Four heads before two tails in a row

Question

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

Accepted Answer

Current-run states: H1, H2, H3 (one, two, three heads) and T1 (one tail). A fourth head wins; a second tail loses. Let a1, a2, a3 be win probabilities from the head states and b from T1. a3 = 	frac12(1) + 	frac12 b, \quad a2 = 	frac12 a3 + 	frac12 b, \quad a1 = 	frac12 a2 + 	frac12 b, \quad b = 	frac12 a1 + 	frac12(0). So b = 	frac12 a1. Then a3 = 	frac12 + 	frac12 b = 	frac12 + 	frac14 a1, a2 = 	frac12 a3 + 	frac12 b = 	frac14 + 	frac18 a1 + 	frac14 a1 = 	frac14 + 	frac38 a1, a1 = 	frac12 a2 + 	frac12 b = 	frac18 + 	frac{3}{16} a1 + 	frac14 a1 = 	frac18 + 	frac{7}{…

Four heads before two tails in a row

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