Two heads before three tails in a row

Question

You flip a fair coin repeatedly until one of two things happens first: you see two heads in a row (you win) or three tails in a row (you lose).

What is the probability you win, meaning two heads in a row appear before three tails in a row?

Accepted Answer

After any flip you are in a "current run" state. Because a head wipes out tail progress and a tail wipes out head progress, the only states are: H1 (one head so far), T1 (one tail), T2 (two tails). Reaching a second head wins; a third tail loses. Let a, b, c be the win probabilities from H1, T1, T2. a = 	frac12(1) + 	frac12 b, \qquad b = 	frac12 a + 	frac12 c, \qquad c = 	frac12 a + 	frac12(0). From the third, c = 	frac12 a. Then b = 	frac12 a + 	frac12 \cdot 	frac12 a = 	frac34 a. Then a = 	frac12 + 	frac12 \cdot 	frac34 a = 	frac12 + 	frac38 a, so 	frac58 a = 	frac12 and a…

Two heads before three tails in a row

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