And now for the follow-up.
The question was, if someone has two children, and tells you that (at least) one of them is a boy born on a Tuesday, what is the probability that the other child is also a boy? And the intended answer, of course, is 13/27. Why? Well, including the birth day (of week) as a variable, at the outset for a two-child family there are 14
2 = 196 possible and equiprobable pairs such as (B-Mon, G-Wed) where each child has a gender and birthday (of week) and the ordering denotes birth order. The vast majority of these pairs don't have a Tuesday boy and can be ignored. Of those that do, 7 of them look like (B-Tue, G-any) and another 7 are (G-any, B-Tue). Similarly, there are 7 pairs (B-Tue, B-any) and 7 (B-any, B-Tue). But wait! The case (B-Tue, B-Tue) has been counted twice! In fact there are only 13 cases where one child is a Tuesday boy and the other is also a boy. So that makes the probability 13/(13+14) = 13/27 where one child is a Tuesday boy and the other is also a boy.
However, there's a sting in the tail. Let's say you take a bunch of people with two children, and ask them if one of their children is a boy. Pick such a parent at random. We have already seen that in 1/3 of cases where the answer was yes, the other child will also be a boy. Now ask this parent for the day of the week that their son was born on (if they have a choice, they can flip a coin and pick one son at random, preferably out of sight so as not to give the game away). You get the reply..."Tuesday". Or some other day. Maybe they will say Saturday. In this case, hearing the day of the week on which a son was born does *not* change the probability that the other child is also a boy. So anyone who interpreted the original situation as meaning "a parent of two children, at least one of whom is a boy, tells you the day of the week on which their son (or a random son if applicable) was born, and their answer was `Tuesday'" would be completely justified in concluding that the probability of the other child being a boy was still 1/3. This solution was mentioned in
More or Less on 11 June, and it was (re)listening to this old podcast (probably no longer available, but
here's a related web page) that prompted me to blog it at last. What Tim Harford could have gone on to point out, but didn't (IIRC), is that even the original "two children" problem is typically under-determined in the way that it is presented. If a parent of two children is asked for the gender of one randomly-selected child, then irrespective of their reply, the probability of their other child being a boy (or alternatively, being the same gender as the one they gave) is...50%. So the solution to this problem also depends on how this "one child is a boy" parent is found.
Similar weaknesses can be found in most statements of the Monty Hall problem too. If all that is reported is the observed actions of the game show host on one occasion, then we don't really have enough information to generalise to a rigorous frequency-based calculation. Maybe Monty only opens another door on the occasions that the player originally picked the car, in which case swapping will lose. Maybe Monte opens a random door (and might expose the car)...in which case the player should still swap, and in fact the probabilities are unchanged from the original problem, in fact.
Wikipedia discusses several alternative interpretations in some detail.
Contrary to Gary Foshee's statement on
this web page, I don't think this sort of ambiguity is particularly controversial, it's just the result of trying to dress up a mathematical problem in natural-sounding (but slightly imprecise) English. When the problem is stated unambiguously, it's not particularly difficult. At least for pigeons.
And to any pigeons still reading, all I can say is: coo.