Lottery Jackpot Sharing Parardox
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.
Does that argument hold water?
Let's look at it another way. Suppose the host of game says he is organizing a 10-player game. He will put up $1000 and split it among all the participants who correctly guess the toss of a coin. Let's say each player writes down heads or tails secretly. Then the $1,000 is split among all the winners. If nobody wins (1 in 1024 chance) then the host keeps the money. He says the price to play this game is $95 do you do it?
I think I could easily show that the expected return is $99.90. Ultimately $1,000 will be split somehow among 10 people, with each having and equal chance, so a value of $100 each. However, there is a small chance nobody will win, thus the $99.90 instead of $100. That makes it seem like a good value at $95.
However, if you win, and before the other results are announced, wouldn't you predict there would be 5.5 total wins. In other words, 1 for yourself and 9*0.5=4.5 others, for a total of 5.5. By that logic, a fair price to pay would be 50%*(1000/5.5) = $90.91, making $95 seem like a bad price. What is wrong with that argument, if anything?
Would you play? I would.
So, I still say that I was correct on the air, for purposes of assessing the expected value of that lottery. I think I know how I would explain why he is wrong, but I'll torture you guys with this little paradox for a while first.
3.6 winners is based upon the total number of tickets sold.
That listener's argument is that after the first person reports a winning ticket, that there are still x million minus one tickets out there, and therefore, still expects 3.6 other winners. But there is no way to know how many losers are already in the trash, so the original 3.6 - 1 = 2.6 is correct.
Regarding the coin toss. I want to invoke the "prior results do not affect future outcomes" rule.
EXCEPT...
In poker, when one player is all in, and there is a showdown with more cards to come, the rule is for other players to not announce what they folded. Letting the underdog know that one of his 'outs' was folded changes the odds of him getting that card.
I never really bought that argument, but feel that, somehow, it plays into this coin toss dilemma.
Here is how it works out, given that you won the coin toss.
| Number of winners | Probability | Payout | Return | Winners*Prob |
|---|---|---|---|---|
| 1 | 0.001953 | $1000.00 | $1.95 | 0.001953 |
| 2 | 0.017578 | $500.00 | $8.79 | 0.035156 |
| 3 | 0.070313 | $333.33 | $23.44 | 0.210938 |
| 4 | 0.164063 | $250.00 | $41.02 | 0.656250 |
| 5 | 0.246094 | $200.00 | $49.22 | 1.230469 |
| 6 | 0.246094 | $166.67 | $41.02 | 1.476563 |
| 7 | 0.164063 | $142.86 | $23.44 | 1.148438 |
| 8 | 0.070313 | $125.00 | $8.79 | 0.562500 |
| 9 | 0.017578 | $111.11 | $1.95 | 0.158203 |
| 10 | 0.001953 | $100.00 | $0.20 | 0.019531 |
| Total | 1 | $199.80 | 5.5 |
Your method of figuring that the expected return is $99.90 is correct, but your method implies nothing about the expected number of winners given that you win, so it is no contradiction that the above method gives the same result while correctly calculating an expected 5.5 winners given that you win (of course, 1+9/2 is an easier way to calculate this).
Therefore, I say that the listener is correct that the expected number of lottery winners, given that you win, is 4.6. However, dividing the jackpot by either 3.6 or 4.6 tells you nothing meaningful. You need to calculate the probability of splitting it 1 way, 2 ways, 3 ways, etc., and add up the expected return. (Probably truncating somewhere in the range of 10 to 20 winners.) However, in doing so, it is important to realize that the probability of splitting it 1 way (not splitting), given that you win, is almost exactly the same as the probability that 0 people win; that the probability of splitting it 2 ways, given that you win, is almost exactly the same as the probability that 1 person wins, etc.
This is just like the joke about the statistician bringing a bomb on a plane. In reality, if there is a 1 in a million chance that someone brings a bomb onto the plane, then there is approximately a 1 in a million chance of there being a second bomb on that plane, NOT a 1 in a trillion chance. He has not affected the probability of anyone else bringing a bomb.
Method 1: Repeatedly deal random 5-card hands behind the scenes until a hand containing at least one Ace is found.
Method 2: Deal an Ace to the hand first, and then deal 4 random cards from the remaining 51 cards.
Each method produces different probabilities.
Wouldn't asking the lottery/coin-toss question be like using the probabilities from Method 1 when Method 2 should be used instead? The 3.6 figure was based on a Method 1 type of analysis, whereas if you already know that you're a winner, you would need to perform a Method 2 type of analysis because you have additional information available.
Is this a correct parallel, or am I way off track as usual?
