Coupon Collector Prob Variation I a programmer not a mathematician so please excuse my ignorance and please dumb it down for me.
My research indicates that this is a variation of the Coupon Collector problem but I really don't know how to get the answer.
Assuming $n$ equally likely desired selections (Probability for each=$1/m$) how many selections with replacement must be made to have an $X$ probability of getting at least one of each of the $n$ desired objects?

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