Probability of winning a game of craps The dice game craps is played as follows: The player throws 2 dice, and if the sum is 7 or 11, he/she wins. If the sum is 2, 3, or 12, he/she loses. If the sum is anything else, then he/she continues throwing until that number appears again, or throws a 7, where the game ends in a loss. 
What I do know is that $P(7) = 6/36$, $P(11)=2/36$. So P(winning) in first roll is $8/36$. Furthermore, the probability of having to roll again will be $1-[P($winning 1st roll$) +P($losing 1st roll$)]=24/36$. It's what happens if the game doesn't end in the 1st roll that's got me a bit confused, since it could go on and indeterminate number of rolls. But since it can't be 7,11,2,3,or 12, it depends on if they roll a 4,5,6,8,9, or 10.
Note, this is not a homework problem, but an intriguing one I found in a different book "Probability Models, Sheldon Ross"
 A: The probability of winning directly is, as you calculated, $8/36$, and the probability of losing directly is $(1+2+1)/36=4/36$.
For the remaining cases, you need to sum over all remaining rolls. Let $p$ be the probability of rolling your initial roll, and $q=6/36=1/6$ the probability of rolling a $7$. Then the probability of rolling your initial roll before rolling a $7$ is $p/(p+q)$, and the probability of rolling a $7$ before rolling your initial roll is $q/(p+q)$. Thus, taking into account the probability of initially rolling that roll, each roll that doesn't win or lose directly yields a contribution $p^2/(p+q)$ to your winning probability.
For $p=5/36$, that's
$$
\frac{\left(\frac5{36}\right)^2}{\frac{5+6}{36}}=\frac{25}{11\cdot36}\;,
$$
and likewise $16/(10\cdot36)$ and $9/(9\cdot36)$ for $p=4/36$ and $p=3/36$, respectively. Each of those cases occurs twice (once above $7$ and once below), so your overall winning probability is
$$
\frac8{36}+\frac2{36}\left(\frac{25}{11}+\frac{16}{10}+\frac99\right)=\frac{244}{495} = \frac12-\frac7{990}\approx\frac12-0.007\;.
$$
A: Suppose you throw a $4$ and let $p(4)$ your winning probability. At your next roll you have a probability $3/36$ of winning (you throw a 4), a probability $6/36$ of losing (you throw a 7) and a probability $27/36$ of repeating the whole process anew (you throw any other number). Then:
$$
p(4)={3\over36}+{27\over36}p(4),\quad
\hbox{so that}\quad p(4)={1\over3}.
$$
Repeat this reasoning for the other outcomes and then compute the total probability of winning as:
$$
p_{tot}={8\over36}+{3\over36}p(4)+{4\over36}p(5)+\ldots
$$
