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Why HTT Beats TTH Head-to-Head

There are two different questions that are easy to accidentally mix up:

Explanatory diagram for Why HTT Beats TTH Head-to-Head.
Local explanatory diagram
  1. How long do you wait for HTT by itself?
  2. If you race HTT vs TTH, which pattern appears first?

Individually, they are symmetric in an important sense: neither pattern has a proper prefix that is also a suffix, so each has expected waiting time

E=2³=8 flips.

So HTT and TTH do not differ in their individual expected waiting times.

But in a direct race, HTT appears before TTH with probability 3/4.

The reason is overlapping partial progress. Once an H has appeared in the right state, the sequences of future flips that build toward HTT interact differently with the partial matches for TTH. The race is a small finite-state Markov process, not two independent eight-flip timers.

A particularly clean way to see it is to track the longest current suffix that is a prefix of either pattern. Solving the resulting states gives:

P(HTT before TTH)=3/4,
P(TTH before HTT)=1/4.

Big idea: Equal standalone waiting times do not imply equal head-to-head chances. Pattern races depend on how prefixes and suffixes overlap with each other.

This is the same family of phenomenon behind Penney's game, where a carefully chosen length-3 pattern can beat another pattern despite every specific three-flip string having probability 1/8 on any fixed block.

sources

Stanford/standard probability treatment of Penney's game; Wikipedia is acceptable for orientationFeller, *An Introduction to Probability Theory and Its Applications*, pattern waiting/renewal methods.