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 $S$?



Randomly picking $k$ members of $\{1,\ldots,n\}$
If $X$ is a random variable with non-negative integer values, the expectation of $X$ equals $$\mathbb{E}(X)=\sum\_{m=1}^\infty {\rm Prob}\,(X\geqslant m).$$
In our situation the event $X\geqslant m$ means that after $m-1$ days there remains a not taken element. By exclusion-inclusion it equals $$\sum\_{i=1}^n (-1)^{i-1}\binom{n}i\left(\frac{\binom{n-i}k}{\binom{n}k}\right)^{m-1}.$$
(the $i$-th summand corresponds to the choice of $i$ not-covered elements and the probability that they are not covered.) When we sum up by $m$, we get
$$
\sum\_{i=1}^n (-1)^{i-1}\frac{\binom{n}i}{1-\frac{\binom{n-i}k}{\binom{n}k}}.
$$
Is such an answer ok for you?


For small $k$ and large $n$ the following further identical transform of the above sum may be helpful. Choose a polynomial $h(x)$ of degree at most $k-1$ for which the polynomial $1-h(x)\cdot (1-x)(1-x/2)\dots(1-x/n)$ is divisible by the polynomial $1-\binom{n-x}k/\binom{n}k$. The ratio which we denote $f(x)$ is a polynomial of degree at most $n-1$. Thus $\sum\_{i=0}^n (-1)^{i-1}\binom{n}i f(i)=0$. But for $i=1,2,\dots,n$, we have $f(i)=(1-\binom{n-i}k/\binom{n}k)^{-1}$. Thus our sum equals $f(0)$.
For example, if $k=1$, we have $h(x)=1$ and $f(x)=nx^{-1}(1-(1-x)\dots(1-x/n))$, $f(0)=n(1+1/2+\dots+1/n)$ (that is known as [Coupon Collector's Problem](https://en.wikipedia.org/wiki/Coupon_collector%27s_problem)). For $k=2$ we get $h(x)=1+ax$ with exponentially small (for large $n$) value $a$ and $f(0)=(1+\dots+\frac1n-a)/(\frac1n+\frac1{n-1})$.


