= 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> gives
$$
E_0=1+\tfrac12E_0+\tfrac12E_1,\qquad E_1=1+\tfrac12E_0.
$$
Solving gives $E_1=4$ and \b[$\mathbb E[F]=E_0=6$]. 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 $\mathbb P(F>2m)\leq(3/4)^m$. 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 $1/4$.
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