Pearls

A puzzle from Chance News. The solution involves a combination of simple probability and and optimization ideas.

[The statement of the problem as given below could be better. Essentially we are asked if it is possible to to distribute 50 white pearls and 50 black pearls in two boxes (not necessarily 50 in each box), so that if a box is chosen at random and then a pearl drawn at random from that box, the chance that it is black is more than 1/2.]

I get 50 of these black pearls and I put them in a cigar box. And I get 50 faux white pearls. So, I've got 50 black pearls from the Seychelles and then I have 50 white pearls in another cigar box. And I tell my wife, "Look, I'm going to put these cigar boxes in front of you. You will be blindfolded and you will instruct me to open one or the other, either A or B. You will then pick a pearl out and, if it's a black one, you get the black pearls, and, if it's a white one, you get the cheap pearls that I had intended to buy you in the first place". So it's obvious, since there are 50 of each pearl, her chances are 50/50.

But she can mix them up. She can put all the pearls in one box, she can put half the white ones in one box and half the black ones in the same box so she has 25 of each color in one box. Is there any way she can mix up these pearls to improve her chances beyond 50/50? That's the question.

Note the simple and elegant solution:

put one black pearl in box A and the rest in box B.

Proof:

If we put an equal number of pearls in each box, the probability of getting a black pearl is only 1/2. Since putting one black pearl in box A and the rest in box B does better than 1/2, the optimal solution must have more black pearls than white in one box and more white pearls than black in the other. Show that, the proposed solution maximizes the probability of getting a black pearl in both the box with more black pearls than white and the box with more white pearls than black.