Flips until two heads in a row on a lucky coin

Question

A biased coin lands heads with probability 2/3 on each flip. You flip until you see 2 heads in a row; any tail resets your current head streak to zero. On average, how many flips does that take?

Accepted Answer

Let e0 and e1 be the expected remaining flips with a current head-streak of 0 or 1, and e2 = 0. Each flip is a head with probability p = 2/3 (advance the streak) or a tail with probability 1/3 (crash to 0): e0 = 1 + 	frac23 e1 + 	frac13 e0, \qquad e1 = 1 + 	frac23 e2 + 	frac13 e0 = 1 + 	frac13 e0. Rearrange the first equation: e0 - 	frac13 e0 = 1 + 	frac23 e1, so 	frac23 e0 = 1 + 	frac23 e1, which gives e0 = 	frac32 + e1. Now substitute e1 = 1 + 	frac13 e0: e0 = 	frac32 + 1 + 	frac13 e0 = 	frac52 + 	frac13 e0 \;\Longrightarrow\; 	frac23 e0 = 	frac52 \;\Longrightarrow\; e0 = \…

Flips until two heads in a row on a lucky coin

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