Lottery Jackpot Sharing Parardox

Quote: weaselman

You do realize, for that to be true, there must NEVER be any other winning tickets at all (because there cannot be a negative number, if in any drawing there is just one other winner, then the average will not be 0), don't you?

No, because there could be drawings with zero winners and drawings with multiple winners, bringing the average to 1.

I don't know how to describe the following in simpler terms:

Since the probability of winning is 1 in 175,711,536, when there are 175,711,536 tickets in play, there will be an average of 1 winner.

So, I give up.

If JB does not buy a ticket, will it improve my chances ? Or must I buy another ticket in case he would have been the winner ?

Quote: JB

No, because there could be drawings with zero winners and drawings with multiple winners, bringing the average to 1.

But you said, the average would be zero, not 1.

Quote:

Since the probability of winning is 1 in 175,711,536, when there are 175,711,536 tickets in play, there will be an average of 1 winner.

This is absolutely true, but it is not the question we are trying to answer. Our question is: provided, that there is at least one winning ticket, how many other winners will there be on average, not counting the first one?
What I am telling you is that the answer to that question cannot be zero, because if it was, that would necessarily mean that there is never more than one winner, which is obviously false.

Quote: weaselman

Quote: JB

No, because there could be drawings with zero winners and drawings with multiple winners, bringing the average to 1.

But you said, the average would be zero, not 1.

I said that the average number of OTHER winners would be zero when it is known that the first ticket verified is a winner. You are trying to say that I said the average number of winners PERIOD is zero, which is obviously wrong.

Quote: weaselman

Quote: JB

Since the probability of winning is 1 in 175,711,536, when there are 175,711,536 tickets in play, there will be an average of 1 winner.

This is absolutely true, but it is not the question we are trying to answer. Our question is: provided, that there is at least one winning ticket, how many other winners will there be on average, not counting the first one?
What I am telling you is that the answer to that question cannot be zero, because if it was, that would necessarily mean that there is never more than one winner, which is obviously false.

And I am telling you that you are wrong, because on average, there is 1 TOTAL winner per drawing. If you know you are a winner, then on average, there are zero OTHER winners.

You are looking at a "zoomed-in" part of the picture, whereas I am looking at a "zoomed-out" view of the entire picture, and we're both trying to describe to each other what we see, which is why we disagree.

Quote: weaselman

Here is a similar question,...

I think your "similar question" is almost completely unrelated. In the lottery, there is one random event. In your "similar question" there are 1000 independent random events. I think it would be extremely difficult to draw a meaningful analogy.

Quote: weaselman

You are playing blackjack, the dealer has a ten, and peaks for BJ. He now knows his second card, but you still don't. Would you argue it is now incorrect to discuss the probability of him having a seven in the hole?

In this case, there is one random event (for the dealer's down card), to be determined by the deal. When he has the opportunity to peek, that random event has occurred and its actual outcome has already been established (though I don't know it), so I don't consider it appropriate to discuss a probability of what will occur. Now as far as what action I take, that should be determined by prior studies of probabilities independent of what card was actually dealt down to the dealer.

Now one complicated direction that this discussion could go is in the area of trying to figure out what happened some time in the past that lead to a later, now-known condition. Something like: George places $10 worth of wagers on each spin of the roulette wheel. After a spin that ends up with 36, George has $36 left on the table, a combination of his wagers and his winnings. Now what do you think he "probably" wagered? There are multiple possibilities, and some people might make guesses or even perform some kinds of calculations to get their best estimate of what George did. They could not be certain. Now you might consider variant of this in which you would try to establish certain "probabilities" on George's possible wagers. You might even try to calculate expected values of the amount of money George bet on each possible wager. I, personally, don't view that as appropriate -- unless George is making random wagers. If George is making deliberate choices, I don't consider those to be random events for which expected value is a meaningful calculation. I consider them deterministic events with George doing the determining. Our reconstructing what might have been is likely meaningless.

