Drawing two times without replacement, what is the probability of seeing m matching characters

Question

Assume that we have a set $S=(s_1,...,s_L)$ or cardinality $L$. We now draw 2 times seperatly without replacement and ignoring the order from it. I.e. once we draw $k_1$ times wihtout replacement from $S$ to obtain a Set $S_1$ and once we draw $k_2$ times without replacement from $S$ to obtain set $S_2$. How likely is it now to observe $m$ matching characters between $S_1$ and $S_2$?

A small example, if $S=(a, b, c)$ and we set $k_1$=1 and $k_2$=2 then we obtain the following possibilities for $S_1=\{(a), (b), (c)\}$ and $S_2=\{(a, b), (a, c), (c, b)\}$. Now the Probability $$P(k_1,k_2,L,m)$$ would be

$P(m=0| k_1=1,k_2=2,L=3)=1/3*1/3+1/3*1/3+1/3*1/3=1/3$ $P(m=1| k_1=1,k_2=2,L=3)=1/3*2/3+1/3*2/3+1/3*2/3=2/3$ $P(m=2| k_1=1,k_2=2,L=3)=0$ (of course)

but how to generalized this to a general formual for $k_1$, $k_2$, $L$ and $m$?

Answer 1

From the first draw you have $k_1$ sampled items and $L-k_1$ non-sampled items. If you take another draw using simple-random-sampling without replacement then the number of matches $M$ is just the number of those previous $k_1$ items that are drawn in the second draw. This random variable follows a hypergeometric distribution, with probability mass function given by:

$$\mathbb{P}(M=m|k_1,k_2,L) = \text{Hyper}(m| L, k_1, k_2) = \frac{{k_1 \choose m} {L-k_1 \choose k_2-m}}{{L \choose k_2}}.$$