Tricky Coin Game

For $j=1,2,3$ , let $A_j$ be the event that I win given that I currently have $j$ coins and it is my turn to play. Let $p_j= \mathbb{P}[A_j]$ . We calculate probabilities conditional on the outcome of the next one or two tosses: $\begin{aligned} p_1 & = \; \mathbb{P}[A_1|T]\tfrac12 + \mathbb{P}[A_1|HH]\tfrac14 + \mathbb{P}[A_1|HT]\tfrac14 \; = \; \tfrac14p_1 + \tfrac14p_2 \\ p_2 & = \; \mathbb{P}[A_2|TT]\tfrac14 + \mathbb{P}[A_2|TH]\tfrac14 + \mathbb{P}[A_2|HT]\tfrac14 + \mathbb{P}[A_2|HH]\tfrac14 \; = \; \tfrac14p_1 + \tfrac12p_2 + \tfrac14p_3 \\ p_3 & = \; \mathbb{P}[A_3|TT]\tfrac14 + \mathbb{P}[A_3|TH]\tfrac14 + \mathbb{P}[A_3|HT]\tfrac14 + \mathbb{P}[A_3|HH]\tfrac14 \; = \; \tfrac14p_2 + \tfrac12p_3 + \tfrac14 \end{aligned}$ Solving these equations, $p_1 = \tfrac17$ , $p_2 = \tfrac37$ , $p_3 = \tfrac57$ . Thus the required expected number of won games is $1400p_2 = \boxed{600}$ .