Probability of picking all elements in independent trials Let's say:


*

*I have $N$ unique balls in the urn.

*I pick $K$ balls (with same probability and without replacement) each try.

*I try $S$ times.

*Trials are independent

*Before each try, there are all $N$ balls in the urn.


What is the probability of picking each ball at least once during all $S$ trials ?
 A: Start with the opposite question - what is the probability $p$ that there is a ball which isn't picked throughout all $S$ trials. 
We can consider an inclusion-exclusion way of solving this: consider the event $E_j$ that the $j^{\text{th}}$ ball is never picked in any of the $S$ trials, and more generally let $E_{j_1,j_2,\ldots,j_r}$ be the event that the $j_1^{\text{th}}$, $j_2^{\text{th}}$, ... ,$j_r^{\text{th}}$ balls are never picked in any of the $S$ trials. Note that these events aren't disjoint: $E_1$ and $E_2$ both contain $E_{1,2}$ since the event $E_1$ that the 1st ball isn't picked doesn't specify what happens to the other balls. Thus, the probability of interest is computed using inclusion-exclusion:
$\begin{align}
p &= \sum_{j=1}^N \mathbb{P}(E_j) -\sum_{j_1<j_2} \mathbb{P}(E_{j_1,j_2}) + \sum_{j_1<j_2<j_3} \mathbb{P}(E_{j_1,j_2,j_3}) + \ldots + (-1)^{N-1} \mathbb{P}(E_{1,2,\ldots,N})\\
  &= N\mathbb{P}(E_1) - {N\choose 2} \mathbb{P}(E_{1,2}) + {N\choose 3} \mathbb{P}(E_{1,2,3}) + \ldots + (-1)^{N-1}\mathbb{P}(E_{1,2,\ldots,N}) 
\end{align}$ 
since the subsets of given "order" have the same probabilities due to the balls being basically equivalent.
So now we need only compute $\mathbb{P}(E_{1,2,...,m})$ for each $m \in \{1,\ldots,N\}$.
$\begin{align}
\mathbb{P}(E_{1,2,...,m}) &= \mathbb{P}(\text{first $m$ balls are not picked in any of the $S$ trials}) \\
                          &= \left[ \mathbb{P}(\text{first $m$ balls are not picked in a single trial}) \right]^S
\end{align}$
and now I leave the computation of $\mathbb{P}(\text{first $m$ balls are not picked in a single trial})$ to you (see hypergeometric distribution). 
The answer that you require is then $1-p$.  I hope that my reasoning is correct - it's 2.30 am here...
