11. A and B repeatedly play a game. In each game, the players take turns tossing a fair coin, and the first player to toss heads wins. In the first game, A tosses first, and in subsequent games, the loser of the previous game starts first. The probability that A wins the 6th game is $\qquad$.
11. $\frac{1}{2}\left(1-\frac{1}{3^{6}}\right)$ The probability of Player A winning the first time is $\frac{1}{2}\left[1+\frac{1}{4}+\left(\frac{1}{4}\right)^{2}+\cdots\right]=\frac{2}{3}$, it is easy to induct that the probability of Player A winning the $n$-th time is $\frac{\left(1+\frac{(-1)^{n+1}}{3^{n}}\right)}{2}$.