Powerball Odds and Statistics

[Arcana]

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.

[Arcana]

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.

[avelworldcreator]

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. 

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
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?

[Tubbius]

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

[Aggelakis]

Bingo - FAQ
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

[Vee]

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.

[avelworldcreator]

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.

[Codewalker]

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.

[avelworldcreator]

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.

[Vee]

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.

[Victoria Victrix]

*Watches all the math/statistics talk*

[Twisted Toon]

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*

Oh no.

 Who's going to write more of my favorite Valdemar books?!? 

[Arcana]

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.

[Arcana]

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.

[avelworldcreator]

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
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!) 

[Arcana]

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.

[Vee]

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.

[Arcana]

*Watches all the math/statistics talk*

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.

[Joshex]

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.

[Vee]

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.