Quote: WizardOn my radio show I was talking about jackpot sharing in the big MegaMillions draw. 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.
Does that argument hold water?
Yes, it does.
The expected number of winners was 3.6, assuming about 630 million tickets sold. But, if you win, you should expect another 3.6 winners.
Using Bayes theorem: P(A|B) = P(B|A)*P(A)/P(B)
Let's assume A is you winning the lottery and B is n-other people winning the lottery.
P(A|B) = P(A) The tickets are independent, so the probability that you win doesn't change depending on other winners.
Therefore, P(B|A) = P(B). So, for any number of other winners, the probability is identical whether you win or not.
Given that I did not win, I expected there to be 3.6 winners.
First, a quick recap of the big MegaMillions draw. Let's forget all about taxes and the annuity, to keep things simple.
Tickets sold = 651,915,940
Probability of winning = 1 in 175,711,536
Expected winners (in my opinion)= 651,915,940/175,711,536 = 3.71
Jackpot = 656 million
Expected return of consolation prizes = 18.20%.
It is rather crude to divide the jackpot by the expected number of winners, because there is a 2.47% chance nobody will win, in which case the money stays in the jackpot. However, to keep the math sample, and avoid confusing the issue, I'm going to estimate the jackpot per winner by dividing by 3.71.
Expected return = 18.20% + (650,000,000/3.71)*(1/175,711,536) = 118.20%
Correct me if I'm wrong, but the guy who wrote to me, and CM are saying that IF you win then you'll have to share the jackpot with 3.7 other people, for a total of 4.7 winners.
By that argument the expected return would be 18.20% + (650,000,000/4.71)*(1/175,711,536) = 96.97%.
Again, ignoring taxes, the annuity, and the utility of money, was MegaMillions a good bet?
I would say yes.
I will grant that IF you win, and know of no other results, you can expect 3.7 other winners, for a total of 4.7. However, it is dangerous to talk about expected value for an event that already happened. The reason you shouldn't divide by 4.7 in the original expected value calculation is because you using the estimation of somebody who was cherry picked to win.
Going back to my coin flip example (see first post). Suppose I do that experiment and then cherry pick somebody who won to estimate the total number of winners. He would say 5.5. Then I cherry pick a loser and ask him the same thing. He would say 4.5. Although they give different answers, they are both right. It just goes to show you shouldn't do the original calculation based on the perspective of somebody who was cherry picked to win or lose.
Some other posters brought up the Monty Hall problem and Ace on the Deal. I'm not saying there aren't connections, but I think the better comparison is to the two-envelope problem, which was debated extensively in the early days of the forum.
The defense rests.
Quote: WizardCorrect me if I'm wrong, but the guy who wrote to me, and CM are saying that IF you win then you'll have to share the jackpot with 3.7 other people, for a total of 4.7 winners.
By that argument the expected return would be 18.20% + (650,000,000/4.71)*(1/175,711,536) = 96.97%.
No. Under the condition "IF you win", the expected return is (650,000,000/4.71)*1 = 13,800,424,628.45%
Seriously, you're mixing conditional and unconditional probabilities in the same equation. That's incorrect. Under the condition that you win, you can eliminate the 18.20% from the consolation prizes (you can't win a jackpot and a consolation prize on the same ticket), and you can also eliminate the odds of winning the jackpot because that's given in the conditional. What's left is dividing the pool by you and the other expected winners. The real question is "if you win, how many other winners are there likely to be", and the answer for the lottery is "almost the same as if you didn't win", in which case, the expected return "if you win" is just the jackpot divided by the number of expected winners, plus one for you.
Quote: WizardHowever, if you win, and before the other results are announced, wouldn't you predict there would be 5.5 total wins. In other words, 1 for yourself and 9*0.5=4.5 others, for a total of 5.5. By that logic, a fair price to pay would be 50%*(1000/5.5) = $90.91, making $95 seem like a bad price. What is wrong with that argument, if anything?
As a follow-on to my last post, conditional upon you winning and estimating another 4.5 winners, you've still won an expected 1000/5.5 = $181.82. That's an even better return on your $95 than $99.90. Eliminate the 50% chance of winning because, under your condition, the chance is now 100%.
If, after-the-fact, you find that you hold a winning ticket, you should not adjust the numerator by merely subtracting 1 from it, you should subtract 175711536 from it because that's how many tickets it takes, on average, to produce a winner. In other words, for every 1 winning ticket, there are an average of 175,711,535 corresponding losing tickets, which should also be factored into the equation.
