So it is about devising an optimal strategy to find out which is which with minimum expected number of draws, and calculate the minimum number of draws.
This is how I attempted to solve the problem. Since the combinations are symmetric, what I mean by symmetric is that how each socks are distributed comes up with unbiased probability, I can just start with drawing the first sock from any drawer, and assume any color(let's say Green) because color of the first sock wouldn't change the probability.
- So if the first sock is Green, the leftover drawers combination would be (GB, BR) or (GR, BR). I would divide the second draw event as having same/different color with first color.
2-1. The probability of choosing Green for second would be $\frac{1}{4}$. In that case, we definitely know what the third drawer would have.(It should be BR.) So one more draw from either of 1st or 2nd drawer would determine each drawers combination. So for this case the minimum draw would be 3, with $\frac{1}{4}$ probability.
2-2. Probability of choosing different color, but would be $\frac{3}{4}$. We definitely need more information so we draw another from third drawer. If the third one matches with either of 1st of 2nd sock, then we just need one more draw. (For example, 1st draw G, 2nd draw B, 3rd draw G, then we know 2nd drawer has RB, so we draw one from either of 1st or 3rd drawer.) The probability of third draw matching with either of 1st of 2nd would be $\frac{2}{3}$ so the minimum draw for this case is 4, and probability is $\frac{3}{4} * \frac{2}{3} = \frac{1}{2}$.
2-3 Continue with 2-2, but if third one does not match with either of 1st or 2nd, nothing is conclusive and you need to draw 2 more from either 3. So for this the minimum is 5 and expected value is $\frac{3}{4} * \frac{1}{3} = \frac{1}{4}$.
So the minimum expected number of draw should be $\frac{1}{4} * 3 + \frac{1}{2} * 4 + \frac{1}{4} * 5 = 4$
I'm not quite sure I got the probabilities right especially for 2-2 and 2-3. Could anyone confirm if this method is legitimate? Or someone can thankfully come up with refined method?
Thanks in advance.