Example 5: A and B take turns rolling a fair 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 players always follow this rule. Then the probability that the $n$-th roll is made by A, $p_{n}=$ $\qquad$ [2]
$(2014$, National High School Mathematics League Shandong Province Preliminary Competition)
【Analysis】The probability of A rolling a 1 and B rolling a 3 is $\frac{1}{6}$ each, and the probability of A not rolling a 1 and B not rolling a 3 is $\frac{5}{6}$ each.

Let the probability of A rolling on the $k$-th turn be $p_{k}$. Then the probability of B rolling is $1-p_{k}$.
Obviously, $p_{1}=1$.
If A rolls first, then the probability of A rolling again on the second turn is $p_{2}=\frac{1}{6}$, and the probability of B rolling is $\frac{5}{6}$, so the probability of A rolling on the $(k+1)$-th turn is
$$
\begin{array}{l}
p_{k+1}=\frac{1}{6} p_{k}+\frac{5}{6}\left(1-p_{k}\right)=\frac{5}{6}-\frac{2}{3} p_{k} \\
\Rightarrow p_{k+1}-\frac{1}{2}=-\frac{2}{3}\left(p_{k}-\frac{1}{2}\right) .
\end{array}
$$

Therefore, $\left\{p_{n}-\frac{1}{2}\right\}$ is a geometric sequence with the first term $p_{1}-\frac{1}{2}=\frac{1}{2}$ and the common ratio $-\frac{2}{3}$.
Hence $p_{n}-\frac{1}{2}=\frac{1}{2}\left(-\frac{2}{3}\right)^{n-1}$ $\Rightarrow p_{n}=\frac{1}{2}+\frac{1}{2}\left(-\frac{2}{3}\right)^{n-1}$.