Having first solved this with the meanings of p & q interchanged (since p is traditionally the prob. of stepping to the right in a random walk on Z), I flubbed the denominator. The following is corrected: ------------------------------------------------------------------------ Bill Cordwell asks for a derivation of the probability that a random walk on Z, starting from 0, ever reaches a negative number, where the probability of any one step's going to the left = p. Let L be the desired probability: that the walk ever reaches -1. Then L = p + q*L^2 (q = 1-p), since if the first step is to the right, the probability of then ever reaching -1 requires first ever reaching 0 (prob = L by translation-invariance) and then ever reaching -1 from there (L again). Which gives L = (1+-sqrt(1-4pq)) / 2q. The + sign would give probabilities > 1, so the answer must be L = (1-sqrt(1-4pq)) / 2q. ------------------------------------------------------------------------