A fair coin is tossed repeatedly until the three consecutive outcomes head, tail, tail (in this order) appear for the first time. The expected number…

Set up a Markov chain on the useful progress. Let $E$ be the expected remaining tosses from the start (no progress), $H$ the expectation after the latest toss is a head that could begin the pattern, and $HT$ after the last two tosses read head, tail. Then $E=1+\tfrac12H+\tfrac12E$, $H=1+\tfrac12H+\tfrac12HT$ (a head keeps us at $H$, a tail moves to $HT$), and $HT=1+\tfrac12H+\tfrac12\cdot0$ (a tail completes the pattern, a head returns us to $H$). From the third equation $HT=1+\tfrac H2$. Substituting into the second gives $H=1+\tfrac H2+\tfrac12+\tfrac H4$, so $\tfrac H4=\tfrac32$ and $H=6$. Then $E=2+H=8$.