TITLE: How many trial picks expectedly sufficient to cover a sample space?
QUESTION [5 upvotes]: Consider a sequence of independent events where an $r$ element subset of an $n$ element set is picked uniformly randomly (ie. any of the $\begin{pmatrix}n\newline r\end{pmatrix}$ possibilities being equally likely).
What is the expected number of subsets one has pick to cover the whole set?
Here the terminology means: a sequence of picks $A_1,A_2,\ldots,A_n$ covers the whole set if
 $|A_1 \cup \cdots \cup A_n| = n$. A sequence $A_1, A_2,\ldots$ succeeds to cover the whole set in $n$ steps, if $A_1,\ldots,A_n$ covers the whole set but $A_1,\ldots, A_{n-1}$ does not.
The expected numbers seems to be much higher than one would imagine. But I could not quite come up with a closed form. But chances are, its always a rational number.

REPLY [2 votes]: The expected number of picks needed equals the sum of the probabilities that at least $t$ picks are needed, which means that $t-1$ subsets left at least one value uncovered. We can use inclusion-exclusion to get the probability that at least one value is uncovered.
The probability that a particular set of $k$ values is uncovered after $t-1$ subsets are chosen is 
$$\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1}$$
So, by inclusion-exclusion, the probability that at least one value is uncovered is 
$$ \sum_{k=1}^n  {n \choose k}(-1)^{k-1}\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg) ^{t-1} $$
And then the expected number of subsets needed to cover everything is 
$$ \sum_{t=1}^\infty \sum_{k=1}^n {n \choose k}(-1)^{k-1} \Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1} $$
Change the order of summation and use $s=t-1$:
$$ \sum_{k=1}^n {n \choose k}(-1)^{k-1} \sum_{s=0}^\infty  \Bigg( \frac{n-k \choose r}{n \choose r}\Bigg)^s$$
The inner sum is a geometric series.
$$ \sum_{k=1}^n {n \choose k} (-1)^{k-1}\frac{n \choose r}{{n \choose r}-{n-k \choose r}}$$ 
$$ {n \choose r} \sum_{k=1}^n (-1)^{k-1}\frac{n \choose k}{{n \choose r}-{n-k \choose r}}$$
I'm sure that should simplify further, but at least now it's a simple sum. I've checked that this agrees with the coupon collection problem for $r=1$.
Interestingly, Mathematica "simplifies" this sum for particular values of $r$, although what it returns even for the next case is too complicated to repeat, involving EulerGamma, the gamma function at half-integer values, and PolyGamma[0,1+n].