8. A certain intelligence station has four different passwords $A, B, C, D$, and uses one of them each week. Each week, one of the three passwords not used in the previous week is chosen with equal probability. If the first week uses password $A$, what is the probability that the seventh week also uses password $A$? (Express your answer as a simplified fraction)

Fill in the blank: $\qquad$ .
Solution: $\frac{61}{243}$.
Let $P_{k}$ represent the probability of using the $A$ type codebook in the $k$-th week, then the probability of not using the $A$ type codebook in the $k$-th week is $1-P_{k}$. Therefore, we have $P_{k+1}=\frac{1}{3}\left(1-P_{k}\right), k \in \mathbf{N}^{*}$, which means $P_{k+1}-\frac{1}{4}=-\frac{1}{3}\left(P_{k}-\frac{1}{4}\right)$.

Given $P_{1}=1$, we know that $\left\{P_{k}-\frac{1}{4}\right\}$ is a geometric sequence with the first term $\frac{3}{4}$ and common ratio $-\frac{1}{3}$.
Thus, $P_{k}-\frac{1}{4}=\frac{3}{4}\left(-\frac{1}{3}\right)^{k-1}$, which means $P_{k}=\frac{3}{4}\left(-\frac{1}{3}\right)^{k-1}+\frac{1}{4}$.
Therefore, $P_{7}=\frac{61}{243}$.