There are two bears - white and dark. We may reasonably ask several questions:

Copyright © 1996-2009 Alexander Bogomolny

The above and the next two sections imitate the reasoning from [Gardner, p. 104].

Male bears, first solution to the last question

Since it's now given that the lighter bear is male there are only two possible outcomes (mf, mm). Thus the probability that both are male goes up to 1/2. Note how each additional piece of information changed the number of possibilities and, hence, the probability of the outcome.

Male bears, second solution to the last question

The sequence of three question is supposed to lead one on to wondering what difference does it make to specify that the white bear is male. And, in my experience, the trick works too. But since it's now known that the white bear is male, its sex is removed from the realm of random. All that matters is the sex of the dark bear who is believed to be male with the probability of 1/2. A short way to express the same idea is as follows:

where P(A|B) means the (conditional) probability of A provided B is known to take place.

Male bears, second solution to the second question

Since the third outcome mm may have occurred in two ways depending on which of the two bears was acknowledged to be male, the space of the possible outcomes is, in fact, (mf, fm, mm, mm) giving two favorable outcomes out of four equally probable. The probability of the second bear to be male is then 1/2.

This solution is quite unorthodox. To make it more digestable it may perhaps be helpful to denote the acknowledged male bear M and the other male (if any) m. Then indeed the sample space of the problem is (Mf, fM, Mm, mM). (Note a serendipitously acquired symmetry between unacknowledged male and female bears.)

Male bears, third solution to the second question

At the outset, there are four equally probable events, ff, mf, fm, mm. Assuming any of these, the probabilities of one of the bears being a male equal 0, 1/2, 1/2, 1. Once, one of the bears was found to be male, the probabilities of the four possibilities change but remain proportional to the above. Since the latter probabilities ought to add to 1, they become 0, 1/4, 1/4, 1/2. Only in the last case, the second bear is a male, making probability of this event 1/2.

M. Gardner appears to make an error in his solution (the first one). The progression of probabilities from one question to the next looks appealing, but the reasoning is faulty. The space of possible outcomes indeed changes, but the probabilities in the last two questions remain the same.

Reference

1. M. Gardner, aha! Gotcha. Paradoxes to puzzle and delight, Freeman & Co, NY, 1982

Thanks are due to my son David for the two beautiful pictures.

Copyright © 1996-2009 Alexander Bogomolny