Conditional Probability for an Infinite Sum series (modified Craps game) I am having trouble understanding a modified version of the Craps game. There are two dice as in the original Craps game (no requirement to understand that game's rules here). The probabilities of Sum on first roll come out to be like this. Probabilities of Sum on First Roll. The rules of win/lose/re-roll are as follows.


*

*If the sum in any roll is $10$ (probability $\dfrac3{36}$), we win and game ends.

*If the sum in any roll is $7$ (probability $\dfrac6{36}$), we lose and game ends.

*If I roll any other number, the game continues. (There is no requirement to get the same number again to win in this modified Craps. We only win when we get a sum of $10$).


Example:
$5$ rolls: $11,12,5,6,7$ (We lost)
$6$ rolls: $5,6,6,3,4,10$ (We won).
I need to find two things.
1. Probability of player winning.
2. Expected number of rolls until game ends
So far I have done this:
I read here (Probability of winning a game of craps) but in that (original version), we win if we roll a point ($4,5,6,8,9,10$) on the first roll, and then get the same number on a subsequent roll. In my question, this is not the same. We can roll the same number again but we don't win. We only win when we get a 10, or lose if we get a $7$.
I tried using the method in the above mentioned stack exchange post ("each roll that doesn't win or lose directly yields a contribution $\dfrac{p^2}{p+q}$ to the winning probability") but I am sure that is a wrong approach.
I know that $P(A\mid B) = \dfrac{P(A \cap B)}{P(B)}$. 
Win : $10$ : $\dfrac3{36}$ (in craps it is $7,11$ i.e. $\dfrac6{36} +\dfrac2{36} = \dfrac8{36}$)
Lose: $7$ : $\dfrac6{36}$ (in craps it is $2,3,12$ i.e. $\dfrac1{36} + \dfrac2{36} + \dfrac1{36} = \dfrac4{36}$)
Re-Roll : $\dfrac{27}{36}$ (of not W or L on first roll)   (in craps it is $4,5,6,8,9,10$ i.e. $= \dfrac{24}{36}$)
I know that the probability of winning directly is $\dfrac3{36}$ (getting a $10$ on first roll), which I will need to add to the rest of the probabilities. However, I am stuck on the latter part. I will appreciate any help. Thank you.
 A: Your case is easier because you do not have to track all the rolls to wait for pairs which might appear. Indeed, every roll is independent of the previous. For a fixed number of rolls you could write this down as a tree (win, continue, lose). The win or lose branches stop the game, the "continue" branch again branches in (win, continue, lose) with the same probabilities. 
Starting with the second question: This shows directly that the number of rolls until the game ends (either win or lose) is geometrically distributed with parameter $p=\frac{3}{36}+\frac{6}{36}=\frac{1}{4}$. The expected value of the geometric distribution is just $p^{-1}$. 
Concerning the first question: To win after $n$ rolls you need to roll anything but a 7 or a 10 exactly $n-1$ times and then roll a 10. So 
$$\mathbb{P}[\text{win after n rolls}] = \left(1-\frac{1}{4}\right)^{n-1}\frac{3}{36}$$ 
and therefore
$$\mathbb{P}[\text{win}]= \mathbb{P}[\exists n\in\mathbb{N}:\text{win after n rolls}]=\sum_{n=1}^{\infty}\mathbb{P}[\text{win after n rolls}]=\sum_{n=1}^{\infty} \left(1-\frac{1}{4}\right)^{n-1}\frac{3}{36}=\frac{12}{36}=\dfrac{1}{3}.$$
I hope that helps.
