8. A certain intelligence station has four different passwords $A, B, C, D$, and uses one of them each week, with each week's password being randomly selected with equal probability from the three passwords not used in the previous week. If the first week uses password $A$, then the probability that the seventh week also uses password $A$ is $\qquad$ (expressed as a simplified fraction).


Let the probability of using password $A$ in the $n$-th week be $P_{n}$, then the probability of not using password $A$ in the $n$-th week is $1-P_{n}$. Therefore, $P_{n+1}=\frac{1}{3}\left(1-P_{n}\right)=-\frac{1}{3} P_{n}+\frac{1}{3}$. Suppose $P_{n+1}+\lambda=-\frac{1}{3}\left(P_{n}+\lambda\right) \Rightarrow P_{n+1}=-\frac{1}{3} P_{n}-\frac{4}{3} \lambda$, comparing with $P_{n+1}=-\frac{1}{3} P_{n}+\frac{1}{3}$ we get $\lambda=-\frac{1}{4} \Rightarrow$ the sequence $\left\{P_{n}-\frac{1}{4}\right\}$ is a geometric sequence with a common ratio of $-\frac{1}{3}$.
Since $P_{1}=1 \Rightarrow P_{n}-\frac{1}{4}=\left(P_{1}-\frac{1}{4}\right) \cdot\left(-\frac{1}{3}\right)^{n-1}=\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n-1}$ $\Rightarrow P_{n}=\frac{1}{4}+\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n-1} \Rightarrow P_{7}=\frac{1}{4}+\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{6}=\frac{61}{243}$.