TITLE: Expected number of trials before I get one of each type
QUESTION [5 upvotes]: Say I have a population that I sample from without replacement. In this population there are $a_1$ members of type 1, $a_2$ members of type two and so forth up to $a_k$ of type $k$. 
What is the expected number of trials until I collect at least one member out of each type? An example here would be the expected number of cards drawn from a deck until I get one of every valor.

REPLY [3 votes]: For $k=1$ you clearly have an expectation of $1$.
For $k=2$ you seem to have an expectation of $$\displaystyle \sum_{n=0}^{a_1+a_2} \frac{{a_1 \choose n}+{a_2 \choose n} - {0 \choose n}}{{a_1 + a_2 \choose n}}=  1+\dfrac{a_1}{a_2+1}+\dfrac{a_2}{a_1+1}$$ taking ${m \choose n}=0$ when $0 \le m \lt n$
For $k=3$ you seem to have an expectation of $$\displaystyle \sum_{n=0}^{a_1+a_2+a_3} \frac{{a_1+a_2 \choose n}+{a_1+a_3 \choose n}+{a_2+a_3 \choose n}-{a_1 \choose n}-{a_2 \choose n}-{a_3 \choose n} + {0 \choose n}}{{a_1 + a_2 +a_3\choose n}} \\=1 + \frac{a_1+a_2}{a_3+1}+\frac{a_1+a_3}{a_2+1}+\frac{a_2+a_3}{a_3+1}-\frac{a_1}{a_2+a_3+1}-\frac{a_2}{a_1+a_3+1}-\frac{a_3}{a_1+a_2+1}$$ 
and I would expect a similar inclusion-exclusion type calculation to apply for larger $k$.