Therefore the estimated number of OTHER winners is reduced to (651915940 - 175711536)/175711536 = 2.71
——————————————————————————————————————————————————————————————
Applying the same logic to the coin-toss game: before the fact, if each participant writes down heads or tails randomly, then on average there will be 10/2 = 5 winners. The "flip-side" of this is that there will also be an average of 5 losers. So on average there will be 1 loser for each 1 winner.
Now, if I find out that I won before finding out about anyone else, I need to first assume that there was 1 corresponding loser for my 1 win. Therefore the average number of remaining winners is (10 - 2)/2 = 4.
By not taking the average corresponding losers for each winner into consideration, you are introducing an imbalance in the equation.
Quote: FleaStiffHow does one have 3.06 winners? Its either 3 or 4, but it ain't 3.6.
The average die roll is 3.5 but you can't roll that either.
You should see the dice I try to switch into a game!!!Quote: JBThe average die roll is 3.5 but you can't roll that either.
The modal average sure ain't 3.5.
Quote: WizardI will grant that IF you win, and know of no other results, you can expect 3.7 other winners, for a total of 4.7. However, it is dangerous to talk about expected value for an event that already happened. The reason you shouldn't divide by 4.7 in the original expected value calculation is because you using the estimation of somebody who was cherry picked to win.
Going back to my coin flip example (see first post). Suppose I do that experiment and then cherry pick somebody who won to estimate the total number of winners. He would say 5.5. Then I cherry pick a loser and ask him the same thing. He would say 4.5. Although they give different answers, they are both right. It just goes to show you shouldn't do the original calculation based on the perspective of somebody who was cherry picked to win or lose.
So, you agree that the expected number of winners, given that you win, is 4.7. This is good. However, I disagree with your statement about why you shouldn't divide by 4.7. The reason is that there is no inherent meaning to dividing by an expected value. E[c/X] is not c/E[X]. In general, E[f(X)] is not f(E[X]). You have to do E[f(X)]=Sum[f(x)*p(x)], where p(x) is the probability mass function of f (or an integral if the distribution is continuous).
NEITHER dividing by 4.7 NOR dividing by 3.7 has any inherent validity in general. However, dividing by 3.7 is a decent approximation in this MegaMillions example, so let's investigate that.
Quote: WizardFirst, a quick recap of the big MegaMillions draw. Let's forget all about taxes and the annuity, to keep things simple.
Tickets sold = 651,915,940
Probability of winning = 1 in 175,711,536
Expected winners (in my opinion)= 651,915,940/175,711,536 = 3.71
Jackpot = 656 million
Expected return of consolation prizes = 18.20%.
It is rather crude to divide the jackpot by the expected number of winners, because there is a 2.47% chance nobody will win, in which case the money stays in the jackpot. However, to keep the math sample, and avoid confusing the issue, I'm going to estimate the jackpot per winner by dividing by 3.71.
Expected return = 18.20% + (650,000,000/3.71)*(1/175,711,536) = 118.20%
Correct me if I'm wrong, but the guy who wrote to me, and CM are saying that IF you win then you'll have to share the jackpot with 3.7 other people, for a total of 4.7 winners.
By that argument the expected return would be 18.20% + (650,000,000/4.71)*(1/175,711,536) = 96.97%.
To avoid confusion, I will restate your estimate of the expected return where you divide by 3.71, but using the jackpot value from your original assumptions:
Expected return = 18.20% + (656,000,000/3.71)*(1/175,711,536) = 118.83%
The actual expected return is:
18.20% + 656,000,000*E[1/N]*(1/175,711,536)
where N is the random variable giving the number of winners, given that you won (so it is never 0).
Using the binomial distribution (i.e., N=Binomial(651915940, 1/175711536)+1, since it is a given that you won) we get E[1/N]=0.2629344, which in turn gives us an exact result for the return of 116.36%. Thus, dividing by 3.71 is pretty close, far closer than dividing by 4.71. Why is this?
Well, for this binomial distribution, p is very small and n is very large, so a Poisson distribution with lambda=np=3.71 will be a very good approximation. Specifically, N=Poisson(lambda)+1.
The Poisson pmf is (lambda^n / n!) * e^-lambda, so
E[1/N] = Sum[n=0..Infinity, 1/(n+1) * (lambda^n / n!) * e^-lambda]
= Sum[n=0..Infinity, lambda^n / (n+1)!] * e^-lambda
= Sum[n=0..Infinity, lambda^(n+1) / (n+1)!] * e^-lambda / lambda
= Sum[n=1..Infinity, lambda^n / n!] * e^-lambda / lambda
= (Sum[n=0..Infinity, lambda^n / n!] - 1) * e^-lambda / lambda
= (e^lambada - 1) * e^-lambda / lambda
= (1 - e^-lambda) / lambda
Obviously, 1/lambda becomes a good approximation for E[1/(1+Poisson(lambda))] for lambda of even moderate size. Certainly e^-3.71 is small. However, if you multiply by (1-e^-3.71)/3.71 rather than 1/3.71, you get 116.36%. In fact, the result first differs from the binomial distribution result in the 8th most significant digit.
