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You roll a fair $6$-sided die. For each roll, you're paid the face value. The game stops when you roll a $1,2,3$. If you roll a $4,5,6$, you can roll again and keep accumulating payments.

There are several ways to solve this problem. I tried the geometric variable approach, but I feel that my answer, while correct, is not complete, but I don't know what I'm missing.

Essentially, define $N$ as a geometric random variable for the number of rolls until the game terminates. $E[N] = 2$.

Define $X$ as the pay off per roll. $E[X] = 3.5$.

Each roll is independent. Define $P$ as the payoff. So $P = N * X$ So the expected pay off should be $E[P] = E[N] * E[X] = 7$. This is the step that I feel like I'm missing some justification on, and I don't feel it's complete.

Is it valid to define $P = N * X$? If so, then is this answer complete, or is there something missing?

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3 Answers 3

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The result is correct, but it's hard to justify along the lines you proposed.

As an alternative method:

Consider the possible outcomes of the first roll, and note that, if the game does not end, then it restarts (and, of course, will have the same expectation, $E$, going forward).

We see that $$E=\frac 16 \times \left(1+2+3+(4+E)+(5+E)+(6+E)\right)\implies 6E=21+3E\implies E=7$$

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  • $\begingroup$ What is wrong with my justification? I also used this conditional expectation approach, but I found the geometric random variable approach intuitive as well, but I just don't feel it's complete $\endgroup$ Apr 30, 2021 at 13:52
  • $\begingroup$ I added a short part to my answer. The problem with your approach is that $E(X)$ is not $3.5$. $\endgroup$
    – YJT
    Apr 30, 2021 at 14:04
  • $\begingroup$ The problem with the approach is that the number of games is not independent of the payoff. $\endgroup$
    – lulu
    Apr 30, 2021 at 14:07
  • $\begingroup$ I did some reading on Wald's equality this afternoon, and I think it applies in this case. From Wald's equality we have $E[X_1 + \ldots + X_N] = E[N] * E[X_i]$. $\endgroup$ May 1, 2021 at 0:45
  • $\begingroup$ @student010101 That requires work though. The issue, again, is that $N$ and the $X's$ are dependent. If we know that $N$ is large then we know that all the $X's$ (except for the last one) are also large. Now, it is true that Wald sometimes holds for dependent variables, but it also sometimes fails in that case. this provides a list of everything that must be checked. Worth doing! Though, as I have said, this problem is easily done by other means. $\endgroup$
    – lulu
    May 1, 2021 at 10:00
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While $P = NX$ is a valid definition, $N$ and $X$ are not independent, so you cannot say that $\mathbb E(P) = \mathbb E(N)\mathbb E(X)$, which is generally false for dependent variables. Additionally, I do not think that $\mathbb E(X) = 3.5$ -- but to get the right answer for your question, I think you'll want to take a different approach regardless.

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  • $\begingroup$ $N$ and $P$ aren't independent, but how does this affect $E[P] = E[N]E[X]$, which only requires independence between $N, X$? $\endgroup$ Apr 30, 2021 at 13:52
  • $\begingroup$ Sorry, typo -- I meant to say that $N, X$ are not independent. (Note that longer runs of rolls will have higher averages, because they include more digits from 4 to 6.) Will fix now. $\endgroup$ Apr 30, 2021 at 13:53
  • $\begingroup$ Hmm I see what you're saying about the first $N - 1$ rolls having higher averages, but isn't $X$ here defined to just be the outcome of a roll, which always has expectation 3.5? $\endgroup$ Apr 30, 2021 at 13:55
  • $\begingroup$ If $X$ is just a variable for a generic single die roll, then $P = N*X$ is no longer valid. I took your definition as valid essentially because I understood it as $X = P/N$. $\endgroup$ Apr 30, 2021 at 13:57
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    $\begingroup$ Identically distributed variables are not identical. $X_i \neq X_j$ in that sum (note that in the right-most expession $i$ is suddenly a free variable!). But they're not even identically distributed: by the setup $X_1, \ldots, X_{N-1}$ lie in $\{4,5,6\}$ while $X_N$ lies in $\{1,2,3\}$. $\endgroup$ Apr 30, 2021 at 14:03
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To make your approach more formal, let $N$ be the number of rolls until the game stops, and $X_i$ the value of the $i$th roll ($i=1,2....$). The total payoff $P$ is $\sum\limits_{i=1}^\infty X_i$. Given $N$: $$E(P\vert N)=E(\sum\limits_{i=1}^\infty$X_i\vert N)=\sum\limits_{i=1}^N E(X_i\vert N)=5\cdot (N-1)+2=5N-3$$ since the expected value of each roll that didn't end the game is $5$ (options are $4,5,6$) and the terminal roll's expectation is $2$. From the tower law, $$E(P)=E(E(P\vert N))=E(5N-3)=7$$

You can use this method to compute $E(X)=E(E(\tfrac{P}{N}\vert{N}))=5-3E(\tfrac{1}{N})$. Wolfram says the inside expectation is $0.69$ so $E(X)=2.92$ which is not $3.5$.

This makes sense in the following way: $E(NX)=7$ while $E(N)E(X)<3.5\cdot 2$ so $Cov(N,X)>0$ and indeed, higher values of $N$ tend to go with higher values of $X$, as there were many rolls of $4-6$ before the terminal $1-3$.

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  • $\begingroup$ Shouldn't the total payoff be defined as $\sum_{i=1}^N X_i$ instead of going summing to infinity? $\endgroup$ Apr 30, 2021 at 13:56
  • $\begingroup$ Since $X_i=0$ for $i>N$, both are fine. $\endgroup$
    – YJT
    Apr 30, 2021 at 14:00

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