8. Two people, A and B, take turns rolling a die, with A starting first. The rules are: if A rolls a 1, A continues to roll; otherwise, B rolls. If B rolls a 3, B continues to roll; otherwise, A rolls. The two always follow this rule. What is the probability $p_{n}$ that the $n$-th roll is made by A? $p_{n}=$ $\qquad$
8. $\frac{1}{2}-\frac{1}{3}\left(-\frac{2}{3}\right)^{n-2}$.

If Player A rolls a 1 (Player B rolls a 3), the probability is $\frac{1}{6}$; if Player A does not roll a 1 (Player B does not roll a 3), the probability is $\frac{5}{6}$.

Let the probability that Player A rolls on the $k$-th turn be $p_{k}$. Then the probability that Player B rolls is $1-p_{k}$.

The first roll is by Player A, so the probability that Player A rolls on the second turn is $p_{2}=\frac{1}{6}$. Thus, the probability that Player A rolls on the $(k+1)$-th turn is
$$
\frac{1}{6} p_{k}+\frac{5}{6}\left(1-p_{k}\right)=\frac{5}{6}-\frac{2}{3} p_{k},
$$

i.e., $p_{k+1}=\frac{5}{6}-\frac{2}{3} p_{k}$.
Since $p_{2}=\frac{1}{6}$, we have
$$
p_{k+1}-\frac{1}{2}=-\frac{2}{3}\left(p_{k}-\frac{1}{2}\right) \text {. }
$$

Therefore, $p_{n}=\frac{1}{2}-\frac{1}{3}\left(-\frac{2}{3}\right)^{n-2}$.