Suppose that I have a multiset containing multiple copies of two types of items, such as ${A, A, A, A, B, B}$. Suppose I start drawing random pairs from this set without replacement, such as $AA, AB, AB$, until I have used all of the elements of the set. How do I calculate the expected number of a certain type of pairing (say $AB$)?
This seems like this would be a problem that would come up in genetics (ie, the probability of getting a certain allele pairing, given separate frequencies of each allele in a large population), but I cannot find anything along those lines.
Let $E_{AB}(a,b)$ be the expected number of $AB$ pairs if the urn contains $a$ $A$'s and $b$ $B$'s. So your problem is to compute $E_{AB}(4,2).$ There are $\binom{6}{2} = 15$ ways to draw two letter. There is 1 way to draw $BB$, 6 ways to draw $AA$ and 8 ways to draw $AB$. So $$E_{AB}(4,2)= \frac{1}{15}E_{AB}(4,0) + \frac{6}{15}E_{AB}(2,2) + \frac{8}{15}(1+E_{AB}(3,1)).$$ Of course, $E_{AB}(4,0) = 0,$ but you need to continue, computing the remaining terms by the same method, until you get the final answer.
Of course, similar calculations work for the expected number of other pairs. As a sanity check, if you compute them all, they have to add up to 3, since we always get 3 pairs.
EDIT Here are the formulas you asked for. In what follows, we let $$p_{AA} = \frac{\binom{a}{2}}{\binom{a+b}{2}},p_{BB} = \frac{\binom{b}{2}}{\binom{a+b}{2}},p_{AB} = \frac{ab}{\binom{a+b}{2}}$$ Then $$\begin{align}E_{AA}&=p_{AA}(1+E_{AA}(a-2,b))+p_{AB}E_{AA}(a-1,b-1) +p_{BB}E_{AA}(a,b-2)\\E_{BB}&=p_{AA}E_{BB}(a-2,b)+p_{AB}E_{BB}(a-1,b-1) +p_{BB}(1+E_{BB}(a,b-2))\\E_{AB}&=p_{AA}E_{AA}(a-2,b)+p_{AB}(1+E_{AB}(a-1,b-1)) +p_{BB}E_{AB}(a,b-2)\\ \end{align}$$
Now we need some ending conditions. All the expectations are defined to be $0$ if $a<0, \text{ or } b<0 \text{ or } a+b<2.$ We also have the usual convention for binomial coefficients: if $a<2 \text{ then } \binom{a}{2} = 0.$