Probability of Gambler's Ruin with Unequal Gain/Loss I've spent some time reading other questions about Gambler's Ruin, but couldn't find the answer I was looking for. In most questions, it is assumed the Gambler wins \$1 or loses \$1. I'm curious how one could approach the problem with a payoff or loss that is not equal.
For example, what is the probability of ruin on a gambler who starts with \$1000.00 who wins \$41.00 with probability 0.6 and loses \$43.00 with probability 0.4? Is it possible to generalize such a solution?
Cheers,
Josh
 A: The following paper treats your question: Gambler’s ruin probability - a general formula by Guy Katriel
In this paper you will find a formula giving the exact probability of ruin for any payoff distribution. This formula
is expressed in terms of the solutions in the complex unit disk of the equation p(z)=1, where p(z) is the generating function of the payoff distribution. In the example you mention 
the function p(z) would be p(z)=0.6*z^41+0.4*z^(-43). The equation p(z)=1 will have
43 solutions in the unit disk of the complex plane (out of 84 solutions in all, since it is equivalent to a polynomial equation of degree 84), and the ruin probability can be 
expressed in terms of these solutions by the formula given in the paper. Finding
these roots will require numerical computation using a mathematical software (for example wolfram alpha will do it).
A: Perhaps you can solve it using a Markov process with infinitely many states


*

*state 0 = ruined 

*state 1 = \$0

*state 2 = \$1.00 left  

*state 3 = \$2.00 left


The transition matrix is something like
$$ \begin{bmatrix}1 & 0.4 & 0.4 & ... & 0 & ...\\ 0 & 0.4 & 0.4 & ... & 0 & ...\\ ... & ... & ... & ... & ... & ...\end{bmatrix}$$
The initial state is
$\begin{bmatrix} 0\\ 0\\ ...\\ 1\\... \end{bmatrix}$
Now you should try to compute
$$ \lim_{n \to \infty} \begin{bmatrix}1 & 0.4 & 0.4 & ... & 0 & ...\\ 0 & 0.4 & 0.4 & ... & 0 & ...\\ ... & ... & ... & ... & ... & ...\end{bmatrix}^n \cdot \begin{bmatrix} 0\\ 0\\ ...\\ 1\\... \end{bmatrix}$$
The first element in the resulting column vector is the solution. But it will require a lot of effort.
