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:
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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.
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