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You decide to make a lottery with n tickets, where each ticket is numbered between 1 and n, and each ticket is unique. Each ticket costs $5, and the lottery works in the following manner. Once all n tickets have been purchased, a number x is selected at random between 1 and n and all the money is divided equally between people with tickets less than x. That way, if 1 is selected, you(as the organizer) get to keep the prize pool.

Everyone's number is randomized, the only case where the organizer wins the prize pool is if x=1, as no ticket number is less than 1, so no person wins anything and so the organizer automatically wins the prize pool. Also, the organizer does not hold the number 1, and does not win any money if the number is greater than 1.The only way where they earn the prize pool is if the chosen number is 1, other than that, there is no way they can win any money.

The first part is to calculate the expected winnings per lottery as the organizer. So there is only one case where you can win the prize pool, I multiplied the case where you win and the prize pool, so I get:

$$\frac{1}{n} \cdot 5n = 5$$

Now, the second part is to calculate the expected winnings if somebody purchases a ticket. I am not able to figure out the probability where somebody wins. For the prize pool, I am confused if it is 5n or 5p (where p is the person's ticket number).

PsychBit
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    The rules are not clear. Does everyone get to select their number or is it randomized? You appear to suggest that at least the organizer gets to select their number. Also: you say the prize is distributed to those holding less than $x$, but in that case, if $x=1$ is selected, nobody should get the prize (as no value is less than $1$). Or did you mean to say $≤x$? Also: How big is the prize pool? – lulu Mar 23 '24 at 15:34
  • Everyone's number is randomized, the only case where the organizer wins the prize pool is if x=1, as no ticket number is less than 1, so no person wins anything and so the organizer automatically wins the prize pool. – PsychBit Mar 23 '24 at 15:37
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    Please edit your post to include those details. Don't leave critical information for the comments. – lulu Mar 23 '24 at 15:39
  • Your computation doesn't appear to be complete. Sure, the organizer has a chance at a big payout, if the number $1$ is chosen, but they might make money with other selections, depending on the ticket they hold. For instance, if they hold $1$ and $2$ is selected, they get the entire pool. If they hold $2$ and $3$ is selected, they get half the pool and so on. Again, you have not stated the size of the pool. – lulu Mar 23 '24 at 15:41
  • Or is it the case that the organizer doesn't even retain a ticket? It's hard to guess at the rules here. – lulu Mar 23 '24 at 15:42
  • I think the organizer doesn't purchase a ticket – LegNaiB Mar 23 '24 at 15:44
  • @LegNaiB, Yes that is what I mean, I have made some more edits to my question. – PsychBit Mar 23 '24 at 15:45
  • Ok, So a contestant's expected prize is clearly dependent on the ticket they draw. If they draw $n$, then no scenario results in a payout for them (right?). Since the value on the ticket is unknown, you have to average over the $n$ possible expected payouts. – lulu Mar 23 '24 at 15:49
  • @lulu, Exactly, but I am not able to figure out the probability that a person wins and the total prize pool, as I did in the first part. – PsychBit Mar 23 '24 at 15:52
  • Symmetry is enough, here. Before tickets are drawn, each contestant has the same expected payout. See my posted solution. – lulu Mar 23 '24 at 16:10

2 Answers2

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A quick way to solve the problem:

As you have noted, the probability that no contestant gets any payout is $\frac 1n$ (as only the selection $1$ achieves that result). Thus there is a $\frac {n-1}n$ chance that the contestants (collectively) will get the entire pool of $5n$. But, of course, a priori, each contestant has the same expected gain here so the answer must be $$\frac {n-1}{n^2}\times 5n=\frac {5(n-1)}n=5-\frac 5n$$

Phrased differently: the expected payout to the organizer is $5$, and the expected total payout is $5n$ (that's the guaranteed payout). thus each contestant must expect to get $$\frac {5n-5}n=5-\frac 5n$$ which, of course, is the same result.

Sanity check: Let's do this in the case $n=2$. In that case a contestant either gets a $1$ or a $2$. If they get a $2$, they can not make any money. If they get a $1$, they have a $\frac 12$ chance of getting $10$ and a $\frac 12$ chance of getting $0$. Thus the answer is $\frac 12\times \frac 12\times 10=\frac 52$, as predicted.

