Gambler's ruin when the coin is biased against you

Question

You start with \10 and bet \1 per round, but now each round you win \1 with probability p = 0.4 and lose \1 with probability q = 0.6. You stop at \20 or at \0. What is the probability you reach \20 before going broke? Compare it to the fair-coin case.

Accepted Answer

For a biased random walk with win probability p 
e 	frac12, absorbed at 0 and N, the probability of reaching N from bankroll i is P = \frac{1 - r^{\,i}}{1 - r^{\,N}}, \qquad r = \frac{q}{p}. Here r = \dfrac{0.6}{0.4} = 1.5, i = 10, N = 20. Compute the powers: r^{10} = 1.5^{10} \approx 57.67, \qquad r^{20} = (1.5^{10})^2 \approx 3325. Then P = \frac{1 - 57.67}{1 - 3325} = \frac{-56.67}{-3324} \approx 0.017. So you reach \20 only about 1.7\% of the time, despite starting exactly halfway to the target. Compare to the fair coin. With p = 	frac12 the answer would be the linear 	frac{i}{N} = 	…

Gambler's ruin when the coin is biased against you

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