What is the probability of getting the sequence HHT before HTH when flipping a fair coin repeatedly?

Free worked solution · Quant Blueprint

2/3.

1. Track the longest suffix that is a prefix of either target: empty, H, HH or HT. Let their probabilities of eventually winning with HHT be p₀,pH,pHH,pHT.
2. T from empty stays empty, so p₀ = pH. From H, a head moves to HH and a tail to HT, so pH = (pHH+pHT)/2.
3. From HH, a tail completes HHT and a head leaves suffix HH. Thus pHH = (1+pHH)/2 = 1.
4. From HT, a head completes HTH and loses; a tail resets to empty. Hence pHT = p₀/2. Substitute: p₀ = (1+p₀/2)/2, yielding p₀ = 2/3.

Common mistake

Equal pattern lengths and equal single-window probabilities do not make their first-occurrence probabilities equal. Overlapping prefixes determine the race.

Follow-up

The reverse race has probability P(HTH before HHT) = 1/3. Neither pattern can complete simultaneously, and eventually one completes with probability one.

Original educational explanation for this practice question. Reviewed 2026-10-13. Firm-style practice; the label describes the collection’s preparation context.