8. Given $a_{1}, a_{2}, \cdots, a_{n}$ these $n$ people correspond to $A_{1}, A_{2}, \cdots, A_{n}$ these $n$ cards $\left(n \geqslant 2, a_{i}\right.$ corresponds to $\left.A_{i}\right)$. Now $a_{1}$ randomly picks a card from the deck, and then each person takes a card in sequence. When taking a card, if the deck contains their corresponding card, they take that card. Otherwise, they take any card. Then the probability that $a_{n}$ can get $A_{n}$ is $\qquad$ .
8. $\frac{1}{2}$.

Let the probability be $P_{n}$.
When $n=2$, $\left\{\begin{array}{l}a_{1} \rightarrow A_{1}, a_{2} \rightarrow A_{2}, \\ a_{1} \rightarrow A_{2}, a_{2} \rightarrow A_{1},\end{array}\right.$ $P_{2}=\frac{1}{2}$ $P_{3}=\frac{1}{2}$.

Thus, conjecture $P_{n}=\frac{1}{2}$.
It is obvious that it holds for $n=2,3$.
Assume it holds for $n \leqslant k(k \geqslant 3)$.
If $a_{1} \rightarrow A_{1}, a_{k} \rightarrow A_{k}, P=1-\frac{1}{k}$,
$$
a_{1} \rightarrow A_{k}, a_{k} \nrightarrow A_{k}, P=0,
$$

then $a_{1} \rightarrow A_{m}(m \neq 1,2, k)$.
For $1, m, m+1, \cdots, k$, apply the induction hypothesis (1) to get
$$
P=\frac{1}{2} \cdot \frac{k-3}{k} \text {. }
$$

Thus, $a_{1} \rightarrow A_{2}$,
$$
\begin{array}{l}
P=\frac{1}{2}-\frac{1}{k}-\frac{1}{2} \cdot \frac{k-3}{k}=\frac{1}{2 k} ; \\
a_{1} \nrightarrow A_{2}, \\
P=\frac{1}{2}-\frac{1}{2 k}=\frac{k-1}{2 k} .
\end{array}
$$
When $n=k+1$,
If $a_{1} \rightarrow A_{1}, a_{k+1} \rightarrow A_{k+1}$,
$P=1-\frac{1}{k+1}$;
If $a_{1} \rightarrow A_{k+1}, a_{k+1} \nrightarrow A_{k+1}$,
$P=0$.
Then $a_{1} \rightarrow A_{m}(m \neq 1,2, k+1)$.
For $1, m, m+1, \cdots, k+1$, apply the induction hypothesis (1) to get
$$
P=\frac{1}{2} \cdot \frac{k-2}{k+1} .
$$

Thus, $a_{1} \rightarrow A_{2}$, either $a_{2} \rightarrow A_{3}$, apply conclusion (2) to 3, $4, \cdots, k+1$, to get $P=\frac{1}{2 k} \cdot \frac{1}{k+1}$; or $a_{2} \nrightarrow A_{3}$, apply conclusion (3) to $2,3, \cdots, k+1$, to get
$$
\begin{aligned}
P= & \frac{k-1}{2 k} \cdot \frac{1}{k+1} . \\
& \text { Hence } P_{k+1} \\
& =\frac{1}{2} \cdot \frac{k-2}{k+1}+\frac{1}{k+1}+\left(\frac{1}{2 k}+\frac{k-1}{2 k}\right) \frac{1}{k+1} \\
& =\frac{1}{2} .
\end{aligned}
$$