10 Four people, A, B, C, and D, are practicing passing a ball. The ball is initially passed by A. Each person, upon receiving the ball, passes it to one of the other three people with equal probability. Let $p_{n}$ denote the probability that the ball returns to A after $n$ passes. Then $p_{6}=$ $\qquad$
$10 \frac{61}{243}$ Hint: After one pass, the probability of the ball landing in the hands of B, C, or D is $\frac{1}{3}$, while the probability of it landing in A's hands is 0, so $p_{1}=0$. After two passes, the probability of the ball landing in A's hands is
$$
p_{2}=\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{1}{3}=\frac{1}{3} .
$$

Now consider the recursion. For the ball to be in A's hands after $n$ passes, it must not be in A's hands after $n-1$ passes, so $p_{n}=\frac{1}{3}\left(1-p_{n-1}\right), n=1$, $2,3,4, \cdots$, therefore
$$
\begin{array}{c}
p_{3}=\frac{1}{3}\left(1-p_{2}\right)=\frac{1}{3} \times \frac{2}{3}=\frac{2}{9}, \\
p_{4}=\frac{1}{3}\left(1-p_{3}\right)=\frac{1}{3} \times \frac{7}{9}=\frac{7}{27}, \\
p_{5}=\frac{1}{3}\left(1-p_{4}\right)=\frac{1}{3} \times \frac{20}{27}=\frac{20}{81}, \\
p_{6}=\frac{1}{3}\left(1-p_{5}\right)=\frac{1}{3} \times \frac{61}{81}=\frac{61}{243} .
\end{array}
$$