puzzle

Can someone help me with this puzzle?

There are 2 boxes A and B. Both contain red and green balls. It is known that in one of the boxes, 1/2 of the balls are red and 1/2 are green; and in the other box, 1/4 of the balls are red and 3/4 are green. Let the box in which 1/2 of the balls are red be denoted box W and suppose P(W=A) = 'X' and P(W= = 1 - 'X'. Suppose you may select one ball at random from either box and that, after observing its color, you must decide whether W=A or W=B. Prove that if 1/2 < X < 2/3, to maximize the probability of making a correct decision, you should select the ball from box B. Prove also that if 2/3 < X < 1, then it does not matter from which box the ball is selected.

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Maybe there is a faster route. Here is the long and plodding one...

Let us call P_A(X) the probability of being correct when choosing box A. Correspondingly, P_B(X) = P_A(1-X) is easy to work out. We are checking whether P_A or P_B is bigger.

Suppose we choose box A. In general the strategy could be - If we find a red ball in A, we call A 'W' with probability p, and if we find a green ball in A, we call B 'W' with probability q.

Prob of being correct = Prob (A = W) Prob (Calling A 'W' A = W) + Prob (B = W) Prob (Calling B 'W' B = W)

Prob (A = W) = X, Prob (B = W) = 1-X Prob(Calling A 'W' A = W) = p/2 + (1-q)/2 = (1 + p - q) /2 Prob(Calling B 'W' B = W) = (1-p)/4 + 3q/4 = (1 - p + 3q)/4

=> P_A = X(1+p-q) /2 + (1-X)(1-p+3q)/4 P_A(X) = (p (3X-1) + q (3-5X) + X + 1)/4

We can choose p and q appropriately to maximise P_A. Thus, p = 0 for X < 1/3, and 1 for X > 1/3. ly, q = 1 for X < 3/5 and 0 for X > 3/5 gives the best results for P_A, say P_A*. => P_A* = 1 - X for X < 1/3 (3 - X)/4 for 1/3 < X < 3/5 X for X > 3/5

Now P_B(X) = P_A(1-X) => P_B = (p (2-3X) + q (5X-2) - X + 2)/4 => p must be 1 for X < 2/3, and 0 for X > 2/3 and ly q must be 0 for X < 2/5 and 1 for X > 2/5

=> P_B* = 1 - X for X < 2/5 (X+2)/4 for 2/5 < X < 2/3 X for X > 2/3

Comparing, for the region the problem asks - viz. X in (1/2, 2/3), P_A = (3-X)/4 for 1/2 < X < 3/5 and X for X > 3/5 P_B = (X +2)/4 for 1/2 < X < 2/3 It is easily verified that (X+2)/4 > (3-X)/4 as well as X in the respective regions, and we are better off choosing box B and following the calling strategy determined by p = q=1.

For the other region the problem asks - viz. X in (2/3, 1), we have P_A = P_B = X. So there really isn't any advantage choosing either box.

P.S. for X > 2/3, you are supposed to say box A is W irrespective of what ball and which box - to be correct most of the time. In other words, the results of the experiment itself is not contributing anything more to our knowledge. All the p/q algebra is needed to bring out this (perhaps counter intuitive) strategy.