TITLE: Randomly picking $k$ members of $\{1,\ldots,n\}$
QUESTION [6 upvotes]: Every day, I randomly pick a sample consisting of $k$ members of $\{1,\ldots,n\}$ where $k\leq n$. I stop as soon as every number of $\{1,\ldots,n\}$ has been picked at least once. Let $S$ be the number of days needed to reach my goal. What is the expected value of $S$?

REPLY [4 votes]: I think you should have already a good estimate just comparing with the classical coupon colector's problem (CCP).
Let us consider the CCP and pick the numbers one by one. We define the following stopping times 
$T_i$ = The first $t\geq T_{i-1}$ such that there are $k$ different numbers picked between $T_{i-1}$ and $t$ 
Then the set of number picked at $T_i$ is the same as the set picked in your process after $i$ days. Moreover, the random variable $X_i=T_{i+1}-T_i$ are iid random variables (of same law $X$). And we have $$ t=\sum_{i:T_i\leq t} X_i+(t-T_i)$$
If we call $T$ the time the classical coupon colector has picked all its numbers, we have $$T=\sum_{i:T_i\leq T} X_i+(T-T_i)$$ In your process, $S$ is the first $i$ such that $T_i\geq T$. Therefore
$$\mathbb{E}(T)=\mathbb{E}(\sum_{i<S} X_i)+\mathbb{E}(T-T_{S-1})$$
$X_i$ are independant of $S$ and therefore 
$$\mathbb{E}(T)=\mathbb{E}(S-1)\mathbb{E}(X)+\mathbb{E}(T-T_{S-1})$$
and remark that $\mathbb{E}(T-T_{S-1})\leq \mathbb{E}(X)$ and therefore we have the estimate $$\frac{\mathbb{E}(T)}{\mathbb{E}(X)}\leq\mathbb{E}(S)\leq \frac{\mathbb{E}(T)}{\mathbb{E}(X)} +1 $$ with $\mathbb{E}(T)= \sum_{j=1}^n\frac{n}{j}$ and $\mathbb{E}(X)=\sum_{j=n-k+1}^n \frac{n}{j}$.