Three-Handed Gambler’s Ruin - end of the game Three players start with $a,b$, and $c$ chips, respectively, and play the following game. At each stage, two players are picked at random, and one of those two is picked at random to give the other a chip. This continues until one of the three is out of chips, and quits the game; the other two continue until one player has all the chips.
Okay so let us consider the following stopping time: $T$- end of the game, the moment of time, when one of the players has all the chips.
How to show that $ET<\infty$? In other words: how do we know that game will end in some time?
 A: Assuming you know the result for the two-player game, it follows quickly.  From the point of view of player A, he face as coalition of B and C; he wants to get all their chips and he doesn't acre how they transfer chips among themselves.
This game almost surely ends in finite time.  Then either A has won, and then game is over, or B and C play as two-person game which again almost surely ends in finite time.
The result extends in the same way to the $n$-person game.
EDIT
For the two-person game, suppose there are $n$ chips in play, and let $E_k$ be the expected time till the end of the game if player A has $k$ chips.  We have the boundary conditions $$E_0=E_n=0$$ and the recurrence relation $$E_k=1+\frac12E_{k+1}+\frac12E_{k-1},\ k=1,2,\dots,n-1\tag1$$ because we always have to make one play, and A gains or loses a chip with equal probability.  $(1)$ is a second-order linear recurrence relation with constant coefficients, and may be solved by standard methods.  One can easily check that the solution is $$E_k=k(n-k),\ 0\leq k\leq n$$
One can extend this result to the three-person game by first computing the probability that the player with $k$ chips wins the two-person game in a manner similar to the way we computed the expectation of the length of the game.
