The Last Man Standing This is my second question following this post.

Three players are playing a game. They all have small amounts of
  money, let say: player 1 has $\$a$, player 2 has $\$b$, and player 3
  has $\$c$, where $a<b<c$. The probability of each player wins each
  turn of the game is $p$ for player 1, $q$ for player 2, $r$ for player
  3, and $s$ for having draw, where $p+q+r+s=1$. The losers will transfer a dollar ($\$1$) to the winner for each turn. The game ends until one
  player has all the money. What is the probability of each player going
  bankrupt? What is the expected number of turns so that only one player
  left as the winner?

Suppose that they play blackjack, if player 1 gets 20 points, player 2 gets 19 points, and player 3 gets 18 points, then the winner of that turn is player 1, so the other two players must pay a dollar to the player 1. If there are two players get, for example, 19 points and the another player gets 18 points, then that turn is considered draw. If they all get 19 points, this is also considered draw. If one player loses all the money, then he will stop playing and only two player will continue the game with probability of winning for each player is $x$ and $y$, also the probability of draw is $z$. Each turn will be repeated until one player has all the money.
To be honest, I can't answer this question and I really don't get it. I left my answer sheet totally empty for this one. (─‿‿─)
Please help me to answer this question and provide a simple explanation about the answer you submit. Every answer would be greatly appreciated.
 A: For fixed, $p,q,r,s$, let $f(a,b,c)$ denote the probability that player 1 is the first to go bankrupt, say. Then we have the recursion
$$f\tag1(a,b,c)=pf(a+2,b-1,c-1)+qf(a-1,b+2,c-1)+\\+rf(a-1,b-1,c+2)+sf(a,b,c) $$
and boundary conditions $f(0,b,c)=1$, $f(a,0,c)=f(a,b,0)=0$ for $a,b,c>0$. 
We must also add $f(a,0,0)=0$, $f(0,b,0)=f(0,0,c)=1$ to account for the possibility that two players get bankrupt together (whereas $f(0,0,0)$ is undefined). Note that the same question for player 2 and player 3 results in the same recurson $(1)$, but different boundary conditions. Also note that $(1) $ can be transformed to 
$$\tag2f(a,b,c)=p'f(a+2,b-1,c-1)+q'f(a-1,b+2,c-1)+r'f(a-1,b-1,c+2) $$
with $p'=\frac p{1-s}$ etc. (as was suggested in the comments).
For concrete values, it is enough to use $(2)$ to climb from the boundary cases to the specific values, for example $f(1,1,1)=q'+r'$ and $f(a,1,1)=0$ for $a>1$. For the general (i.e. abstract) solution, it woul dreally be helpful to understand Markov chains, eigenvalues and stuff.
A: For starters, you need to be more specific about that probability of a draw. It's possible that the top two players tie and the bottom-ranked player loses, and it's also possible that all $3$ players tie. Your draw probability doesn't distinguish these different cases, and yet these different cases will affect the result. You need to say how the money is divided when two or more top players tie, and also specify what is the probability that an arbitrary pair of two players tie for the top, and what is the probability that all three players tie for the top. Otherwise the problem is ill-posed.
