Let $x_{TH}$ be the expected number of remaining throws, given that $TH$ are the last two coins (and that no one has yet won). And same for the other alternatives.
Then we can write the system of recursions
$$\begin{align}
x_{TH}&=1 + \frac{1}{2}x_{HT} + \frac{1}{2}x_{HH}\\
x_{TT}&=1 + \frac{1}{2}x_{TT}\\
x_{HH}&=1+\frac{1}{2}x_{HT} + \frac{1}{2}x_{HH}\\
x_{HT}&=1+ \frac{1}{2}x_{TH}\\
\end{align}
$$
The second equation implies $x_{TT}=2$. The rest are solved by $x_{TH}=x_{HH}=6$ , $x_{HT}=4$.
Then the expected number of throws after the first two throws is $\frac{1}{4}(2+ 6 +6 +4)=9/2$
and the total expected number of throws is $$2+\frac92=\frac{13}{2}=6.5$$