Expected flips to two heads in a row with a biased coin

Question

You flip a biased coin that lands heads with probability p = 	frac13 (tails with probability 	frac23). What is the expected number of flips until you first see two heads in a row (HH)?

Accepted Answer

Use states S0 (last flip not a head, or start) and S1 (last flip a head), with expected remaining flips E0, E1. A head has probability p, a tail 1 - p. - From S0: head moves to S1, tail stays: E0 = 1 + p E1 + (1-p) E0. - From S1: head finishes, tail returns to S0: E1 = 1 + p \cdot 0 + (1-p) E0. The first equation gives p E0 = 1 + p E1, so E0 = 	frac1p + E1. Substituting E1 = 1 + (1-p)E0: E0 = \frac1p + 1 + (1-p)E0 \implies p E0 = \frac1p + 1 \implies E0 = \frac{1 + p}{p^2}. Plug in p = 	frac13: E0 = \frac{1 + 	frac13}{(	frac13)^2} = \frac{	frac43}{	frac19} = \frac43 \cdot 9 = 12. So 12 …

Expected flips to two heads in a row with a biased coin

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