In a craps game, is the expected number of rolls to win greater for the house or the player? In a craps game two fair six sided die are rolled. If we have that the sum of the two dice are $2$, $3$ or $12$, the player loses to the house. However, if the player rolls a $7$ or $11$, the player wins. But if the player rolls another number $y$, the player has to roll the dice until the sum is either $y$ or $7$. In the case of $y$, the player wins, and in the case of $7$, the house wins. Once either the house or player wins, the game ends. 
I am trying to determine whether the expected number of rolls given that the player wins is less than the expected number of rolls given that the house wins. I saw this problem in a book on Markov Chains and it uses the optimal stopping theorem. However, I am not quite sure how I can do this at the undergraduate level because my expected value is an infinite sum that doesn't converge. Could anyone give me some guidance on how to approach this problem? Thank you.
 A: The geometric series should converge.
To win the point, you need to re-roll your y before a 7. This can be conditioned, but suppose we have 3 events:A: roll y, B: roll 7, C: other (hence re-roll)
We can assign probabilities to them.
So P(A before B) = PA + P(C)P(A) + CCA + CCCA + ...  (etc, these are all prob)
this becomes A(1 + C + CC + CCC + ...)
this is a geometric series.
A* Sum (i from 0 to infinity) of (C^i)
Since P(C)<1 it falls within radius of convergence
so you get: P(A)(1/(1-P(C))
Since C is the event of everything not A or B
P(C)= 1 - (P(A) + P(B))    ... i.e the complement.
So you get P(win point) =   A / (A + B).
From here you should be able to show that there is an advantage to the house.
i.e. P(win pass line bet) = p7 + p11 + Sum (i all point values) [pi*(pi/(pi+p7))].
A: This reads like a very simple probability question so I'd break it down?
How many outcomes are there for each value from 2 to 12 and how likely is each?
How many of those outcomes result in you winning?
How many of those outcomes result in the house-winning?
Well, I'm not familiar with craps but from what you just said.
I know the ways of rolling
2 is 1
3 is 2
4 is 3
5 is 4
6 is 5
7 is 6
8 is 5
9 is 4
10 is 3
11 is 2
and 12 is 1 
Therefore, from what I infer, we know there is 1+2+3+4+5+6+5+4+3+2+1 possible outcomes, i.e 6^2 as a fair die has 6 equally probable sides it may land on and there's two dice.
You should be able to work it out from here, but I'll help:
There's only 4 ways to roll 2,3 or 12 and there is 8 ways to roll 7 or 11 (this shouldn't come a surprise). Since I don't know the rules of craps but I know the answer should be quite trivial to work out. [Maybe this would give you some clues][1] (although it's not the maths itself, I figure maybe you'd like to work that out).
[1]: http://wizardofodds.com/games/craps/number-of-rolls/ Maybe that helps?