However, I think the Wizard would prefer that we keep the discussion on the topic of the lottery, not 1000 coin flips, not a blackjack game with a ten showing, nor George playing roulette.

Time for a simpler question.

There are two people. Each person flips a coin. If your coin lands on heads, you win.

Now, let's say that you flip your coin and it is heads. According to some, the average number of other winners is 0. This means that it is impossible for the other person to win. Is this right?

Quote: JB

I said that the average number of OTHER winners would be zero when it is known that the first ticket verified is a winner. You are trying to say that I said the average number of winners PERIOD is zero, which is obviously wrong.

No. What I am trying to say, that we definitely agree that the average total number of winners is 1. That was the given, and there is nothing to discuss here.

The question at hand is how many other winners will be there on average, provided that there is one winning ticket. I maintain that this number cannot be zero.

Quote:

And I am telling you that you are wrong, because on average, there is 1 TOTAL winner per drawing. If you know you are a winner, then on average, there are zero OTHER winners.

But how is it possible? There are certainly cases when there is at least one other winner. Summing them all up, there is no way you can get zero.

Quote:

You are looking at a "zoomed-in" part of the picture, whereas I am looking at a "zoomed-out" view of the entire picture, and we're both trying to describe to each other what we see, which is why we disagree.

Well. Let's try and look at the same part then.
First, let me say it again, that I completely agree that the total number of winners on average is 1. Let's just note that we are in agreement on that point and not return to this question any more, ok?
Now, with that out of the way, can we agree that provided that there is at least one winner the average number of other winners cannot be zero (because if average was zero, that would require that there never are any other winners)?

Quote: Doc

I think your "similar question" is almost completely unrelated. In the lottery, there is one random event. In your "similar question" there are 1000 independent random events. I think it would be extremely difficult to draw a meaningful analogy.

Why do you think there is only one random event in the lottery? If we assume that every player selects their combination randomly, there are as many events as there are players. (In fact, if we don't assume that, it is still the case).

Quote:

In this case, there is one random event (for the dealer's down card), to be determined by the deal. When he has the opportunity to peek, that random event has occurred and its actual outcome has already been established (though I don't know it), so I don't consider it appropriate to discuss a probability of what will occur. Now as far as what action I take, that should be determined by prior studies of probabilities independent of what card was actually dealt down to the dealer.

But how can it be independent? For one thing, you know the card in the hole is not an ace. If you simply disregard that fact considering what action to take, your decision will be wrong.

Quote:

However, I think the Wizard would prefer that we keep the discussion on the topic of the lottery, not 1000 coin flips, not a blackjack game with a ten showing, nor George playing roulette.

The first two are exactly the same question (coin flip is exactly equivalent to filling a lottery ticket having 50% chance to win). The blackjack analogy is just an illustration of the faultiness of your aversion to the concept of aposteriori probabilities ... and I don't know who is George :)

Quote: Wizard

Quote: ZPP

..., but dividing by the expected number of winners is not valid. It just happens to be the case when X has a Poisson distribution that E[1/(X+1)] is approximately 1/lambda=1/E[X] (when lambda is greater than 3 or so); fooling you into thinking that ignoring the plus-one is valid reasoning.

Here is another way to calculate equity.

As I stated before, the value of the consolation prizes is 18.2%. Then was 656,000,000 in the jackpot and 651,915,940 tickets sold. Let's just give every ticket an equal share of that, since each ticket has the same chance of winning (assuming everyone did Quick Picks). So the jackpot value per ticket is 656,000,000/651,915,940 = 100.63%. Adding the consolation prizes and jackpot value we get 18.20% + 100.63% = 118.83%.

Here again is my method of dividing by number of winners.

Fixed return + (Probability of winning)*(jackpot)/(expected total number of winners)
= 18.2% + (1/175,711,536.00)*(656,000,000)/3.71 = 118.83%.

Is that just a coincidence?

