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

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.

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.

## 10 comments:

coo coo ca chew?

Coo to you too.

But Ron, are you the walrus?

I love this problem! In the US, if the first child is a boy, that actual data shows that 52.2% of the time the second child is a girl.

http://www.in-gender.com/XYU/Odds/Gender_Odds.aspx

Even though there is a 51/49 Boy/Girl birth ratio. :)

Dallas, think you might have misread that, the page seems to say the probability of the second child being a girl if the first is a boy is actually 50%, though I agree the ratios aren't usually so even...except in maths puzzles!

Yes, the probability is 50% according to the link, the actual results, last table I think, showed the results of their survey as 52.2% chance of mixed (i.e. boy + girl) in a two child situation.

So my "first child boy" was misleading, though, there is a slightly greater probability of a first child boy (51/49 birth ratio).

Their actual results were very close to the normal birth ratio rather than the rounded 50/50 estimate.

So, I my opinion, the 13/27 prediction should actually be, 50/50, if you consider the historic birth ration :)

So far the pigeons are batting 1000.

Exactly :)

Shorter explanation: the problem isn't a probability problem at all, it's a sampling problem. And as the sampling method is never stated (as least not in any formulation I've ever seen), the answer cannot be determined.

You can take it one step further. Pick one parent from the same bunch of people with two children, and ask them to tell you the gender of one of their children. You get the reply "boy." The probability that this parent has two boys in indeed 1/2, not 1/3, for the exact same reason hearing "Tuesday" did not change the answer in your variation. If there are 100 parents in this group, 75 will *have* a boy in this group, but only 50 will *say* that they have a boy; 25 of those that could will say "girl." In either subgroup, 25 will have two boys, so the probability is different in the two cases.

Most versions of this question are ambiguous because they do not distinguish between those two cases. 1/3 (or 13/27) is correct only if you were looking for a boy (or a boy born on Tuesday). If it was a random fact, and the fact appropriate for either child could have been stated, the answer is always 1/2. And if you aren't told which is true, you can assume the latter cases (because it represents no bias), but not the first.

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