I have a set of 60 unique numbers and every round I draw m unique numbers from the set. After every round the m numbers are replaced. I'm looking to find the probability that all 60 numbers have been selected at least once after n rounds.
I have looked at this question: Probability of choosing 54 unique values with replacements after n tries. and the answer there seems to be for a specific variant of my question where m=1.
Is there a more generic solution for any m?
Given a set $S$ of $k$ numbers, the probability that none of these are chosen in any given round is ${60-k \choose m}/{60 \choose m}$ and the probability none are chosen in $n$ rounds is $\left( {60-k \choose m}/{60 \choose m}\right)^n$. By the inclusion-exclusion principle, the probability that every number is chosen at least once in $n$ rounds is $$ 1 - \sum_{k=1}^{60-m} {60 \choose k} (-1)^{k-1} \left(\left. {60-k \choose m}\right/{60 \choose m}\right)^n$$