I'll start off by pointing out that the first method of calculating equity above ignores the possibility that no one wins. You did mention this in a previous post, so I'm just noting it because it explains the discrepancy between your 118.83% and my 116.36%.

Your equity method can be modified:
18.2% + (656,000,000/651,915,940) * (100%-2.45%) = 116.36%
The probability that no one wins, 2.45%, is given by (175,711,535/175,711,536)^651,915,940, and with these huge values e^-(651,915,940/175,711,536) is a very good approximation, so it is no surprise that the factor of (1-e^-lambda) accounts for the possibility that no one wins.

However, now I can take a step back and address the question of whether it is a coincidence that dividing by the expected numbers of winners gives you the expected payout, given that there is at least one winner. Obviously, it is not a coincidence for the Poisson distribution. However, it turns out that this applies more generally than just to the Poisson distribution for games with this structure, which I didn't notice previously.

Consider the following structure for a game. There are M players. If there is at least one winner, a total of exactly J dollars is paid out. Before the game is resolved, all the players are in a completely symmetric situation with respect to each other. (So, a lottery where everyone uses Quick Picks and keeps them secret from the other players, and ignoring the consolation prizes because they aren't affected by, or relevant to, any of this.)

You are a player.
Let W = a random variable for the total number of winners
Let X = a random variable for your winnings
Let f(n) = the probability that exactly n players win

Your equity method is a valid method to calculate your expected return, given that there is at least one winner.
E[X | someone wins] = J / M
By symmetry, every player's expected return, given that someone wins, is the same, and they must all add to J.

Your division by the expected winners method is:
J / E[W] * P[you win]
Does this come to J / M? Yes.

By symmetry, if there are n winners, your probability of winning is n/M, so
P[you win] = Sum[n=1..M, f(n)*n/M] = 1/M * Sum[n=1..M, f(n)*n]
However, E[W] = Sum[n=1..M, f(n)*n], so
P[you win] = E[W] / M
and J / E[W] * P[you win] = J / M = E[X | someone wins]

I still maintain that this sort of reasoning is necessary to justify dividing by E[W]. That is, dividing by E[W] happens to give you J/M, which in turn equals E[X | someone wins], whereas dividing by E[W | you win] does not. Neither division is intrinsically valid without structure to the game or probability distribution that justifies it. E[W | you win] is still approximately 1+E[W] for the lottery, and you saw that when I did the calculations with a concrete probability distribution I had to use the plus-one to get the same result as we get with (J / E[W] * P[you win]) * (1 - P[no one wins]).

Thus, none of this makes your original statement correct that you can expect to share with 2.7 other people if you win. True, you divide by 3.7, but you don't divide by E[W | you win].

In the coin toss game, suppose you offer a side bet where, if I lose the coin toss, the bet is a push, and if I win, you pay me $1 for each winner, including myself (and keep my original bet). The fair price for that bet is $5.5, not $5 (and I'm sure you know this).

This is a very confusing thread but I find myself agreeing with JB. My university math is from 40 years ago so I can't provide any mathmatical reasoning but think the following scenerio is somewhat relevant.

When playing texas holdem one calculates the pots odds only from the card you know, typically your own cards and the community cards. The fact that the other players each know their own cards doesn't change the odds that you can fill your hand. This is true even if the completing cards for your hand are held by other players and actually not available. This seems to be somehow related but I am not sure how.

While keyboarding this I also wondered if this is not the gamblers fallacy and in that case I must switch my allegiance to Crystalmath. From JB's post if the first 10 tickets looked at are all winners than the next ones couldn't possibly be a winner since we have used up all the winners.

Not that I am completely confused again I think I will go to bed and sleep on it.

Wiz, on the night of the drawing would you value your WINNING ticket at: $656,000,000/3.7 winners = $177,297,297 ?

CM, on the night of the drawing would you value your WINNING ticket at: $656,000,000/(3.7+1) winners = $139,574,468 ?

