Bill Cordwell asks: << I'm looking for a simple derivation of the probability that a random walk starting at 0, with unit steps, will ever end up to the left of 0. Probability that the person goes to the left (towards the negative) at any step is p.

...

Here's a stab at it: 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))/(2p). The + sign would give probabilities > 1, so the answer must be L = (1-sqrt(1-4pq))/(2p). --Dan