15. A and B take turns tossing a fair coin, and the one who tosses heads first wins. They play several rounds, and it is stipulated that the loser of the previous round tosses first in the next round. If A tosses first in the 1st round, what is the probability that A wins in the 6th round?
15. The probability of the first player to tie winning is
$$
\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}=\frac{2}{3} \text {. }
$$

The probability of the second player to tear winning is $1-\frac{2}{3}=\frac{1}{3}$.
$\{$ Player A wins the $k$-th game $\}=\{$ Player A wins the $(k-1)$-th game and Player A wins the $k$-th game $\} \cup\{$ Player B wins the $(k-1)$-th game and Player A wins the $k$-th game \}.
Let $P_{k}$ be the probability that Player A wins the $k$-th game, then we have
$$
\begin{array}{l}
P_{1}=\frac{2}{3}, \\
P_{k}=\frac{1}{3} P_{k-1}+\frac{2}{3}\left(1-P_{k-1}\right)=\frac{2}{3}-\frac{1}{3} P_{k-1} \\
\Rightarrow P_{k}-\frac{1}{2}=-\frac{1}{3}\left(P_{k-1}-\frac{1}{2}\right) \\
\Rightarrow P_{k}=\frac{1}{2}+\frac{(-1)^{k-1}}{3^{k-1}}\left(P_{1}-\frac{1}{2}\right)=\frac{1}{2}+\frac{(-1)^{k-1}}{2 \times 3^{k}} .
\end{array}
$$

Therefore, $P_{6}=\frac{364}{729}$.
Hence, the probability of Player A winning the 6th game is $\frac{364}{729}$.