Gambler's ruin with a small edge in your favor

Question

You start with \50 and bet \1 repeatedly; each bet wins with probability p = 0.52 and loses with probability q = 0.48, independently. You stop at \100 or at \0. What is the probability you reach \100?

Accepted Answer

The general biased-ruin win probability is hi = \frac{1 - (q/p)^i}{1 - (q/p)^N}, derived from hi = p\,h{i+1} + q\,h{i-1} with roots x = 1 and x = q/p. Numbers. Here r = q/p = 0.48/0.52 = 12/13 \approx 0.9231, and i = 50, N = 100. Since r < 1, r^{50} = e^{50\ln(0.9231)} \approx e^{-4.00} \approx 0.0183, \qquad r^{100} = (r^{50})^2 \approx 0.000334. Then h{50} = \frac{1 - 0.0183}{1 - 0.000334} = \frac{0.9817}{0.99967} \approx 0.982. So with a 2\% per-bet edge in your favor, your chance of doubling 50 into 100 is about 98%, up from the fair-coin 50\%. Compare the mirror-image case (a 1\% edge ag…

Gambler's ruin with a small edge in your favor

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