Coupon Collector Prob Variation

Solution 1:

We can formulate this as a ball-and-urns problem: we have $k$ distinct balls (number of extractions) that we can place with equal probability in $m$ urns, of which $n<m$ are marked as "desired". We want to compute the probability of event $E$: the $n$ desired urns are non-empty. Let $j=0 \cdots k$ be the total number of balls that fall in the desired urns.

Then

$$P(E \mid j) = \frac{j! \,S_{n,j}}{n^j}$$ where $S_{n,j}$ is the Stirling number of the second kind. Further, $P(j)={k \choose j} n ^j /m^k$. Then

$$P(E ) =\frac{1}{m^k}\sum_{j=0}^k {k \choose j} j! \,S_{n,j} = \frac{k!}{m^k} \sum_{j=0}^k \frac{1}{(k-j)!}S_{n,j} $$

Perhaps this can be simplified a little, I doubt it. From this you could get your desired $k$ numerically.

An approximation using Poissonization: $P(E) \approx (1 - e^{-k/m})^n$.