A Ball Passing Game

Method #1

In a game of $n$ players, let $p_k$ be the probability that Daniel has the ball after the $k$ -th turn.

We have $p_k = \displaystyle\frac{1}{n-1}\left(1-p_{k-1}\right)$ , and obviously $p_0=1$ and $p_1 = 0$

we can solve the recursive relation by subtracting both sides by $\displaystyle \frac{1}{n}$

$\begin{aligned}\displaystyle p_k - \frac{1}{n} &= -\frac{1}{n-1}p_{k-1} +\frac{1}{n-1}-\frac{1}{n} \\&= -\frac{1}{n-1}p_{k-1} + \frac{1}{n(n-1)} \\&= -\frac{1}{n-1}\left(p_{k-1} - \frac{1}{n}\right) \end{aligned}$

and thus $\displaystyle p_k - \frac{1}{n}$ is a geometric progression with common ratio $\displaystyle -\frac{1}{n-1}$

$\displaystyle p_k - \frac{1}{n} = \left(p_1 - \frac{1}{n}\right)\left(-\frac{1}{n-1}\right)^{k-1}$

and we have $\displaystyle p_k = \frac{1}{n} +\frac{n-1}{n}\left(-\frac{1}{n-1}\right)^k$

substituding $n = 5$ and $k = 7$ yields our answer $\displaystyle p_7 = \frac{1}{5} + \frac{4}{5}\left(-\frac{1}{4}\right)^7 = \boxed{\displaystyle \frac{819}{4096}}$

Method #2

The game can be visualised by a graph with the adjacency matrix

$A = \begin{pmatrix} 0&1&1&1&1\\ 1&0&1&1&1\\ 1&1&0&1&1\\ 1&1&1&0&1\\ 1&1&1&1&0 \end{pmatrix}$

Computing $A^7$ would give us the number of ways the ball passing can happen

$A^7 = \begin{pmatrix} 3276&3277&3277&3277&3277\\ 3277&3276&3277&3277&3277\\ 3277&3277&3276&3277&3277\\ 3277&3277&3277&3276&3277\\ 3277&3277&3277&3277&3276 \end{pmatrix}$

The probability is therefore $\displaystyle \frac{3276}{3277\times4 + 3276} = \boxed{\displaystyle \frac{819}{4096}}$