The probability of the intersection set of $L$ independent sets is empty Denote the set $\{1,2,\ldots,N\}$ as $\mathcal{N}$, In round-$i$, I randomly choose $R< \frac{N}{2}$ different values from $\mathcal{N}$, and then collect the $R$ numbers as a new subset $\mathcal{A}_1$. Through independent $L$ rounds, I obtain $L$ subsets, then how to calculate $ {\rm Pr} ( \cap_{i=1}^L \mathcal{A}_i) = \emptyset$, where ${\rm Pr}(\cdot)$ is the probability operator.
Any comments would be appreciated!
 A: The intersection is non-empty if at least one element is in all subsets. So, using inclusion-exclusion, you want to compute the complement to
$$\sum_{i=1}^R (-1)^{i-1} \sum_{U\subseteq {\cal{N}}, |U|=i } P(U \text { is contained everywhere}).$$
This seems to be equal to
$$\sum_{i=1}^R (-1)^{i-1} {N\choose i} \frac{{N-i\choose R-i}^L}{{N\choose R}^L}
$$
Not sure how to simplify nicely.
A: After multiple mis-starts, I have what I think is an OK argument.
Thanks to Kris and Peter for nudging me on the right path.
Let $E_x$ be the event that $x\in\bigcap_{j=1}^L A_j$, that is, $E_x=\bigcap_j[x\in A_j]$, so that $\bigcup_{x=1}^N E_x$ is the event that $\bigcap_{j=1}^L A_j$ is not empty. It is easy to see that $P(E_x)=\left(\binom {N-1} {R-1} \big/ \binom N R\right)^L$, since there are $\binom {N-1}{R-1}$ ways of picking the other $R-1$ elements of $A_j$ other than $x$,  out of $\binom N R$ ways to pick $A_j$ in all;   since the sets $A_j$ are chosen independently of each other, the ratio of binomial coefficients gets raised to the $L$ power.  Similarly, $P(E_x\cap E_y)=\left(\binom {N-2} {R-2} \big/ \binom N R\right)^L$, and by the principle of inclusion and exclusion,
$$
1-P\left(\bigcap_{j=1}^L A_j=\emptyset\right) = P(\bigcup_{x=1}^NE_x) =\sum_{k=1}^n (-1)^{k-1} 
\left(\frac{\binom {N-k} {R-k}} {\binom N R}
\right)^L.
$$
This is ugly and not very useful-looking, but is a formula.
