7. Two people, $A$ and $B$, are playing a dice-throwing game with two dice, following these rules: If the sum of the numbers rolled is a multiple of 3, the person who rolled the dice continues to roll; if the sum of the numbers rolled is not a multiple of 3, the other person takes the next turn. The first roll is made by $A$. If the probability that the $n$-th roll is made by $A$ is $P_{n}$, then $P_{n}=$ $\qquad$
7. $\frac{1}{2}+\frac{1}{2}\left(-\frac{1}{3}\right)^{n-1}$ Detailed explanation: It is easy to know that the probability of the sum of the dice being a multiple of 3 is $P=\frac{12}{36}=\frac{1}{3}$. The event "the $(n+1)$-th time is controlled by $A$" includes the event "the $n$-th time is controlled by $A$, and the $(n+1)$-th time continues to be controlled by $A$" and the event "the $n$-th time is controlled by $B$, and the $(n+1)$-th time is controlled by $A$". The probabilities of these two events are $\frac{1}{3} P_{n}$ and $\left(1-\frac{1}{3}\right)\left(1-P_{n}\right) \cdot P_{n+1}=\frac{1}{3} P_{n}+$
$$
\begin{aligned}
\left(1-\frac{1}{3}\right)\left(1-P_{n}\right) & =-\frac{1}{3} P_{n}+\frac{2}{3}, \\
\therefore P_{n+1}-\frac{1}{2} & =-\frac{1}{3}\left(P_{n}-\frac{1}{2}\right) . \text { Since } P_{1}=1, \therefore P_{n}=\frac{1}{2}+\frac{1}{2}\left(-\frac{1}{3}\right)^{n-1} .
\end{aligned}
$$