Like all existing gamble's ruin problem, assume P(i) represents the probability that the gambler wins N dollars given that his current wealth is i dollars (he has i dollars at the moment). In most of the existing solutions, it is assumed that the gambler bets on 1 dollars. It means, if the gambler wins the current step, his state is changed to (i+1) and if he loses, the state is changed to (i-1). In this case, P(i) is calculated using the iterative equation below: P(i+1) - P(i) = (q/p)(P(i) - P(i-1)) where p is the probability of winning in each single game and q = 1-p. Moreover, P(0) = 0 and P(N) = 1
Here I have a little different assumption. In a single gamble, assume the gambler wins 2 dollars with the probability of p and loses 1 dollar with the probability of q = 1-p (I just changed the bet in case he wins). In this especial case, if the gambler wins, then the current state i is changed to i+2. If the gambler loses, the current state is changed to i-1.
Does anybody know how to calculate P(i) in such an especial case? I have tried the mentioned iterative equation like below: P(i+2) - P(i) = (q/p)(P(i) - P(i-1)) But I cannot achieve a general solution for P(i). The result through the iterative method will be quite complicated. Using the iterative solution, you should calculate P(i) according to P(i-1). Then, since P(0) = 0, P(1) is calculated easily and other P(i) would be appeared. But, I could not find a general equation to calculate P(i) directly. Please note that P(N+1) = P(N) = 1 and P(0) = 0