A gambler starts with an initial fortune of i dollars. On each successive game, the gambler wins $1 with probability p, 0<p<1, or loss $1 with probability q=1-p. He will stop if he either accumulates N dollars or loses all his money. What is the probability that he will end up with N dollars?Assume at certain stage, the gambler gets n dollar, then it is easy to prove that when p is not equal to q, (q/p)^n is a martingale. Assume the probability to get N dollar is X, then(q/p)^i=X*(q/p)^N+(1-X)*(q/p)^0 Equation (1)So, X = [1- (q/p)^i]/[1 - (q/p)^N].Can we use Equation (1) to solve X?Also if p = q, can we use the martingale approach?Thank you in advance.

A way to solve directly the problem.You have P(0)=0 and P(N)=1.P(i)=p*P(i+1)+q*P(i-1).-> p*[P(i+1)-P(i)]=q*[P(i)-P(i-1)]Call A(i)=P(i+1)-P(i)So A(i)=q/p*A(i-1) and Sum(i=0 to N-1)A(i)=1.If q/p=x<>1 -> A(i)=(1-x)/(1-x^N)*x^i If q/p=x=1 -> A(i)=1/N So P(i)=(1-x^i)/(1-x^N) what you have found if x<>1 and P(i)=i/N if x=1

EdisonCruise, yes, eq (1) works. and if p=q, just use n which is martingale, i.e., i = X*N + (1-X)*0 with your notation.one follow-up question, is it possible to write n as the limiting case of some equivalent form of (q/p)^n?

Not sure how to do that, but at least we have [1- (q/p)^i]/[1 - (q/p)^N] --> i/N as p --> q.

ymous, the martingale could be[$]\displaystyle \sum_{m=0}^{n-1}\left(\frac{q}{p}\right)^m[$],which is [1-(q/p)^n]/[1-(q/p)] when p!=q and simply n when p=q

Nice. The answer was right there!