Fair chip game, reaching the jackpot before broke

Question

You sit down with 15 chips. Each round is a fair game: you either win 1 chip or lose 1 chip, each with probability $1/2$ , independently. You keep playing until you either build up to 60 chips (jackpot) or drop to 0 chips (broke).

What is the probability you reach 60 chips before going broke?

Accepted Answer

Let hi be the probability of reaching the target N = 60 before hitting 0, starting from i chips. The two ends are pinned down: h0 = 0 (already broke) and h{60} = 1 (already won). After one round you are equally likely to be at i+1 or i-1, so hi = 	frac12 h{i+1} + 	frac12 h{i-1}. Rearranging, h{i+1} - hi = hi - h{i-1}: every rung of the ladder shifts the probability by the same amount. A sequence with constant gaps is a straight line, so hi = i/N. Plugging in, h{15} = \frac{15}{60} = \frac14. Another way to see it: a fair game leaves your average chip count unchanged, so your expected final …

Fair chip game, reaching the jackpot before broke

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