Four people drawing five of seventeen balls overlap

An urn holds $n$ balls numbered $1$ to $n$. Each of $k\ge 2$ people in turn draws $r$ balls ($1\le r\le n$) without replacement, records them, and returns all of them to the urn before the next person draws. Let $p(k,n,r)$ denote the probability that at least one ball is drawn by every one of the $k$ people. Compute $p(4,17,5)$ to three decimal places.

About $0.123$. Apply inclusion–exclusion to the set of balls shared by all $k$ draws. A fixed group of $j$ balls lies inside one person's draw with probability $q_j=\binom{n-j}{r-j}\big/\binom{n}{r}$, hence $p=\sum_{j=1}^{r}(-1)^{j+1}\binom{n}{j}q_j^{\,k}$. Plugging in $k=4$, $n=17$, $r=5$ gives $\approx 0.1233$, which a simulation confirms.