A coin shows heads with probability 2/3 on each independent toss. It is tossed repeatedly until the pattern H, T, H (heads, tails, heads in…

Set up a Markov chain on the progress towards HTH: state $0$ (no progress), state $1$ (last toss H) and state $2$ (last two tosses HT). Let $E_i$ be the expected remaining tosses with $p=\frac23$, $q=\frac13$. Then $E_0=1+pE_1+qE_0$, $E_1=1+pE_1+qE_2$ and $E_2=1+q E_0$ (a head completes the pattern, a tail resets to state $0$). From the first, $E_0=\frac{1}{p}+E_1=\frac32+E_1$. Substituting $E_1=E_0-\frac32$ in the second: $q E_1=1+qE_2$ so $E_1=3+E_2$. With $E_2=1+\frac{E_0}3$ this gives $E_0-\frac32=4+\frac{E_0}3$, hence $E_0=\frac{33}4=8.25$. This agrees with the formula $\frac1{p^2q}+\frac1p=6.75+1.5$.