HHT Before HTT

Neither pattern can be completed until a head has appeared. So any initial tails can be ignored, and we may start in the state “last flip $H$, no further progress”. Track the relevant suffix of the sequence:

• $S_H$: the last flip is $H$, and it is not preceded by a useful $H$;
• $S_{HH}$: the last two flips are $HH$;
• $S_{HT}$: the last two flips are $HT$, following a reset or a start.

Let $a$, $b$, $c$ be the probabilities that $HHT$ appears first from $S_H$, $S_{HH}$ and $S_{HT}$ respectively.

From $S_{HH}$, a head leaves us in $S_{HH}$ and a tail completes $HHT$. Two consecutive tails can never occur from here without first producing $HHT$, so $b=1$.

From $S_H$, a head moves to $S_{HH}$ and a tail moves to $S_{HT}$:

From $S_{HT}$, a tail completes $HTT$, a loss. A head gives the suffix $HTH$. Its last $H$ is a fresh start, because neither pattern can use the preceding $HT$, so we return to $S_H$:

Substituting gives $a = \tfrac12 + \tfrac14 a$, so $a=\tfrac23$.

Key idea. Once $HH$ appears, $HHT$ is certain to win. Thus $HHT$ wins exactly when $HH$ occurs before $HTT$.