2026-04-25

Problem — 2026-04-25

Hard

Hardprobability

A fair coin is flipped repeatedly. What is the expected number of flips until the pattern HTHHTH appears for the first time?

Enter a number

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Use a Markov chain with states: SS (start), HH (last flip was H, no progress beyond that), HTHT (last two flips were HT). State HTHHTH is absorbing. Transitions: from SS, heads → HH, tails → SS; from HH, tails → HTHT, heads → HH; from HTHT, heads → HTHHTH (done), tails → SS. Let ES,EH,EHTE_S, E_H, E_{HT} be expected flips from each state. ES=1+12EH+12ESE_S = 1 + \frac{1}{2}E_H + \frac{1}{2}E_S, EH=1+12EHT+12EHE_H = 1 + \frac{1}{2}E_{HT} + \frac{1}{2}E_H, EHT=1+12(0)+12ESE_{HT} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_S. Solving: EHT=6E_{HT} = 6, EH=8E_H = 8, ES=10E_S = 10.