Similarly, let's handle $n=3$. A contestant's outcome is determined by the pair $(i,j)$ where $i$ is the ticket value and $j$ is the chosen value. There are $9$ possible outcomes, each with probability $\frac 19$. The only pairs that have positive payouts are $(1,2)$ which has a payout of $15$ and $(2,3), (1,3)$ each of which has a payout of $\frac {15}2$. Thus the expected payout is $\frac 19\times \left( 15+\frac {15}2+\frac {15}2\right)=\frac {10}3$ as predicted.

lulu
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  • The contestant will also not get a payout if the number on their ticket is greater than the chosen number and yes, if the chosen number is 1, there is no payout for any of the contestants. – PsychBit Mar 23 '24 at 16:18
  • I understand, but that's taken into account. The rules apply to each of the contestants equally. – lulu Mar 23 '24 at 16:18
  • Check my explicit computation for the case $n=2$. That computation is, I believe, consistent with my understanding of the rules. If I still have the rules wrong, please clarify what (if anything) is wrong with that explicit computation. – lulu Mar 23 '24 at 16:19
  • To make it even more explicit: in the case $n=2$ there are exactly $4$ scenarios: $(1,1), (1,2), (2,1),(2,2)$ where $(i,j)$ means the contestant has the value $i$ and the value $j$ was chosen. Each of those is equally likely (probability $\frac 14$). the only one in which a contestant makes any money is $(1,2)$ and in that case the contestant gets a payout of $10$, making the expected profit for a contestant $\frac {10}4=\frac 52$. – lulu Mar 23 '24 at 16:23
  • Also if you sum all the possibilities up (see in my answer) you also get $5-5/n$ for everybody – LegNaiB Mar 23 '24 at 16:52
  • @LegNaiB sure, but that's a lot of work. Symmetry tells you that all the contestants have the same expectation, and that's all you need. – lulu Mar 23 '24 at 16:53
  • Your answer makes sense for calculating the chance that the contestants (collectively) will get the entire pool of 5n, but what if I want to calculate for just one specific person, in that case will it be (n-p)/n where p is the ticket number and I multiply it by 5p? . – PsychBit Mar 23 '24 at 16:56
  • I don't understand the question. I computed the expected payout for a contestant. Are you asking for the expected payout to the contestant who holds ticket number $p$? – lulu Mar 23 '24 at 16:59
  • Yeah I know, it is easier to use symmetry. I just wanted to point out that if you calculate it manually using all possibilities, you still get the same value – LegNaiB Mar 23 '24 at 16:59
  • If so, just sum over the cases. This is the method employed by @LegNaiB . It's the best way to handle the single ticket case. – lulu Mar 23 '24 at 17:00
  • So you want to know the expected payout if you already know which ticket the contestant has gotten? Then you just need to average the payout over all possible cases with which you can win – LegNaiB Mar 23 '24 at 17:00
  • @LegNaiB, Yes I also want to calculate it in this way – PsychBit Mar 23 '24 at 17:03
  • Then just take the part of my solution but remove the first sum. As the payout depends on the chosen ticket $q$ and is then divided by $q$, you get a sum like the harmonic series. I adjust my answer accordingly – LegNaiB Mar 23 '24 at 17:04
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So if a person has ticket $p$, the likelihood of winning the lottery is $(n-p)/n$ right? And then if $q>p$ is the chosen number, one wins $5n/(q-1)$. So you need a double sum, once over the possible ticket purchases and over the possible chosen ticket $q$. In sum you will get the following expected outcome:

$$ E[X] = \sum_{i=1}^n\sum_{j=1}^nP(p=i)P(q=j)1_{i<j} \frac{5n}{j-1} \\ =\sum_{i=1}^{n}\sum_{j=i+1}^n\frac{1}{n^2} \frac{5n}{j-1} \\ =\frac{5}{n}\sum_{i=1}^{n} \sum_{j=i}^{n-1} \frac{1}{j} \\ =\frac{5}{n}\sum_{j=1}^{n-1}\frac{j}{j} = \frac{5(n-1)}{n} = 5 - \frac{5}{n} $$

Which also makes sense: If there wouldn't be an organizer, on average everyone should get back his $5$. But in $1/n$ of cases you 100% don't get back your input as the organizer wins it. Thus you need to subtract $5/n$.

If you want to know the expected result given that the contestant already has chosen ticket $p$, you just get the second sum: $$ E[X|\text{contestant}=p]=\sum_{j=p+1}^n \frac{1}{n}\frac{5n}{j-1} = 5\sum_{j=p}^{n-1}\frac{1}{j} $$

The average probability to win is a bit easier as you don't need the $5n/(j-1)$ part: $$ P(p<q)=\sum_{i=1}^{n}\sum_{j=i+1}^n\frac{1}{n^2} = \frac{1}{n^2}\sum_{i=1}^{n} (n-i) = 1 - \frac{1}{n^2}\sum_{i=1}^n i \\ =1- \frac{n(n+1)}{2n^2} = 1-\frac{n+1}{2n} = \frac{2n-n-1}{2n} = \frac{n-1}{2n} $$

LegNaiB
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