A race between two patterns

Idea. Only the last few tosses matter, so the game is a Markov chain on the prefixes of the two patterns. One state turns out to be a guaranteed win for A, which makes the system easy to solve.

Step 1: states. Let $p_s$ be the probability that A wins from state $s$, where $s$ is the longest end of the toss sequence that is the start of a pattern: $\varnothing$, $\text{H}$, $\text{HH}$ or $\text{HT}$.

Step 2: the state HH. After $\text{HH}$, a head keeps us in $\text{HH}$ (the last two tosses are again $\text{HH}$), and a tail completes $\text{HHT}$. B can never win from here, because B needs $\text{HT}$ followed by $\text{H}$, and the first $\text{T}$ ends the game in A's favour. So $p_{\text{HH}}=1$.

Step 3: the other states.

• From $\text{HT}$: a head completes $\text{HTH}$ (B wins); a tail leaves nothing useful. So $p_{\text{HT}}=\frac{1}{2}\cdot 0+\frac{1}{2}{p}_{\varnothing }$.
• From $\text{H}$: a head gives $\text{HH}$, a tail gives $\text{HT}$. So $p_{\text{H}}=\frac{1}{2}\cdot 1+\frac{1}{2}{p}_{\text{HT}}$.
• From $\varnothing$: a tail changes nothing, so the game really starts at the first head, and $p_{\varnothing }=p_{\text{H}}$.

Step 4: solve. With $x=p_{\text{H}}$: $p_{\text{HT}}=\frac{x}{2}$, so $x=\frac{1}{2}+\frac{x}{4}$, which gives $x=\frac{2}{3}$.

A wins with probability $\frac{2}{3}$.

Why this is not obvious. On its own, $\text{HHT}$ takes $8$ tosses on average and $\text{HTH}$ takes $10$, but waiting times do not decide a race. What matters here is that B's partial progress ($\text{HT}$) can be wiped out by a tail, while A's partial progress ($\text{HH}$) can never be lost.