I've written an explanation, and will post it whenever you'd like me to, Brian. (O'Henry or Norm may not be around this weekend, and they may not be interested in thinking about this.)

Added later: or I could PM you but not post the explanation until we see if O'Henry wants to think about this at all.

Problem: I have 20 red and 10 blue stones in a box. I pick two at random, and then pick one of those at random and tell you it is blue. What is the probability that the other is blue?

1. If you have a set of n objects and you select k of them, a combination is the number of ways you can select the objects if order does not count: that is, AB and BA are the same combination. The number of combinations is abbreviated nCk (pronounced n choose k). The formula for it is n!/(k! • (n–k)!) – you can find this explained on the internet, and many scientific calculators have a key combination for it.

In simple situations, one can figure a combination without a formula. For instance, 5C2 = 10, for the following reason. There are 5 ways to pick the first object, and then 4 ways to pick the second object from the remaining set, which leaves 20 ways to pick the two objects in order. However, every combination can be picked in two different ways (AB = BA), so 20/2 = 10 combinations.

(Numbers of the form nCk are called combinatorials, and also binomial coefficients, and are integral to that fascinating object, Pascal’s Triangle. They are very important in mathematics.)

2. In our situation, we have 30 stones and we are going to choose 2 of them. The number of ways to do this is 30C2 = (30 x 29)/2 = 435.

3. Question: of those 435, how many of them will be both blue? Since there are 10 blues and we are choosing 2 of them, 10C2 = (10 x 9)/2 = 45. Therefore the probability of choosing two blues is 45/435 = 10.3%. This could be written as 10C2/30C2.


4. Likewise the probability of choosing 2 reds is 20C2 = (20 x 19)/2 = 190, and P(two reds) = 190/435 = 43.7%. This could be written as 20C2/30C2

5. We can calculate the probability of choosing one red and one blue. First, since the only three possibilities are 2 blue, 2 red, or 1 red and 1 blue, we can just subtract the answers in 2, and 3. From 100%: P(1 red and 1 blue) = 100% – 10.3% – 43.7% = 46.0%.

We can also calculate P(1 red and 1 blue) in a better, more general way: there are 20C1 = 20 ways to choose a red, and 10C1 = 10 ways to choose a blue, so the number of ways to choose a red and a blue is 20 x 10 = 200. Therefore, P(1 red and 1 blue) = 200/435 = 46.0%. This could be written as (20C1 x 10C1)/30C2.

6. Now to answer our question. You are told that one of the stones is blue. There are 200 ways that only one is blue, and 45 ways that both are blue, so there are 200 + 45 = 245 ways that at least one is blue. However, there are only 45 ways the second is blue. Therefore, the probability that the second stone is blue, given that the first stone is blue, is 45/245 = 18.4%

7. In general, the answer was (number of ways to choose 2 blues)/(number of ways to choose 2 blues + number of ways to choose 1 blue.)

If we apply this general solution to O’Henry’s original problem, with four stones and two blues, we have:
number of ways to choose 2 blues = 2C2 = 1
number of ways to choose 1 blue = (2C1 x 2C1) = 4
Therefore, the probability that the second stone is blue, given that the first stone is blue, is 1/(1 + 4) = 1/5.