Expected flips to get two heads in a row

Question

You flip a fair coin repeatedly, stopping the moment you see two heads in a row. What is the expected number of flips until you stop?

Accepted Answer

Use two states: A = last flip was not a head (or the start), B = last flip was a head. Let eA, eB be the expected additional flips. Heads has probability 	frac12. eA = 1 + 	frac12 eB + 	frac12 eA, \qquad eB = 1 + 	frac12 (0) + 	frac12 eA. From the first, eA = 2 + eB. Substitute: eB = 1 + 	frac12(2 + eB) = 2 + 	frac12 eB, so eB = 4 and eA = 6. Starting fresh (state A): eA = 6. Takeaway. Two heads in a row averages 6 flips, matching \frac{1+p}{p^2} = \frac{1.5}{0.25} with p = 	frac12, three times the 2 flips needed for a single head.

Expected flips to get two heads in a row

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