This is a variation of the colored socks in a drawer problem. Suppose that instead of having one drawer, you have two drawers.  Each drawer has some socks that are white and some that are black.  Drawer 1 has w black socks and x white socks.  Drawer 2 has y black socks and z white socks.  w+x=y+z.  If you take out all the socks randomly, 1 each from each drawer to make pairs, until both drawers are empty, what is expected (or average) number of times you pulled out a black sock from drawer 1 and 2?  Is it [{w/(w+x)}{y/(y+z)}](w+x)?
 A: We change notation a little. Let $n$ be the number of socks in each drawer, and suppose the first drawer has $a$ black and the second has $b$.
On the $i$-th draw we pull out a sock from each drawer. For $i=1$ to $n$, let $X_1=1$ if both are black, and let $X_i=0$ otherwise. We want $E(Y)$, where
$Y=X_1+\cdots +X_n$. By the linearity of expectation this is $E(X_1)+\cdots+E(X_n)$.
Finally, $E(X_i)=\Pr(X_i=1)=\frac{a}{n}\cdot\frac{b}{n}$. So $E(Y)=\frac{ab}{n}$. 
Remark: If I am reading your formula correctly, you came to the same conclusion.
A: A drawer contains 4 red socks and 4 blue socks. Find the lowest number of socks that must be drawn from the drawer to be assured of having a pair of red socks.
This is not a probability question so I'm not sure how to approach it. I am thinking that I would need to pick at most 6 because the first 4 could all the blue. Then the only socks left are red. After two more pulls, I am guaranteed to have a pair of red. I am wondering if I am overthinking this though.
