Probability & Statistics

HHH Before THH

HHH beats THH only when the first three tosses are all heads, probability 1/8. Any earlier tail arms THH, which then must complete before any later HHH can, an absorbing argument requiring no algebra.

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You flip a fair coin until either the pattern HHH or the pattern THH appears as three consecutive tosses. What is the probability that HHH appears first?

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#Where can HHH appear?

Completing HHH means three heads in a row. Look at the toss immediately before the first of those three heads. Either it is the very start of the sequence, or it is a genuine earlier toss. If that earlier toss were itself a head, the run would have started sooner, so slide back to the first head of the maximal run; the toss before that one is again either the start or a tail.

#A tail seeds THH

If a tail sits just before the run of heads, then that tail followed by the first two of those heads already spells THH, which therefore lands before HHH is finished. So HHH can win only when its three heads occupy the very first three tosses.

#Count it

P(HHH first)=P(first three tosses are HHH)=(12)3=18.(1)\PP(\text{HHH first}) = \PP(\text{first three tosses are HHH}) = \left(\tfrac{1}{2}\right)^{3} = \frac{1}{8}. \tag{1}
HHHHHH wins
THHTHH wins firstthe only winning startany earlier tail seeds this
Completing HHH needs two heads in a row, and the toss just before them is either the very start or a tail. A tail there already spells THH, so HHH can win only from the pristine start, with probability one eighth.

The other 78\tfrac{7}{8} go to THH, which needs only a single tail anywhere to arm the run of heads that completes it. The two patterns are far from symmetric.