. So the first pack you buy is guaranteed to get you an enhancement you don't have yet, because you don't have any yet. So you have to buy one pack to get the first one. Now that you have one, the odds of getting a *different* one on the next buy is 4 in 5, or 4/5. You could get any random one of the five, but only four of them are ones you want now. So that means that the statistical average number of packs you have to buy to get unique enhancement number two is 5/4 or 1.2 packs. Then the next one would be 5/3, then 5/2, and then 5/1. The sum of all of those is about 11.4. That's the *average* number of packs you'd have to buy. If you have a thousand players going after it, the odds are good there will be players that take twice as many, and even three times as many. Those odds can be calculated.
But if you don't calculate them, as a game designer you could simply guestimate, and guestimate wrong. You could design a system where out of ten thousand players there are hundreds of them that take so long to get something relative to the average that the cost becomes oppressive. There are ways to design safety values for that, but those options aren't always used (for example, if you are okay with about 12 packs being necessary but you want to make sure it doesn't take more than 20, you can change the pack design so that the players get something such that twenty of them guarantees they will be able to get all the enhancements; that sets a floor on the worst case scenario. Allowing players to trade four unwanted cards for a wanted one would be another way to guarantee the floor).
The lottery is actually an extremely simple problem statistically. Real world design problems have a much higher chance of getting the terms and conditions confused, and then the math misapplied. That's why statistics is one of the most error-prone math disciplines. In fact there have been arguments going on for a decade now about how statistics is misused in scientific papers that actually goes back to a problem (or rather a class of them) I first realized and discussed with my college professors thirty years ago. The problem has been around that long, and presumably smart people have failed to address it for at least that long. Variations of it even cropped up on the City of Heroes forums. I would often have to confront players that decided to use statistical computations improperly and try to explain why their calculations were inappropriate, and would usually get a textbook read back to me. Even when the real world ended up disagreeing with their calculations, they would stubbornly presume the problem somehow had to be with the game and not their calculations.
** I think most developers assume that the number of packs you have to buy in this situation is roughly proportional to the number of things, while it it actually roughly proportional to the number of things times the log of the number of things. In other words, it increases as n log(n), not just n. That extra log(n) factor is what makes developer intuition seem reasonably close for small numbers like 5, but get quickly out of whack when it comes to bigger numbers. Star Trek Online has a collection game like this where the number of things you need to acquire to complete the set is over a hundred, and you can only draw about one thing per day. I would bet the devs thought it would take something slightly longer than a hundred days to collect them, when it will actually take over 500.
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Well, I posited that point of confusion as an explanation as it seemed the most likely explanation for the ongoing discussion. It would also explain why, for example, the issue of winning the lottery being the matching of two random variables (the combination on the ticket and the winning combination) was raised as being relevant, when it actually has no effect on the probability of a win (since the probability of the winning combination matching the one on the ticket is constant across all possible combinations). I believe you're right in saying this has been specifically addressed, but after a while it's hard to remember. :)
Though this does make me curious as to density of particular combinations selected. I would guess, for example, that the numbers up to 12 (and to a lesser extent up to 31) are over-represented as people play dates as part of their selection.
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Though this does make me curious as to density of particular combinations selected. I would guess, for example, that the numbers up to 12 (and to a lesser extent up to 31) are over-represented as people play dates as part of their selection.
Maybe, maybe not. Some people pick dates. Some people have lucky digits, and pick numbers ending in 7 (including 7) or 8 (if you're Asian). Some low numbers might be avoided more than higher ones, like 13. Even if you pick dates, some people use the year as part of the date, so numbers from 32 to 69 would still be in play, particularly at the higher end of the range. And of course if you're picking numbers at all as opposed to letting the computer pick a random sequence, you might be thinking about deliberately trying to pick uncommon numbers to avoid what you believe the common numbers are.
The winning sequence for the recent mega jackpot powerball was 4, 8, 19, 27, 34 and powerball 10. That's an overrepresented set of low numbers but you had to have 34 in there as well. The jackpot increased by about $600-$700 million which suggests about 900 million tickets purchased, approximately. There were 26,110,646 winners total and you'd expect to see about that many winners out of about 600 million tickets sold (there's about a one in 25 chance of winning something in general). There were three jackpot winners which is what you'd expect out of about 876 million tickets entered. That's relatively good agreement between the numbers given the very rough nature of these estimates, although it is possible that the discrepancy between the monetary estimate and the winning frequency estimate suggests the winning numbers drawn were statistically less likely to be selected by players than average. I'd be careful about concluding that from these very rough numbers though.