In what I hope is a final attempt to clear up the Mr. Smith probability problem, I’ve got another explanation and a new problem involving a coin, a die and $100.

First, the explanation, from Craig Fox, the U.C.L.A. professor who got this argument started:

There seems to be some debate among your readers concerning the interpretation of the Mr. Smith problem as formulated by Martin Gardner (“Mr. Smith says, ‘I have two children and at least one of them is a boy.’ Given this information, what is the probability that the other child is a boy?” ). I agree that this problem is misleading by design—it focuses your attention on an arbitrary child (“at least one of whom is a boy”) and then focuses on an arbitrary “other” child (“the other is a boy”) which gives rise to the illusion that we are talking about specific children.

This is why our intuition, guided by the partition-edit-count strategy that I described Thursday, leads most of us to say ½, whereas the canonical interpretation of the problem yields a “correct” answer of 1/3. Of course the main point in the paper I described in Thursday’s post was not to defend Gardner’s wording but to illustrate how the partition-edit-count strategy can lead to predictable responses that vary with the wording of a problem.

In order to determine a “correct” probability for any problem we need to map the problem onto a focal outcome and a randomizing “experiment” that produced it. One way to do this for the Mr. Smith problem is to think of the sex of his children like two successive coin flips, with heads interpreted as a boy (B) and tails interpreted as a girl (G). If we flip the coin twice then the following possibilities are equally likely: {BB, BG, GB, GG}.

Thus, when we first learn that Mr. Smith has two children, we know that he has a 1/4 chance of having two boys, a 1/4 chance of having two girls, and a 1/2 chance that he has one of each (if you don’t believe me try flipping two coins several times and you will see that about half the time you get one head and one tail). Learning that “at least one” of Mr. Smith’s children is a boy tells us only that we can rule out GG as a possibility.

Assuming we interpret the condition “other child is a boy” as meaning “both children are boys” then it becomes clear that the possibility of two boys (given that at least one is a boy) is one out of three remaining possibilities {BB, BG, GB} yielding a probability of 1/3. When Jonathan Levav and I revised the wording of the problem (“Mr. Smith says, ‘I have two children and it is not the case that both are girls.’ Given this information what is the probability that both are boys?”) it helped our participants think about the problem in this way and led them to get the right answer almost ten times as often.

And now the new problem, also from Dr. Fox:

To test readers’ understanding of these principles, here is a less ambiguous variation of Mr. Smith that I developed with my friend Yuval Rottenstreich (a management professor at N.Y.U.):

Suppose that I offer you a prize of $100 if a fair coin lands “heads” and a fair six-sided die also lands “1”. After I flip the coin and roll the die, you ask me whether at least one of them is “good.” I say “yes.” Would you rather (A) continue with the bet and receive $100 if both are “good” or (B) start over with a bet that pays $100 if a new roll of a six-sided die lands “1”?

I’ll give the answer tomorrow. I’ll also give another explanation that — I again hope — will clear up the confusion over the three-card problem posed Friday.

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