How to optimally label three boxes. The problem:
I have three infinite boxes. One contains both Apples and Bananas, one contains Apples and Cherries and the other contains Bananas and Cherries. All fruit in all boxes are in equal proportion, that is the probability of receiving one of the two available fruit is $\frac{1}{2}$. You do not know which of the boxes are which but to correctly label them you may randomly draw fruit from them in order order you wish to. What is the optimal (with respected to expected draws) way to do this. What is the expected number of draws?
My Result:
I only managed to solve this nuts and bolts by manual analysis of combinations. I get $4.75$. Is there a nice and clean way to solve this. Do you get a different answer?
 A: Draw from one box (doesn't matter which one). Take note of which fruit it is. Draw from one of the other two boxes. Either you find the same fruit or you find a different fruit. If it is the same fruit, you instantly know what is in the last remaining box. So, keep drawing from one of the two boxes until you get a different fruit and you know all three.
If it is a different fruit, draw from the third box. If it is the same as one of the first two boxes, it tells you the third. Now, keep drawing from one of the two unknown boxes until you find a different fruit, and that will tell you the remaining boxes. If it is different, then there is one of two possible ways that can happen. Keep drawing from the third box to figure out which of the two it is.
So, let's calculate expectation:
You draw from the first box. Whatever fruit you get, there is a $\dfrac{1}{4}$ chance that a draw from a different box will be that same fruit, and a $\dfrac{3}{4}$ chance it is a different fruit.
Case 1: You get the same fruit when picking from the second box. This tells you what the third box contains (it contains the other two fruit). So, keep drawing from the second box until you get a different fruit, and then you know what all the boxes hold. The expectation for this case is:
$$\sum_{k=1}^\infty \left(\dfrac{1}{2}\right)^k(2+k) = 4$$
Case 2: The second box is different. Regardless if the third box is the same as one of the first two or different, it either tells you the contents of the one box that had a different fruit or it tells you that you are in one of two possible cases where you picked different fruit from all three boxes, and once you know the second fruit of one of them, you can figure out the second fruit of all of them. Now, you keep drawing from the third box until you get a different fruit. The expectation for this case is:
$$\sum_{k=1}^\infty \left(\dfrac{1}{2}\right)^k(3+k) = 5$$
So, the total expectation is the probability of being in case 1 times the expectation of case 1 plus the probability of being in case 2 times the expectation of case 2:
$$\dfrac{1}{4}\cdot 4 + \dfrac{3}{4}\cdot 5 = \dfrac{19}{4}$$
