Craps Two Dice Question In the game of craps, a player throws a pair of dice. The game ends with the first throw if a 2, 3, 7, 11, or 12 is thrown. Otherwise, the player throws the dice repeatedly until either the number obtained on the first throw occurs again or a 7 is thrown. What is the expected number of throws to complete the game?
 A: Use the Law of Iterated Expectation.   Where $N$ is the count of trials, and $X_n$ is the result of the $n$-th trial, we have:
$$\mathsf E(N) = \mathsf P(X_1{\in}\{2, 3, 7, 11, 12\})\cdot 1 + {\sum}_{k\in\{4,5,6,8,9,10\}}\mathsf P(X_1{=}k)\cdot\big(1+\mathsf E(N{-}1\mid X_1{=}k)\big)$$
Also: $(N{-}1)\mid (X_1{=}k) \;\sim\; \mathcal{Geo}_1\Big(\mathsf P\big(X_\ast\in\{k,7\}\big)\Big) \qquad : k\in\{4,5,6,8,9,10\}$
That is: The expected number of trials after the first, will be conditionally Geometrically Distributed conditioned on the result of the first throw.
Now calculate the relevant probabilities and evaluate.
A: Ok.  Let E denote the answer. there's a $\frac 13$ chance the game is over on the first roll.  Assume now that we have gotten past the first roll.
Note:  The method below should be correct but it is a bit error prone (especially if the writer is tired).  I advise checking the calculation.
Whatever the target number is, the probability of getting it can be written as $\frac n{36}$ for some integer $n$ ($n\in3,4,5$ as it happens). To be precise, if the target is $\{4,5,6,8,9,10\}$ then $n=\{3,4,5,5,4,3\}$ respectively. Let $F_n$ denote the expected number of rolls it will take to end the game if the target has probability $\frac n{36}$. (to be clear: at this stage we are ignoring the first roll, the one that got us the target number in the first place.) Of course the probability of throwing a $7$ is $\frac 6{36}$, thus the probability of ending the game on a given toss is $\frac {n+6}{36}$.  Not hard to see that this implies that $$F_n=\frac {36}{n+6}$$
Now, put this together.  We see that $$E=\frac 13 1+\frac {6}{36}(F_3+1)+\frac {8}{36}(F_4+1)+\frac {10}{36}(F_5+1)\sim 3.376$$