What would be a fair price to buy the winning ticket from a winner (before results are revealed the next morning)?

Could I expect to buy a winning ticket for $160,000,000?

Edit: Ignore the fact that this is life changing money here. Divide the #s by like 1,000 or 10,000 if it helps.

Quote: Wizard

For purposes of estimating the value of a ticket, I said there would be an expected 3.6 winners, so if you won you would have to share it with 2.6 other people.
Somebody wrote to me saying that I was in error. That if I won, I could still expect 3.6 other people to win, for a total of 4.6 winners, since my one ticket does not diminish the total number of other tickets in any significant way.
...
Here again is my method of dividing by number of winners.
Fixed return + (Probability of winning)*(jackpot)/(expected total number of winners)

Hate to say this, but the other person was right. The mistake here isn't even a matter of conflating pre-event and post-event probabilities, Monty's Hall or Gambler's Fallacy, but one of elementary math - the average result of splitting $X by an average of N winners does not equal X/N.

If there can be 1, 2 or 3 winners with equal probability, for an average of 2, the result is not 1/2, but rather (1/1+1/2+1/3)/3=11/18. For (0,1,2), it's (0+1+1/2)/3=1/2.

If you play a game of 3 players putting $1 into the pot, and the pot is split between everyone flipping heads, the expected number of winners is 3/2=1.5. Your formula would give ER=0+0.5*3/1.5=1 - which is not the case.
The real ER is as follows: for outcomes 000...011 (binary) it's 0, then for 100 it's 3, for 101 and 110 it's 1.5, for 111 it's 1, for a total of 3/8+1.5/8+1.5/8+1/8=7/8, correctly accounting for the probability of everyone flipping tails. The average win, when you win, is 7/4.

Quote: Wizard

= 18.2% + (1/175,711,536.00)*(656,000,000)/3.71 = 118.83%.
Is that just a coincidence?

As for this one, it's not a coincidence, but a miscalculation. This formula only seems to work when you consider the probability of no one winning small enough to ignore, and the amount of error is similar to that probability. As you know, small magnitude of the error does not confirm the validity of a method and its assumptions.

If you win the coin-flip game with 3 players, the expected number of other winners is 1, not 0.5. If you win the lottery with 656M players (and 3.6 average winners), you can still expect to share it with 3.6 other people, not 2.6. However, your average share is not 1/4.6, neither is it 1/3.6, but rather a product of a more complex calculation based on winner count distribution.



P.S. Also, now I notice ZPP just made a similar point, but with more generalized math, and he also did that more complex calculation. Well, hopefully my post will at least be easier to understand for less math-inclined readers.

Quote: Lucyjr

Wiz, on the night of the drawing would you value your WINNING ticket at: $656,000,000/3.7 winners = $177,297,297 ?

CM, on the night of the drawing would you value your WINNING ticket at: $656,000,000/(3.7+1) winners = $139,574,468 ?

What would be a fair price to buy the winning ticket from a winner (before results are revealed the next morning)?

Could I expect to buy a winning ticket for $160,000,000?

Edit: Ignore the fact that this is life changing money here. Divide the #s by like 1,000 or 10,000 if it helps.

As shown in my post on page 2, on the night of the drawing I would value my winning ticket at $656,000,000*(1-e^-3.71)/3.71 = $172.5 million (ignoring the annuity, taxes, etc.). CrystalMath and I appear to be in agreement about this.

Probably my posts were too math-heavy. The core point is that dividing by the expected number of winners is not inherently valid, so any argument about 3.7 vs 4.7 misses the point when it comes to valuing the ticket. You need to do a more complex analysis, which I did.

It isn't that I'm ignoring the thread. I've written lots of responses, but always end up not posting them or quickly busting them. The last two days I have spent several hours thinking about this problem. For now, I claim that if it were not for the chance of 0 winners my formula would be right. Using examples like the 3-coin one are not useful to debunk it, because of the high chance of 0 winners. I definitely think that dividing by 4.7 winners is not right.

