Gambler's ruin, reach 80 from 30, and how long it takes

Question

You start with \30 and repeatedly bet \1 on a fair coin flip (win \1 or lose \1, each with probability 	frac12). You stop when you reach \80 or go broke. What is the probability you reach \80 before hitting \0? And how many bets do you expect to make before stopping?

Accepted Answer

Probability of success. With fair bets your bankroll St is a martingale, so optional stopping gives E[ST] = S0 = 30. At stopping, ST is 80 with probability p and 0 otherwise: 80p + 0(1 - p) = 30 \implies p = \frac{30}{80} = \frac{3}{8}. Equivalently, for a fair game the success probability is just your fraction of the distance to the target: 	frac{	ext{start}}{	ext{target}} = 	frac{30}{80} = 	frac38. You go broke with probability 	frac58. Expected number of bets. For a fair random walk absorbed at 0 and N, the expected duration starting from i is E[T] = i\,(N - i). (This solves Di = 1 +…

Gambler's ruin, reach 80 from 30, and how long it takes

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