2 added 117 characters in body

I think I've worked out

EDIT: While the $r=1$ case - is the easiest, I thought it would be helpful to work it out anyway. I get that the expected number of picks necessary for $r=1$ is $nH_n$, where $H_n$ is the $n$th harmonic number. , which is in line with Tal K's answer since $H_n\approx\ln(n)$.

Suppose the total number of elements covered by our picks so far is $k$. If we calculate the expected number of picks it will take to get to $k+1$, then we simply take the sum of our result from $k=0$ to $k=n-1$. There are $n-k$ elements we still need to hit, so there is an $\frac{n-k}{n}$ probability of having $k+1$ covered after 1 pick, $\frac{n-k}{n}(\frac{k}{n})$ probability of having $k+1$ covered after exactly 2 picks, and in general $\frac{n-k}{n}(\frac{k}{n})^j$ probability of going to $k+1$ after exactly $j$ picks. Thus, the expected number of picks to go from $k$ covered to $k+1$ covered is $(\frac{n-k}{n})\sum_{j=1}^\infty k(\frac{k}{n})^{k-1}$, which by the standard derivative trick we know is $(\frac{n-k}{n})\frac{1}{(1-\frac{k}{n})^2}=\frac{n}{n-k}$. Thus the expected number of picks of 1 element subsets necessary to cover an $n$ element set is $\sum_{k=0}^{n-1}\frac{n}{n-k}=n\sum_{k=1}^n\frac{1}{k}=nH_n$.

1

I think I've worked out the $r=1$ case - I get that the expected number of picks necessary for $r=1$ is $nH_n$, where $H_n$ is the $n$th harmonic number.

Suppose the total number of elements covered by our picks so far is $k$. If we calculate the expected number of picks it will take to get to $k+1$, then we simply take the sum of our result from $k=0$ to $k=n-1$. There are $n-k$ elements we still need to hit, so there is an $\frac{n-k}{n}$ probability of having $k+1$ covered after 1 pick, $\frac{n-k}{n}(\frac{k}{n})$ probability of having $k+1$ covered after exactly 2 picks, and in general $\frac{n-k}{n}(\frac{k}{n})^j$ probability of going to $k+1$ after exactly $j$ picks. Thus, the expected number of picks to go from $k$ covered to $k+1$ covered is $(\frac{n-k}{n})\sum_{j=1}^\infty k(\frac{k}{n})^{k-1}$, which by the standard derivative trick we know is $(\frac{n-k}{n})\frac{1}{(1-\frac{k}{n})^2}=\frac{n}{n-k}$. Thus the expected number of picks of 1 element subsets necessary to cover an $n$ element set is $\sum_{k=0}^{n-1}\frac{n}{n-k}=n\sum_{k=1}^n\frac{1}{k}=nH_n$.