# Bear cubs problem  There are two bears - white and dark. We may reasonably ask several questions:
1. What is the probability that both bears are male? Writing 'm' for male and 'f' for female and counting the lighter bear first we get four possible outcomes (ff, mf, fm, mm) of which only one should be considered favorable. The answer, therefore, is 1/4.

2. Now assume I told you that one of the bears is male. What is the probability that both are males? Of the three possible outcomes (mf, fm, mm) only the last where both bears are male is favorable. The answer is 1/3.

3. The third question. I am telling you that the lighter bear is known to be male. What now is the probability that both of them are males? Please stop for a while and think of the problem. Try to answer the question before looking at the solution.

4. The fourth question. It is happen to be known that one of the bears is male and was born on a Tuesday. What is the probability that the other bear is a male? The surprising solution to this question has been posted in a separate file.

• The Basic Rules of Counting
• Coincidence
• Dependent and Independent Events
• An anecdote: Linus Pauling's Argument
• Multiple of 3 out of the Box
• The above and the next two sections imitate the reasoning from [Gardner, p. 104].

### Male bears, first solution to the third 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 third 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:

P("both are male" | "white is male") = P("dark is male")

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 palatable 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, 1, 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, 1/3, 1/3, 1/3. Only in the last case, the second bear is a male, making probability of this event 1/3.

## 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. • 