However, getting back to the point, the fact that you can approximate E[1/(1+Poisson(lambda))] by 1/lambda is a specific property of the Poisson distribution, not something that will work in general, and the calculations that lead to this approximation were in fact based on adding one to the distribution of the number of winners in the first place. Accounting for the number of winners being one greater because you won is the way to go, 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.
Quote: DocSo, Wizard, given no info about any specific winners, you expect there to be 3.7 winners. Assuming that you are a winner, someone suggests you should expect 3.7 other winners for a total of 4.7. Then if you were to go and meet up with one of those others, the two of you winners together should expect that there are 3.7 others for a total of 5.7. Keep it up, and we should expect 651,915,940.7 winners and maybe I'll get a share of the payout! Doesn't that "logic" follow?
You're right, the logic is confusing.
But, let's say that you won. You expect 3.7 other winners to be out there. You then start interviewing all other 651915939 players to see if they won. On average, you will tally up 175711535 losers and 1 winner on the first leg of your quest. You must take all of this information into account. Then, there will be 476204404 more people for you to interview. From those people, you will expect another 2.7 winners.
On the other hand, if you won, then you walked to your neighbors house and found out that he won, and you have no other information, then you both can expect for there to be 3.7 more winners besides the two of you.
Quote: WizardRecently JB disagreed with CrystalMath about something mathematical. I wrote to JB warning to not take on CrystalMath lightly. So, it is with all due caution and respect that I'm about to do so. At least I think I am. This may just be a disagreement over semantics or confusion of what the issue is.
Thank you for the compliment.
Quote: Wizard
Correct me if I'm wrong, but the guy who wrote to me, and CM are saying that IF you win then you'll have to share the jackpot with 3.7 other people, for a total of 4.7 winners.
I am saying that given you win, you will expect the average pay amount to decrease because of your win. On average, given that you win, there will be 4.7 people in the pool. The more likely outcome is that you don't win, and in this case, you expect 3.7 people to be in the pool, for a higher average award.
Quote: Wizard
I will grant that IF you win, and know of no other results, you can expect 3.7 other winners, for a total of 4.7. However, it is dangerous to talk about expected value for an event that already happened. The reason you shouldn't divide by 4.7 in the original expected value calculation is because you using the estimation of somebody who was cherry picked to win.
Agreed.
I mean Lou proved it with division, multiplication, and addition !
Nareed says Who's on first, just repeating my question Who's on first, instead of telling me Who's on first .
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?
By the way, I wanted to review exactly what I said on the air. Here is a the April 5 show. We start to talk about the MegaMillions at about the 1:35 point. The expected number of winners statement is around 4:55.
Quote: CrystalMath
But, let's say that you won. You expect 3.7 other winners to be out there. You then start interviewing all other 651915939 players to see if they won. On average, you will tally up 175711535 losers and 1 winner on the first leg of your quest. You must take all of this information into account. Then, there will be 476204404 more people for you to interview. From those people, you will expect another 2.7 winners.
I like the interview example. In the base case, 1 out of 1 tickets is known to win, so 3.7 more winners are expected. I assume that more or less than 3.7 winners are expected as the interviews continue and more results become known. If I talk to 10 people and find 3 winners, many more than 3.7 winners are expected. If I talk to 640,000,000 people and find 1 winner, then less than 3.7 winners are expected.
In the Wizard's case, 1 person is interviewed and found to be a winner. That leaves 649,999,999 additional interviews to be conducted, and 3.7 more winners are expected to be identified through that interview process.
Quote: CrystalMathBut, let's say that you won. You expect 3.7 other winners to be out there.
No, I might be mistaken, but I don't expect that. In terms of an expected value, I expect there to be a total of 3.7 winners. If I happen to be one of them, great, but that doesn't mean I expect one more additional winner to have mysteriously appeared. I'm just one of the lucky ones included in the actual count of something that had an expected value of 3.7. The total actual count might be one or ten, but the expected value is still 3.7 so far as I can see.
Quote: DocNo, I might be mistaken, but I don't expect that. In terms of an expected value, I expect there to be a total of 3.7 winners. If I happen to be one of them, great, but that doesn't mean I expect one more additional winner to have mysteriously appeared. I'm just one of the lucky ones included in the actual count of something that had an expected value of 3.7. The total actual count might be one or ten, but the expected value is still 3.7 so far as I can see.