I know this post is not of much help, just giving a progress report on where I'm at.

Quote: Wizard

It isn't that I'm ignoring the thread. I've written lots of responses, but always end up not posting them or quickly busting them. The last two days I have spent several hours thinking about this problem. For now, I claim that if it were not for the chance of 0 winners my formula would be right. Using examples like the 3-coin one are not useful to debunk it, because of the high chance of 0 winners. I definitely think that dividing by 4.7 winners is not right.

I know this post is not of much help, just giving a progress report on where I'm at.

Hey Wizard, thanks for posting this problem! It was a neat one and took me a while to prove that if you win you share your prize with an average of 4.7 winners. And yes, you cannot simply divide the jackpot by 4.7 to get the EV.

You are an actuary (like myself), so if you recall your credibility theory, you will find the answer there. Think of what might be the prior distribution and what could be the posterior distribution. That's what made it clear to me!

Quote: cardshark

You are an actuary (like myself), so if you recall your credibility theory, you will find the answer there.

As a fellow actuary, I'm very happy to have you on the forum. Please stick around.

I'm afraid that "credibility theory" rings a bell, but I finished the exams 17 years ago and haven't used actuarial terms like that since.

Quote: Wizard

For now, I claim that if it were not for the chance of 0 winners my formula would be right. Using examples like the 3-coin one are not useful to debunk it, because of the high chance of 0 winners.

Indeed, for a certain class of lottery-type situations, your formula is right for the expected return given that there is at least one winner. That is what my post on page 6 is about. Specifically, it works at least in situations where you are splitting a jackpot among identically situated players.

Probably a good thing to do is give an example where your formula fails with no chance of 0 winners. Let's do that by violating my criterion of identically situated players.

Consider the following game. You have 6 players numbered 1 through 6. You roll a die for a result N and split $60 equally between players 1 through N. The expected number of winners is 3.5.
What is Player 1's expected return?
It is 1/6 * ($60 + $30 + $20 + $15 + $12 + $10) = $24.50.
It is not $60 / 3.5 = $17.14.
This doesn't even involve the difference between the expected number of winners and the expected numbers of winners given that Player 1 wins: Player 1 is guaranteed to win so they are the same.

And this is why it is pedagogically bad to wily-nilly divide by the expected size of a group of people when you know that you will be splitting money equally among that group. You need to have a justification to do something like that.

ZPP, thanks for your explanation, and all others who contributed mathematically-based responses.

I concede that it seems like terrible math to divide by the expected number of winners to get an average win IF I win. However, it would seem to work in the jackpot sharing situation of a lottery if it were not for the possibility of zero winners.

In fact, it seems to work exactly if we add a rule that if there are 0 winners they redraw until there is at least one winner.

Let's take a look at a smaller scale to make the point. Suppose the lottery has $1000 in the jackpot. There are 3 tickets, each with a 50% chance of winning. If nobody wins there is a redraw, until there is at least one winner.

Assuming I win (in which case a redraw can't happen)...
There is a 25% chance of 3 winners, for $333.33 per winner.
There is a 50% chance of 2 winners, for $500.00 per winner.
There is a 25% chance of 1 winner, for $1000.00 per winner.

Expected win, given that I win = .25*333.33 + .5*500 + .25*1000 = $583.33

Next, calculate the estimated total number of winners, with no assumptions.

Before considering the redraw rule, the odds are:

3 winners = 12.5%
2 winners = 37.5%
1 winner = 37.5%
0 winners = 12.5%.

Given the redraw rule, the odds are:

3 winners = 12.5%/.875=0.142857143
2 winners = 37.5%/.875=0.428571429
1 winner = 37.5%/.875=0.428571429
0 winners = 0

Expected winners = 0.142857143*3 + 2*0.428571429 + 1*0.428571429 = 1.714285714

Jackpot/Expected winners = 1000/1.714285714 = $583.33. Wow, it comes out the same! It works for the 2 and 4 player cases too (I checked).

