# Expected payoff of dice game

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?

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

• 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 Apr 30, 2021 at 13:52
• I added a short part to my answer. The problem with your approach is that $E(X)$ is not $3.5$.
– YJT
Apr 30, 2021 at 14:04
• The problem with the approach is that the number of games is not independent of the payoff.
– lulu
Apr 30, 2021 at 14:07
• I did some reading on Wald's equality this afternoon, and I think it applies in this case. From Wald's equality we have $E[X_1 + \ldots + X_N] = E[N] * E[X_i]$. May 1, 2021 at 0:45
• @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.
– lulu
May 1, 2021 at 10:00

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.

• $N$ and $P$ aren't independent, but how does this affect $E[P] = E[N]E[X]$, which only requires independence between $N, X$? Apr 30, 2021 at 13:52
• 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. Apr 30, 2021 at 13:53
• 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? Apr 30, 2021 at 13:55
• 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 $X = P/N$. Apr 30, 2021 at 13:57
• 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 $\{1,2,3\}$. Apr 30, 2021 at 14:03

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 $$N$$: $$E(P\vert N)=E(\sum\limits_{i=1}^\inftyX_i\vert N)=\sum\limits_{i=1}^N E(X_i\vert N)=5\cdot (N-1)+2=5N-3$$ 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 $$1-3$$.

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