If you win, does that mean that your neighbor didn't win? No. He might have, but he has only 1 in 175711536 chance.
If you didn't win, does that mean that your neighbor didn't win? No. He might have, but he has only 1 in 175711536 chance.
What about all of the other 651915939 tickets sold? Does your ticket influence whether or not they won? No. They still have a 1 in 175711536 chance, and if 651915939 other people played, then you expect 651915939/175711536 of them to win.
Quote: CrystalMathDoes your ticket influence whether or not they won? No. They still have a 1 in 175711536 chance, and if 651915939 other people played, then you expect 651915939/175711536 of them to win.
Absolutely. But if you declare that I did win, then the drawing must be over, and it is no longer a matter of chance whether my neighbor's ticket is a winner or a loser. It is an outcome that has already been determined just like my own, whether I know my neighbor's status or not. There is no meaning to calculating an expected value then.
In the specific drawing, they identified that there were three winning tickets. Once that was determined, what do you believe is the expected value of the number of winning tickets? Does such an after-the-fact "expected value" even have meaning to you?
Quote: DocAbsolutely. But if you declare that I did win, then the drawing must be over, and it is no longer a matter of chance whether my neighbor's ticket is a winner or a loser.
You are saying that it is absolutely impossible for your neighbor to have chosen the same numbers that you did. Do you see any problem with this logic? It is, in fact, a matter of chance until he looks at his ticket and determines whether or not it is a winner. Your ticket did not preclude his ticket from winning. If that were the case, your ticket would preclude any other ticket from winning.
Quote: Doc
In the specific drawing, they identified that there were three winning tickets. Once that was determined, what do you believe is the expected value of the number of winning tickets? Does such an after-the-fact "expected value" even have meaning to you?
Yes, everyone's tickets were evaluated, and three were found. The expected number of winners is 3.
Quote: CrystalMathYou are saying that it is absolutely impossible for your neighbor to have chosen the same numbers that you did. Do you see any problem with this logic? It is, in fact, a matter of chance until he looks at his ticket and determines whether or not it is a winner.
I don't think I am saying at all that it is impossible for his ticket to duplicate mine. His may be the same as mine or it may not be, but its set of numbers and whether it is a winning ticket or not are no longer matters of chance after the drawing is complete. He either has a winner or a loser and which one of those two is true is an absolutely certainty whether he or I personally know the answer or not. There are no probability calculations left to perform.
Quote: CrystalMathYes, everyone's tickets were evaluated, and three were found. The expected number of winners is 3.
I contend that the actual number of winners is 3. Prior to the drawing/determination there was an expected value of 3.7, while after the drawing there is no probabilistic event remaining for an expected value to be calculated.
It seems to me there was a thread about some paradox six months or a year ago, where in the discussions "Bayesian" statisticians was contrasted with some other category of statisticians. I think our disagreement on the post-lottery-drawing question may fall back to that same thing, but my memory is too feeble to recall the specifics. And I don't consider myself a statistician at all.
Quote: DocIt seems to me there was a thread about some paradox six months or a year ago, where in the discussions "Bayesian" statisticians was contrasted with some other category of statisticians. I think our disagreement on the post-lottery-drawing question may fall back to that same thing, but my memory is too feeble to recall the specifics. And I don't consider myself a statistician at all.
Yes, I think you're referring to the Probabilities in Reid-Angle Race thread. One point of that is how Bayesian and Frequentist statisticians interpreted polling results. It my opinion it came down to a heated disagreement over semantics, not math.
Quote: WizardYes, I think you're referring to the Probabilities in Reid-Angle Race thread. One point of that is how Bayesian and Frequentist statisticians interpreted polling results. It my opinion it came down to a heated disagreement over semantics, not math.
I did a quick search, and I think I was referring to the Newcomb Problem thread, well over a year ago. Here is the post were ME suggested a disagreement I was having with weaselman was due to Bayesian vs. Frequentist viewpoints. Until I looked back, I couldn't even remember which of those positions I had held, partly because I didn't remember which was which plus I didn't remember what I had claimed then. To some relief, I think it was the same position that I held in disagreeing with CrystalMath. I rarely maintain consistent positions that long.
Quote: DocI did a quick search, and I think I was referring to the Newcomb Problem thread, well over a year ago.
Hmm. If I get bored I'll read through that angle of the thread. For me it came down to more of question of free will and pre-destination. Anyway, let's try to get back to the lottery question.
