Powerball Odds and Statistics

[Aggelakis]

the 600 million agge just used was as far as i can tell just a large number out of a hat.

Correct. :)

[Aggelakis]

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).

Also: the statistical likelihood of EVERY player picking the winning versus the losing number in an A/B choice is equal. ;)

[avelworldcreator]

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?"

[Vee]

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:

[Vee]

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.

[Aggelakis]

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!

[Aggelakis]

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 ;)

[Tubbius]

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.

[avelworldcreator]

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).

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.

[avelworldcreator]

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.

[Vee]

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.

[avelworldcreator]

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?

[avelworldcreator]

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:

[Vee]

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

[avelworldcreator]

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).

[Arcana]

So much math.  :O  It makes this double-English-major whimper in fear and sadness.

You majored in English twice? 

[Arcana]

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.

[Joshex]

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.

[Tubbius]

You majored in English twice?

Bachelor of Arts, 2001 (Major: English; Minor: Professional Education, grades 9-12).  Master of Arts, 2003 (Major: English).

[avelworldcreator]

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].