4. Including A, 6 people pass the ball to each other. The ball is passed from one person to another each time, starting with A. After 5 passes, the number of ways the ball can return to A is


4.520

Analysis: Let the total number of ways the ball returns to player A after $n$ passes be $a_{n}$. Considering the number of ways the ball returns to A after $n+1$ passes, $a_{n+1}$, this is equivalent to the number of ways the ball is not with A after $n$ passes, plus the number of ways the ball is exactly with A after $n$ passes, which together equals the total number of ways after $n$ passes, $5^{n}$. That is, $a_{n+1}=5^{n}-a_{n}$. Given $a_{1}=0, a_{2}=5$, hence $a_{3}=20, a_{4}=105$,
$$
a_{5}=520 \text { . }
$$