Quote: WizardOn my radio show I was talking about jackpot sharing in the big MegaMillions draw. 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.
Does that argument hold water?
I say "yes," because there are an expected 3.6 winners of all of the players besides you. Whether or not you win is independent of how everybody else does.
Why "besides you" ?Quote: ThatDonGuyI say "yes," because there are an expected 3.6 winners of all of the players besides you. Whether or not you win is independent of how everybody else does.
Does this mean you're assuming you're going to lose? If so, why bother buying the ticket in the first place?
Quote: ThatDonGuyI say "yes," because there are an expected 3.6 winners of all of the players besides you. Whether or not you win is independent of how everybody else does.
I still agree with Wizard's initial post, which says that the number of expected total winners remains constant, even if you know that you are a winner:
1) The probability of a ticket winning is 1 in 175,711,536
therefore
2) Among 175,711,536 tickets there will be 1 winning ticket and 175,711,535 losing tickets (on average)
therefore
3) For every 1 winning ticket there are 175,711,535 losing tickets (on average)
therefore
4) If you possess 1 winning ticket, probability dictates that there are also 175,711,535 losing tickets (on average)
therefore
5) You have accounted for 175,711,536 tickets (in theory)
therefore
6) You should remove these 175,711,536 tickets from the quantity of tickets whose outcomes are still unknown
therefore
7) The number of expected other winning tickets is (651915940 - 175711536) / 175711536 = 2.71, which is exactly 1 less than the figure computed before the drawing.
BTW, to insure that at least one person does win... peek at the answers to your L and R. if both are the same pick the opposite. If both differ you are free to choose either H or T.
Quote: JB4) If you possess 1 winning ticket, probability dictates that there are also 175,711,535 losing tickets
This, I think, is incorrect. There is a chance that there are winners in those 175,711,535 tickets, right?
In fact, there is a 0.367879442 chance of 0 winners in those tickets,
0.367879442 chance of 1 winner in those tickets,
0.18393972 chance of 2 winners in those tickets,
0.061313239 chance of 3 winners in those tickets,
...
Consider the following table which details whether or not you win and the number of other winners:
| You | Other Winners | Total Winners | Probability | Contribution to Total Winners |
|---|---|---|---|---|
| 0 | 0 | 0 | 0.0244738819051037 | 0 |
| 0 | 1 | 1 | 0.0908017428243344 | 0.0908017428243344 |
| 0 | 2 | 2 | 0.168443987184093 | 0.336887974368185 |
| 0 | 3 | 3 | 0.208317417235526 | 0.624952251706579 |
| 0 | 4 | 4 | 0.193222152282524 | 0.772888609130097 |
| 0 | 5 | 5 | 0.143376586089217 | 0.716882930446085 |
| 0 | 6 | 6 | 0.0886580773984832 | 0.531948464390899 |
| 0 | 7 | 7 | 0.0469906652467411 | 0.328934656727187 |
| 0 | 8 | 8 | 0.021792794852812 | 0.174342358822496 |
| 0 | 9 | 9 | 0.00898383455528135 | 0.0808545109975321 |
| 0 | 10 | 10 | 0.00333313627167137 | 0.0333313627167137 |
| 0 | 11 | 11 | 0.00112422103958636 | 0.01236643143545 |
| 0 | 12 | 12 | 0.000347585608847701 | 0.00417102730617242 |
| 0 | 13 | 13 | 0.000099199562985322 | 0.00128959431880919 |
| 0 | 14 | 14 | 2.62889383215185E-05 | 0.000368045136501259 |
| 0 | 15 | 15 | 6.50239147689798E-06 | 9.75358721534698E-05 |
| 0 | 16 | 16 | 1.50780248337236E-06 | 2.41248397339577E-05 |
| 0 | 17 | 17 | 3.29068913392306E-07 | 5.59417152766921E-06 |
| 0 | 18 | 18 | 6.78274791087943E-08 | 1.2208946239583E-06 |
| 0 | 19 | 19 | 1.32447388486274E-08 | 2.5165003812392E-07 |
| 0 | 20 | 20 | 2.45699757678764E-09 | 4.91399515357528E-08 |
| 1 | 0 | 1 | 1.3928443516872E-10 | 1.3928443516872E-10 |
| 1 | 1 | 2 | 5.16765975690409E-10 | 1.03353195138082E-09 |
| 1 | 2 | 3 | 9.58639324299868E-10 | 2.8759179728996E-09 |
| 1 | 3 | 4 | 1.18556483634115E-09 | 4.74225934536458E-09 |
| 1 | 4 | 5 | 1.09965547954791E-09 | 5.49827739773955E-09 |
| 1 | 5 | 6 | 8.15977084766899E-10 | 4.8958625086014E-09 |
| 1 | 6 | 7 | 5.0456606277148E-10 | 3.53196243940036E-09 |
| 1 | 7 | 8 | 2.67430736671563E-10 | 2.1394458933725E-09 |
| 1 | 8 | 9 | 1.24025977308843E-10 | 1.11623379577959E-09 |
| 1 | 9 | 10 | 5.11283141159819E-11 | 5.11283141159819E-10 |
