Pattern race

Idea

Track the game state as the longest suffix of the flip sequence that is a prefix of either target pattern. This yields a small Markov chain whose first-step equations determine the win probability exactly.

States and Transitions

Target patterns: Player 1 wins on $HTH$; Player 2 wins on $HHT$.

The transient states are $\{\varepsilon,\, H,\, HH,\, HT\}$, where $\varepsilon$ denotes no useful suffix (start, or after a progress-resetting tail).

State	Flip $H$	Flip $T$
$\varepsilon$	$H$	$\varepsilon$
$H$	$HH$	$HT$
$HH$	$HH$	P2 wins
$HT$	P1 wins	$\varepsilon$

Remark on $HH \xrightarrow{H} HH$: after any run of heads, the longest suffix that prefixes a target is still $HH$ (the length-2 prefix of $HHT$).

System of Equations

Let $p_s$ denote the probability that Player 1 wins from state $s$.

$$p_{\varepsilon} = \tfrac{1}{2}p_H + \tfrac{1}{2}p_{\varepsilon} \implies p_{\varepsilon} = p_H \tag{1}$$ $$p_H = \tfrac{1}{2}p_{HH} + \tfrac{1}{2}p_{HT} \tag{2}$$ $$p_{HH} = \tfrac{1}{2}p_{HH} + \tfrac{1}{2}\cdot 0 \implies p_{HH} = 0 \tag{3}$$ $$p_{HT} = \tfrac{1}{2}\cdot 1 + \tfrac{1}{2}\,p_{\varepsilon} \tag{4}$$

Equation (3) reflects that $HH$ is a trap: every additional $H$ keeps the game in $HH$, and the inevitable first $T$ completes $HHT$, so Player 1 cannot win from $HH$.

Solution

Substituting $(3)$ into $(2)$:

$$p_H = \tfrac{1}{2}p_{HT}.$$

Combined with $(1)$, we have $p_{\varepsilon} = p_H = \tfrac{1}{2}p_{HT}$. Substituting into $(4)$:

$$p_{HT} = \frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}p_{HT} = \frac{1}{2} + \frac{1}{4}p_{HT}.$$ $$\frac{3}{4}p_{HT} = \frac{1}{2} \implies p_{HT} = \frac{2}{3}.$$ $$p_{\varepsilon} = \frac{1}{2}\cdot\frac{2}{3} = \boxed{\dfrac{1}{3}}.$$

The game begins in state $\varepsilon$, so Player 1 wins with probability $\dfrac{1}{3}$.