Expectation of the black ball pairs

Question

Suppose we have total N balls, there are B black balls and W white balls, so that N = B + W. For simplicity, assume both B and W are even numbers. If we randomly pairing them up to form N/2 pairs, assume random variable L is the number of black ball pairs. What is the expectation of L (i.e. black ball pairs)? Is there any known distribution for this problem? Thanks!

Answer 1

To summarize the discussion in the comments:

This sort of thing is best handled by indicator variables, exploiting the fact that expectation is linear, regardless of any possible dependence between the variables.

Thus, we let $X_i$ denote the indicator variable for the $i^{th}$ pair, so $X_i=1$ if the $i^{th}$ pair is $BB$ and $X_i=0$ otherwise. It is easy to see that the probability that the $i^{th}$ pair is $BB$ is $\frac B{B+W}\times\frac {B-1}{B+W-1}$, which is then $E[X_i]$. Note that, of course, this does not depend on $i$. It follows that the desired result is simply $$E=E\left[ \sum X_i\right]=\sum E\left[X_i\right]=\frac {B+W}2\times \frac B{B+W}\times\frac {B-1}{B+W-1}=\boxed {\frac {B(B-1)}{2(B+W-1)}}$$

and we are done.

Sanity checks: If $B=0$ this gives $0$, as it should. If $W=0$ this gives $\frac B2$ as it should. If $B=W$ and $B$ is large then we expect that roughly $\frac 14$ of the pairs will be $BB$ so the answer should be $\frac B4$ which, in the limit, is confirmed by the formula. It is a good idea to work a few cases for small $B,W$ by hand just to confirm, but I will leave that as an exercise.

As a significantly harder exercise, you might try to compute the variance of the number of $BB$ pairs. That can be done along the same lines, but it is harder since variance itself is not linear in the strong way that expectation is. As a hint, use the fact that the variance of a distribution $X$ is $E\left[X^2\right]-E[X]^2$.