Best Of

Child B.

4 apples. 1 definitely good. The remaining 3 each have a 50% chance of being good. So that is an overall probability of 62.5% of picking a good apple from the bag.

Meanwhile for Child A there is only a 50% probability.

If you take it as pure maths and trust that at least 1 of his apples is good, then you've ruled out the permutation where all 3 are bad and left the other 7 permutations where at least one is good.

I think overall that gives you a 12/21 chance of picking a good apple from child A, so about 57%.

Agreeing on the 62.5% on child B, that still, to me, makes B the better bet.

I think you are right on child A but off on child B. Basically, both children have shown the same thing: they have at least one good apple. That rules out the permutation where all apples are bad. That biases the probability of getting a good apple higher, above the basic 50% of the underlying distribution. The size of that bias goes down as the number of apples goes up. With one apple, the bias is worth 50pp: if you know the one apple in the bag is good then your chance of getting a good apple is 100%. With a million apples in the bag, knowing that at least one is good has no effect because the chances of the bag containing no good apples is zero, and so the bias is effectively zero. So you are better off picking from the bag with fewer apples.

Yes, that's my view. Fascinated to hear if I'm wrong because it would be so counter-intuitive.