Consider the following game: you start with one unit, and in each round, you win one unit with probability p and you loose one unit with probability 1-p.This example of a random walk is listed in basically every book about probability theory, and it is well-known that the ruin probability is < 1 if and only if p > 1/2. The ruin probability is the probability that you reach zero units during the game. But what if the possible gain differs from the possible loss? Specifically, consider the following modification:you start with one unit, and in each round, you win x units with probability p and you loose y units with probability 1-p.Intuitively, if it holds for the expected return per round R = x*p-y*(1-p) that R > 0, then there is a positive drift, and so the ruin probability should be < 1 as well. However, I can not find this example in any book, paper etc. As far as I can see, the proofs for the gain=loss-case can not be straightforward adapted. Thanks for a reply, I need this as an ingredient for a paper.

Are we allowed to assume x,y are integers?

I would agree with you intution, check out Thorp's paper at page 6 where this is mentioned.http://www.bjmath.com/bjmath/thorp/ch2.pdf

First, yes, we can assume that x and y are integers.Concerning the reply of betso:I think that the setting described in the paper you posted is quite different: They assume that a constant fraction of the current fortune is invested in each bet. Hence, there is a "relative" change in each round and not an "absolute" one, i.e., if our current fortune is F and we bet the fraction f, then we have (1+f)*F in case we win and (1-f)*F in case we lose. One could argue that we can move the "absolute movement" to the exponent by writing the current fortune in the form (1+f)^(W - c*L), where W is the number of games we won, L is the number of games we lost and c is some scaling factor with (1-f) = 1/(1+f)^c. However, as they mention, their ruin probability is quite different, since they allow to "break" any lower barrier as long as the probability to come back is big enough. In the classical definition of the gamblers ruin problem, we are dead as soon as we lose all our money. Maybe I am wrong? I am not so familiar with this stuff.

I assume that the gambler is bust if he has negative or zero units. He must also stop playing if his fortune reaches N units (say). Otherwise he goes bust with probability 1. In which case let p_k be the probability he goes bust starting from k units. Then p_k satisfies p_k = pp_{k+x} + (1-p)p_{k-y} with the boundary conditions p_{0}=1, p_{N}=0. So we just have to solve this difference equation. Is that right, or have I missed something?

Concerning the response of pmtpr:Sure it is possible to express this process in terms of such an equation and then let N go to infinity. The following page gives a nice introduction to this topic: http://www.mathpages.com/home/kmath084/kmath084.htm. However, it would be nice to have a general theorem that characterises the parameters with ruin probability < 1. For the gain=loss-case, there is an elementary proof that states that the non-ruin probability is < 1 if and only if p > 1/2. Maybe it is possible to get something like this from some differential equations, but I am wondering if there is not an elementary proof as for the gain=loss case.