Wait for HH vs HT

The problem
flip a fair coin until you see…Game A:HHGame B:HTwhich game ends sooner on average?

A fair coin is flipped repeatedly.

  • Game A: stop the first time you see HH (two heads in a row).
  • Game B: stop the first time you see HT (heads then tails).

Which game ends sooner on average? Are they the same?

Tempting (but wrong)
P(HH on next 2 flips) = ¼ = P(HT)HH¼HT¼⇒ same expected wait? ✗

The patterns HH and HT are equally likely on any specific pair of consecutive flips — each occurs with probability 14\tfrac{1}{4}. So both games should have the same expected stopping time. Right?

This is the "frequency equals waiting time" intuition. It's wrong.