Two players toss a coin until HTT or TTH appears. The first player wins if there is HTT and the second player wins if there is TTH. What is the expected number of throws in the game given that the first player wins?
I have computed the expected number of throws without any conditions by formula 3+$\sum _{n=3}^{\infty }\:\frac{n-1}{2^n}\cdot n = 6.5$, where n-1 is the number of winning combinations at the nth throw, $2^n$ is the number of all possible combinations at the nth throw and 3 is the minimal obligatory throws to win in this game. By the same logic, I have tried to compute expectation given that the first player wins. I exclude the only way for the second player to win (TT..TH for number of T equal to n-1) and have n-2 in the numerator and also exclude all combinations that can give only second's player win from the denominator. Second player wins if and only if there are TT in the beginning so I subtract $2^{n-2}$ from the denominator to exclude all combinations after TT. And finally I have: 3+$\sum _{n=3}^{\infty }\:\frac{n-2}{2^n-2^{n-2}}\cdot n$ = 3 + 10/3 which is wrong answer as I know. What am I doing wrong?
P.S I know that there is a solution of such task in the form of system of equations on Stack but I am interested in sum-solution of this problem