Answer

Draw a tree. Imagine after 1st flip you get H. If on the 2nd flip we get another H, the game stops with an even number of flips (success). Otherwise, we get a T and the game continues. If 3rd flip amounts to another T the game stops with an odd number of flips (loss). However, if 3rd flip is a H, we have a sequence HTH - so if next flip is same as previous we end the game with an even number of flips. Notice this is the same situation we were in after we just flipped the first H: one more of the previous flip would have resulted in a success. In turn, the probabiliy of success (even flips before the first HH or TT) after flipping H is the same as after flippoing HTH. Denoting this P, we get $P = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \cdot P$ and so $P = \frac{2}{3}$. The story is the same had we first flipped a T instead of H, so the probability of winning after the first flip T is also $P = \frac{2}{3}$.