Three heads before four tails in a row

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You flip a fair coin until either three heads in a row appear (you win) or four tails in a row appear (you lose). What is the probability you win?

Accepted Answer

Current-run states: H1, H2 (head runs) and T1, T2, T3 (tail runs). A third head wins; a fourth tail loses. Let a1, a2 be win probabilities from the head states and b1, b2, b3 from the tail states. a2 = 	frac12(1) + 	frac12 b1, \quad a1 = 	frac12 a2 + 	frac12 b1, b1 = 	frac12 a1 + 	frac12 b2, \quad b2 = 	frac12 a1 + 	frac12 b3, \quad b3 = 	frac12 a1 + 	frac12(0). From the bottom: b3 = 	frac12 a1, b2 = 	frac12 a1 + 	frac14 a1 = 	frac34 a1, b1 = 	frac12 a1 + 	frac12 \cdot 	frac34 a1 = 	frac78 a1. Then a2 = 	frac12 + 	frac12 \cdot 	frac78 a1 = 	frac12 + 	frac{7}{16} a1, and…

Three heads before four tails in a row

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