Consider the problem of throwing balls uniformly at random into $q$ bins, each has capacity $q$. What is the expected number of balls we need to throw until one of the bins becomes full?
Clearly the expectation is between $q$ and $q(q-1)+1$. My idea is as follows:
$$ L_n:= \text{event that at time $n$ the first full bin occurs} $$
Then some bin $i$ has $q$ balls and others have at most $q-1$ balls. We want to find the expected total number of balls thrown, i.e. the sum of the number of balls in all bins.
I am assuming you are throwing the balls one by one. So The balls are distinguishable and the bins are distinguishable. So we want to know what is the probability that at step $k$ we make one of the bins full so we pick out of the $k-1$ past balls $q-1$ to put in the bag with the $k-$th ball we can do this in $\binom{k-1}{q-1}$ and then we choose $l$ bags that contain balls and we proceed to fill them with the remaining $k-q$ balls, we can do this using the Restricted Stirling numbers of the second kind defined by ${a\brace b}_{\leq c}$ as the number of partitions of $[a]$ into $b$ blocks each block having at most $c$ elements. The formula ends up being $$\sum _{k=1}^{\infty}\frac{k}{q^{k-1}}\binom{k-1}{q-1}\sum _{l=0}^{q-1}\binom{q-1}{l}l!{k-q\brace l}_{\leq q-1}.$$ Notice that the $q/q^k$ comes from the uniform probability and choosing the bin in which the $k-$th ball is thrown.
I simulated it with the following code
Giving me the following results, where the values are $q,$ the simulation and the value of the function above.