If $S_1$ and $S_2$ are subsets of size $\ell$ of $\{1,\ldots,n\}$ what is the probability that they have at least $m$ elements in common? There are answers for checking if the two sets are disjoint. But I could not find anything about them having a certain number of common elements.
 A: Which elements $S_1$ contains doesn’t matter for this, just which elements $S_2$ contains. I am going to assume that $\ell$ can range from $1$ through $n$ and that a set can contain an element only once.
I am going to draw up an analogous problem that will make this one easier to think through:

There are $n$ cards laying in front of you, facedown, where $\ell$ of them are red and $n-\ell$ of them are black. What is the probability that at least $m$ of the cards you draw are red given that you draw $\ell$ cards without replacement.

Here, our red cards are the elements of $S_1$ and the cards you draw are the elements of $S_2$.
This is a hypergeometric probability distribution, because 


*

*each card drawn is either red or it isn’t

*the probability of selecting a red card changes each time


Let $X$ be the random variable that represents the number of red cards drawn. The formula is that
$$P(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$


*

*$N$ is the population size

*$K$ is the number of success states in the population

*$n$ is the number of draws

*$k$ is the number of successes you end up seeing


For us, our population is the $n$ cards laying facedown in front of us; the success states are the red cards, of which there are $\ell$; we are drawing $\ell$ cards; and we want to see $k$ range anywhere from $m$ all the way up to the maximum, $\ell$. Thus, we want to calculate
$$\begin{align}
P(m\le X\le \ell) = P(X=m) + P(X=m+1) + \cdots + P(X=\ell) = \sum_{k=\ell}^m \frac{\binom \ell k \binom{n-\ell}{\ell-k}}{\binom n \ell} \\
\end{align}$$
Since you haven’t specified, I am going to assume sigma notation is fine for you.