| 1 | 10 | 11 | 1.89693651681511E-11 | 2.08663016849662E-10 |
| 1 | 11 | 12 | 6.39810607531463E-12 | 7.67772729037756E-11 |
| 1 | 12 | 13 | 1.97816044830353E-12 | 2.57160858279459E-11 |
| 1 | 13 | 14 | 5.64559196328926E-13 | 7.90382874860497E-12 |
| 1 | 14 | 15 | 1.49614186237224E-13 | 2.24421279355836E-12 |
| 1 | 15 | 16 | 3.70060592601276E-14 | 5.92096948162041E-13 |
| 1 | 16 | 17 | 8.58112407573218E-15 | 1.45879109287447E-13 |
| 1 | 17 | 18 | 1.87277923098393E-15 | 3.37100261577108E-14 |
| 1 | 18 | 19 | 3.86016086586429E-16 | 7.33430564514215E-15 |
| 1 | 19 | 20 | 7.53777425518897E-17 | 1.50755485103779E-15 |
| 1 | 20 | 21 | 1.39831319371699E-17 | 2.93645770680567E-16 |
Contribution to Total Winners is the probability times the number of other winners. To calculate the average total number of winners, given that you won, add up Contribution to Total Winners for the bottom half of the table and divide by the probability that you won (which is also the sum of the probability on the bottom half of the table). [2.68061E-08/5.69115E-09 = 4.71014876]
Therefore, if you win, you expect a total of 4.71 winners.
If you perform the same calculation for you losing, you get 3.71 for the total expected number of winners .
Quote: 98ClubsThe coin-flip posting by ZPP is flawed as it does not account that zero people will guess correctly.
ZPP seems like a very smart guy to me. And I learned something from his lottery analysis (which is spot on, by the way).
His analysis doesn't include zero, because he is calculating the expected number of winners, given that you won. Because these probabilities are based on that assumption, there must be at least one winner (you).
Quote: JBI still agree with Wizard's initial post, which says that the number of expected total winners remains constant, even if you know that you are a winner:
This cannot be right. Suppose, the number of expected winners is exactly 1. Do you insist, that, if you know you won, then expected number of other winners is 0? What if the initial expectation is 0.5?
Conditional expectation certainly changes, depending on the condition, however, as others have pointed out, it does not make a +EV game negative. I like the way ME explained it - provided that you have won, your EV cannot become worse then that computed unconditionally.
Quote: CrystalMathThis, I think, is incorrect. There is a chance that there are winners in those 175,711,535 tickets, right?
In fact, there is a 0.367879442 chance of 0 winners in those tickets,
0.367879442 chance of 1 winner in those tickets,
0.18393972 chance of 2 winners in those tickets,
0.061313239 chance of 3 winners in those tickets,
...
Contribution to Total Winners is the probability times the number of other winners. To calculate the average total number of winners, given that you won, add up Contribution to Total Winners for the bottom half of the table and divide by the probability that you won (which is also the sum of the probability on the bottom half of the table). [2.68061E-08/5.69115E-09 = 4.71014876]
Therefore, if you win, you expect a total of 4.71 winners.
If you perform the same calculation for you losing, you get 3.71 for the total expected number of winners .
By doing this, aren't you effectively saying that the winning ticket had a probability of winning of 1, even before the drawing? It's like setting aside the ticket and saying that the normal probability rules no longer apply to it.
I agree that its probability of being a winner after the drawing is 1, no question. But because every other ticket is still unknown, you still have to assume that your winning ticket's probability of having won was 1 in 175,711,536. Therefore, there are theoretically 175,711,535 losing tickets to make up for your 1 winning ticket.
What you're describing is like saying this:
You sit down with $1,000 at a table game that offers an even-money bet with a 1% house edge.
You bet the entire $1,000 and win; now you have $2,000.
You then claim that the house advantage on the bet you made was -100%; that the theoretical house edge doesn't apply because you won, it only applies when you lose.
I would argue that that is wrong, you still have to look at the bigger picture, and not set aside one individual result because it was favorable, so the normal rules no longer apply to it.
I probably cannot explain this correctly.
