Expected flips to see the pattern HTH

Question

You flip a fair coin repeatedly. What is the expected number of flips until you first see the pattern HTH? Follow-up: why is HTH slower than HTT?

Accepted Answer

Use states S0 (no progress), S1 (have "H"), S2 (have "HT"), with expected remaining flips E0, E1, E2. - From S0: H moves to S1, T stays at S0: E0 = 1 + 	frac12 E1 + 	frac12 E0. - From S1 (have "H"): another H keeps you at "H" (S1), T advances to "HT" (S2): E1 = 1 + 	frac12 E1 + 	frac12 E2. - From S2 (have "HT"): H completes "HTH" (done); T gives "...HTT", whose useful suffix is nothing, so back to S0: E2 = 1 + 	frac12 \cdot 0 + 	frac12 E0. Solve: E0 = 2 + E1 and E1 = 2 + E2. With E2 = 1 + 	frac12 E0: E1 = 2 + 1 + 	frac12 E0 = 3 + 	frac12 E0, \qquad E0 = 2 + 3 + 	frac12 E0 \implies \…

Expected flips to see the pattern HTH

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