15. A and B take turns tossing a fair coin, and whoever tosses heads first wins, at which point the game ends, and the loser gets to toss first in the next game.
(1) Find the probability that the first person to toss wins in any given game;
(2) Suppose they play a total of 10 games, and A tosses first in the first game. Let the probability that A wins the $k$-th game be $P_{k}$. If $1<3^{9}\left(2 P_{k}-1\right)<81$, find the value of $k$.
15. (1) In any given match, the probability of the first person to throw winning is $\frac{1}{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{5}+\cdots=\frac{2}{3}$.
(2) From (1), the probability of the second person to throw winning is $1-\frac{2}{3}=\frac{1}{3}$.

Given $P_{1}=\frac{2}{3}$, for $2 \leqslant k \leqslant 10$, we have $P_{k}=\frac{1}{3} P_{k-1}+\frac{2}{3}\left(1-P_{k-1}\right)=\frac{2}{3}-\frac{1}{3} P_{k-1}$. Therefore, $P_{k}-\frac{1}{2}=-\frac{1}{3}\left(P_{k-1}-\frac{1}{2}\right)$. Hence, $P_{k}=\frac{1}{2}+\frac{(-1)^{k-1}}{3^{k-1}}\left(P_{1}-\frac{1}{2}\right)=\frac{1}{2}+\frac{(-1)^{k-1}}{2 \times 3^{k}}$. Since $1<3^{9}\left(2 P_{k}-1\right)<81$, we have $\frac{1}{2}+\frac{1}{2 \times 3^{9}}<P_{k}<\frac{1}{2}+\frac{1}{2 \times 3^{3}}$. Substituting $P_{k}$ and simplifying, we get $\frac{1}{3^{9}}<\frac{(-1)^{k-1}}{3^{k}}<\frac{1}{3^{5}}$. Therefore, $k$ is odd, and $5<k<9$. Hence, $k=7$.