Quote: weaselmanThis cannot be right. Suppose, the number of expected winners is exactly 1. Do you insist, that, if you know you won, then expected number of other winners is 0?
On average, why wouldn't it be?
For that scenario to occur, there must have been exactly 175,711,536 tickets in play.
The probability of winning is 1 in 175,711,536.
If the very first ticket verified is a winner, then on average, there will be 0 other winners in that drawing.
Quote: weaselmanSuppose, the number of expected winners is exactly 1. Do you insist, that, if you know you won, then expected number of other winners is 0? What if the initial expectation is 0.5?
To these questions, I would repeat my previous comment:
"Suppose the number of expected winners is exactly 1. If you know that you won ...", then the drawing is over, and there is no more random event to take place for which to calculate an expected value. Same thing if the expected number (before the draw, obviously) is 0.5. In either case, after the draw there is no calculation to perform at all -- the answer is absolutely set, not a matter of probabilities.
If I have $10 in one pocket, and $20 in the other pocket, who's pants do I have on ?
Quote: JBIf the very first ticket verified is a winner, then on average, there will be 0 other winners in that drawing.
While we agree on the answer to the original question, I disagree with this interpretation. If there were ever a second winner in such a drawing, the number of "other" winners could not "average" zero.
I think that in this comment you are (like some others) confusing expected number of winners (which is only meaningful prior to the drawing) with the number of actual winners, which is the non-probabilistic number that was determined at the time of the drawing (whether anyone yet knows that number or not.)
Quote: JBOn average, why wouldn't it be?
Because everybody has a positive probability of winning. A sum of positive numbers cannot be 0.
Quote:
For that scenario to occur, there must have been exactly 175,711,536 tickets in play.
The probability of winning is 1 in 175,711,536.
If the very first ticket verified is a winner, then on average, there will be 0 other winners in that drawing.
Huh? There are 175711535 tickets remaining, every one having a probability of 1/175711536 of winning ... How can the number of expected winners be 0?
Here is a similar question, formulated in, perhaps, more trivial terms. Suppose, you flip a coin for 1000 times. The expected number of heads you get before you start is obviously 500. Now, suppose, you already flipped it once, and it was heads. What is the expected number now? What if you threw it ten times, and got ten heads? Does it sound familiar? :)
Quote: DocTo these questions, I would repeat my previous comment:
"Suppose the number of expected winners is exactly 1. If you know that you won ...", then the drawing is over, and there is no more random event to take place for which to calculate an expected value.
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?
Quote: weaselmanHuh? There are 175711535 tickets remaining, every one having a probability of 1/175711536 of winning ... How can the number of expected winners be 0?
If you repeated a drawing of this magnitude enough times (let's say 1 trillion times), how many winners will there be, on average, per drawing? Assume that all 175,711,536 tickets in play are random, independent quick-picks.
What you are doing is cherry-picking the winning ticket, taking it out of the picture, and saying that because its probability of winning after the numbers are known is 1, that its probability of having won before the numbers were drawn was also one. I say this is wrong; it is merely a frame of the bigger picture.
Quote: JBIf you repeated a drawing of this magnitude enough times (let's say 1 trillion times), how many winners will there be, on average, per drawing? Assume that all 175,711,536 tickets in play are random, independent quick-picks.
175711535/175711536?
Is it a trick question?
Quote: weaselman175711535/175711536?
Is it a trick question?
On average there will be 1 winner, since the probability of winning is 1 in 175,711,536.
Now suppose you are one of the 175,711,536 ticket holders in the 1-trillion-and-1st drawing, your ticket is the first one verified, and it is a winning ticket.
Based on the probability, the AVERAGE number of OTHER winning tickets for that drawing is 0.
Quote: JBNo but that sure sounds like a trick answer. :)
I don't follow... N-1 tickets, each having 1/N probability of winning ... What is your answer?
Quote: JBI
What you are doing is cherry-picking the winning ticket, taking it out of the picture, and saying that because its probability of winning after the numbers are known is 1, that its probability of having won before the numbers were drawn was also one.
I say this is wrong; it is merely a frame of the bigger picture.
Yes, of course it is wrong. And I said no such thing. It's probability of winning before numbers are known is 1/N.
But conditional probability of it winning provided that it won is, of course, 1. But the fact that it won does not affect the probabilities of other tickets winning as well. Those are still 1/N.
Quote: weaselmanI don't follow... N-1 tickets, each having 1/N probability of winning ... What is your answer?
I mis-read your post initially, and edited my response - see above.
Quote: JB
Based on the probability, the AVERAGE number of OTHER winning tickets for that drawing is 0.
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?
