Distribution on a collision variant of Hypergeometric distribution? I have the following scenario:


*

*there is a set $\Omega$ of $N$ elements, among which $K$ are marked — let $M\subseteq \Omega$ be this subset.

*Alice select uniformly at random (i.e., sampling without replacement) a set $A\subseteq \Omega$ of $n$ elements.

*Bob select uniformly at random (i.e., sampling without replacement) a set $B\subseteq \Omega$ of $n$ elements, independently of Alice.


Alice and Bob then gather, and look at the number of marked balls they have in common, denoted $X$ (that is, $X=\lvert A\cap B\cap M\rvert$).

Is the distribution of $X$ known in the literature?

(note that I am not asking for more than this -- only if there is a simple reference on this type of questions.)
To be more precise, for a proof I needed to get tail bounds (concentration around its expected value) of $X$. Now, I could analyze this behavior and get the bounds I wanted by modeling $\lvert A\cap M\rvert$ as a hypergeometric random variable, and $X$ as another (with different parameters) conditioned on $A$. This eventually gave me what I wanted, but it is quite cumbersome. So if there is a one-liner to get this, such as "Since $X$ follows a Q5U8-6 distribution, standard results (see e.g. [Blah'56]) give that [...]", I'd be very eager to know...
 A: I do not know whether the distribution of $X$ has a formal name. From your question, you aims for a concentration bound about the expectation and you may interest in the following easy derivation, though it is not a complete answer to your question (it is too long to put as a comment).

Let the $K$ marked elements be $o_1$, $o_2$, $\cdots$, $o_K$. For $o_i$, define
$$
X_i = \begin{cases}
1\quad \text{if }\ o_i \in A\cap B \\
0\quad \text{otherwise}
\end{cases}
$$
Then
$$
X = X_1 + X_2 + \cdots + X_K
$$
We have
$$
\mathbb{E}[X_i] = \Pr[X_i = 1] = \Pr[o_i \in A]\cdot \Pr[o_i \in B] = \frac{n}{N}\cdot\frac{n}{N}
$$
and
$$
\mathbb{E}[X_i^2] = \mathbb{E}[X_i] = \frac{n^2}{N^2}
$$
and
$$
\mathbb{E}[X_iX_j] = \Pr[o_i \in A \wedge o_j \in A]\cdot \Pr[o_i \in B \wedge o_j \in B] = (\frac{N - 2 \choose n - 2}{N \choose n})^2 = (\frac{n(n-1)}{N(N-1)})^2
$$
We conclude that
$$
\mathbb{E}[X] = \sum_{i}\mathbb{E}[X_i] = \frac{Kn^2}{N^2}
$$
and
$$
\mathbb{E}[X^2] = \sum_{i}\mathbb{E}[X_i^2] + \sum_{i\neq j}\mathbb{E}[X_iX_j] = \frac{Kn^2}{N^2} + \frac{K(K-1)n^2(n-1)^2}{N^2(N-1)^2}
$$
and thus we can compute
$
\mathbb{Var}[X]
$ and the Chebyshev's inequality can be used for bounding the concentration probability.
