Suppose you have 3 drawers. One drawer has infinitely many red and green socks. Another drawer has infinitely many green and blue socks. The last drawer has infinitely many red and blue socks.
What's the expected number of draws you need to make in order to conclude which drawer is which?
I tried a lot of case work and got $4.5$, but I'm not sure if that is right. I am sort of just guessing, so I would appreciate some help on this problem
A preliminary point: After you have seen one color of drawer $D_i$ then the expected number $E$ of additional draws to learn the other color of $D_i$ satisfies $$E=1+{1\over2}\, E\ ,$$ because the first such draw gives you with probability ${1\over2}$ the color that you already know. It follows that $E=2$, and that we should try to avoid this procedure.
Therefore we take one sock from drawer $D_1$ and one sock from $D_2$. These two socks have the same color with probability ${1\over4}$, and have different colors with probability ${3\over4}$. In the first case we know the colors of $D_3$, and have to expect $2$ more drawings to find the the second color of $D_1$, which then also reveils the second color of $D_2$. In the second case we draw one sock from $D_3$ and then have either three different socks or two equal socks, one from $D_3$, and another sock. We have to expect $2$ more drawings to find the other color of $D_3$, and then everything will be determined.
This shows that the expected number $E_{\rm total}$ of necessary drawings is given by $$E_{\rm total}={1\over4}(2+2)+{3\over4}(2+1+2)=4.75\ .$$