A coin with P(head)=3/5 is tossed repeatedly until the pattern head, tail, head (HTH) appears for the first time as three consecutive tosses. Find…

Set up a Markov chain with states by progress: $E_0$ (no useful suffix), $E_1$ (last toss H), $E_2$ (last two tosses HT). With $p=\frac35$ and $q=\frac25$: $E_0=1+pE_1+qE_0$, $E_1=1+pE_1+qE_2$, and $E_2=1+p\cdot0+qE_0$ because a head completes HTH while a tail returns to the start. From the first, $E_0=\frac1p+E_1$. From the third, $E_2=1+qE_0$. Putting into the second: $qE_1=1+qE_2$, so $E_1=\frac1q+E_2=\frac1q+1+qE_0$. Combine: $E_0=\frac1p+\frac1q+1+qE_0$, giving $E_0=\dfrac{1/p+1/q+1}{1-q}=\dfrac{1/p+1/q+1}{p}$, which simplifies to $\frac{1}{p^2q}+\frac1p=\frac{125}{18}+\frac53\approx8.6111$, i.e. $8.61$. The same result follows from the autocorrelation formula: the pattern HTH overlaps itself at lengths 1 and 3, so the expected time is the sum of the reciprocals of the probabilities of those prefixes, namely $\frac{1}{p^2q}+\frac1p$, which equals $8.61$ after rounding.