3 gamblers

三个赌徒聚赌，赌本分别为a,b,c元每轮只有一位赌徒获胜，胜者从另外二人那里各收取一元,赌局持续到某一位赌徒输光为止。假设每轮每位赌徒获胜
概率均为1/3,问期望轮数?

Let An, Bn and Cn be the amount of money the three gamblers hold at time n,

Mn = An * Bn * Cn + n * (a + b + c - 2) 

is a martingale. Let N be the stopping time for the game, we know

AN * BN * CN = 0.

Then based on Doob's and Monotone convergence theorem, we get

EN * (a + b + c - 2) = a * b * c

which implies that 

EN = a * b * c / (a + b + c - 2).

You can generalize this problem to the case where the other two players have
to continue playing. To solve this, you first need to get the probability 
for each player to win and then start from there to derive the final result. 

You can also generalize this problem to a game of more than three players, 
but not sure how far you can go.