 Solution

Let $E_0$ be the expected additional flips when there is no immediately preceding tail, and $E_1$ the expectation when the preceding flip is a tail. A one-step first-step analysis givesSolving gives $E_1=4$ and $\mathbb{E}[F]=E_0=6$. To justify finiteness before manipulating these expectations, split the flips into independent two-flip blocks. Each block has probability $1/4$ of being $TT$; hence $\mathbb{P}(F>2m)\le (3/4{)}^m$. This geometric tail bound from a uniform escape probability makes the expectation finite. The state $E_1$ is essential because successive possible pairs overlap: the waiting time is not itself a geometric distribution with parameter $1/4$.