Again, I admit that my logic would make the hair on the back of the neck of respectable math teachers stand up, but it seems to work.

Getting back to the lottery, my method results in an estimated ticket win of $176,812,317. If we correctly take a dot product the answer would be $172,485,033. I submit that the difference is due only to the possibility of zero winners. In fact, according to my spreadsheet, if there was a redraw rule then my figure would be exactly right.

Quote: Wizard

Assuming I win (in which case a redraw can't happen)...
There is a 25% chance of 3 winners, for $333.33 per winner.
There is a 50% chance of 2 winners, for $500.00 per winner.
There is a 25% chance of 1 winner, for $1000.00 per winner.

Expected win, given that I win = .25*333.33 + .5*500 + .25*1000 = $583.33

Also, the expected winners, given that you win = .25*3 + .50*2 + .25*1 = 2

Would you tell a player "if you win, expect there to be 1 more winner" or do you say "if you win, expect there to be another half a winner, on average"? This, I believe, was the original question.

If you win, you will split with one more player on average, but it wouldn't be wise to expect an average win of $1000/2, although that would seem logical to most people. Your average prize, in fact, is $583.33.

In the lottery problem, if you win, you expect to share it with an average of 4.71 people, including yourself. But, your average pay is still 656000000/(651915940/175711536)/(1-e^(-651915940/175711536)) = $172,485,033.

In fact, the average pay is higher if you don't win. Assuming that someone wins, but not you, there will be 3.803228526 winners who will each receive an average of $233,430,140. This is $656,000,000 / 2.81, which doesn't bear much resemblance to the expected number of winners.

The only thing this thread taught me is that I'm glad I dropped my math major in college.

Suppose the individual RESULT, not winners, is varied. Should the legal drawing be something like 1-2-3-4-5 + 6. Here, the DRAW result has varied in favor of statistically more winners. After all 1-2-3-4-5 + 6 has an equal chance, and is one of the most popular selections (not QP's) made. The expectation formula accounts for this variation, by giving "an average" number of winners. When you buy a ticket with a statistically BIASED expectation, do you consider there to be fewer winners than 2.6?, more than 2.6?, or the same 2.6 others?

This leads to an interesting quantum conundrum... how do you prove your ticket is not statistically biased?

Haven't been on this site in a while, but this one pulled me in.
If p is the probability of winning the jackpot and p is incredibly small, then
p*(1+avg. # of winners) +(1-p)*(avg.# of winners) still approximately equals the avg. # of winners.

In any case, this is like the monte hall paradox, where the winner who poses this queston knows he's holding the winning ticket. Does it change the odds for the others? No, because the person holding the ticket could be any of the other ticket holders. Even if the player decides he;'s going to go on TV and announce it, it doesn't mean anything - not unless the TV station randomly selected a ticket holder and that ticket holder happend to be the winner. Then we can say the average for that one draw would approximately increase by one, but the probability of this happening should be weighted as above. The overall average should be relatively consistent.

Quote: Wizard

Expected winners = 0.142857143*3 + 2*0.428571429 + 1*0.428571429 = 1.714285714
Jackpot/Expected winners = 1000/1.714285714 = $583.33. Wow, it comes out the same!

Yes, it does come out the same.
But while jackpot/expected_winners=$583.33, it's a meaningless metric, because the average winnings are not $583.33.

The average amount of winnings is $333*1/7+$500*3/7+$1000*3/7=$690.43.

Quote: P90

Yes, it does come out the same. But while jackpot/expected_winners=$583.33, it's a meaningless metric, because the average winnings are not $583.33. The average amount of winnings is $333*1/7+$500*3/7+$1000*3/7=$690.43.

I would call that a meaningless metric. As a player, what you care about is how much you get when you win.

The average number of winners != The average number of winners when I'm one of them.