To this day, people get it wrong and defend it fervently.

]]>Peter said “It is reasonable to interpret this puzzle as asking you ‘What is the probability that X has two boys, given that at least one of the children is a boy’ in which case the answer is unambiguously 1/3—given the usual assumptions about no twins and equal gender frequency.” I maintain that this is *not* unambiguous.

In most fields of math, it would be. Information that is “given” to you is meant to be accepted as true. But that is not enough in probability. You need to construct an event in the sample space that represents what outcomes are possible. TO do that, you need to be “given” information that tells you not only what to include in the event, but what to exclude as well. That is, you need both a necessary and a sufficient condition, and being “given” information about one outcome only constitutes a necessary condition.

The error is easy to make in probability, tho, because the word “given” is used formally in similar context. In the usual reading of a conditional probability, P(A|B) means “the probability that an outcome in event A occurred, given that an outcome in event B occurred.” But what is being *given* here is not information about B, as in Peter’s statement above, but the already-defined set itself.

The unambiguous wording Peter intended to make is “What is the probability that X has two boys, given that at X was chosen from the set of families where at least one of the children is a boy.”

]]>I dread getting asked a question like this in an inteview – I can picture the interviewer phrasing it poorly and me stubbornly insisting on correcting him/her…

]]>If someone tells you he has 2 childern and at least 1 of them is a boy he has made a disticntion between the two children. One has had information about it revealed and the other has not. Therefore the two children can be labeled as Revealed and Hidden.

What are the possible Revealed/Hidden combinations that can exist before the information is revealed? Answer: B/B, B/G, G/B, G/G. After the information is revealed the possible combinations are B/B and B/G. G/B is not possible because the revealed child is a boy. Therefore the probability of 2 boys given that at least one is a boy is one half.

The same logic applies to the Tuesday Boy puzzle. The probability of 2 boys given that at least one is a boy born on Tuesday is one half.

I have sent this solution to New Scientist where I first read about this puzzle. ]]>

Now change the probability that the revealed child is a boy from 1/2 to 1 and the probability that the revealed child is a girl from 1/2 to 0. The tree will now calculate the probability of 2 boys given that the revealed child is certainly a boy. The calculated probability of 2 boys changes to 1/2 not 1/3.

Now change the probability that the revealed child was born on Tuesday from 1/7 to 1 and the probability of Not Born on Tuesday from 6/7 to 0. The tree will now calculate the probability of 2 boys given that the revealed child is certainly a boy born on Tuesday. The calculated probability of 2 boys remains at 1/2.

What is wrong with this procedure? ]]>

When I read “This is the first year since at least 1990 that Tuesday wasn’t the biggest birth day” at this page, http://www.babycenter.com/0_22-surprising-facts-about-birth-in-the-united-states_1372273.bc (based on 1996 data), I thought there is just much more to this boy-puzzle-question, and adding in the distribution of identical twins makes it just too silly altogether.

Stephan

]]>Here is a bit of imaginary dialogue:

Math Whiz: ‘… so if someone tells you she has two children and at least one son, the probability is 1/3 that she has two sons!’

AC: ‘…and, likewise, is she tells you she has two children and at least one daughter, the probability is 1/3 that she has two daughters?’

MW: ‘Of course…..’

AC: ‘so, whatever she tells you, the chances are 1/3 that if she has two children, they will be two sons or two daughters? I’ve always been able to convince myself that 1/2 randomly selected 2-child families will have same-sex children….’

==== silence =======

]]>The two usually come together very easily — by supposing that the particular case in one which one finds oneself has been sampled at random from a large number of sets of circumstances, which are equivalent in all relevant respects.

When puzzles like this have an unequivocally correct answer, it is because they have been carefully framed according to the frequency or random sampling model.

The trouble is that in this form, they may make good examples for a class test in Probability 101, but they don’t make good _puzzles_: ‘Who cares?’ would be commonest response to ‘Given that a child picked at random from a family picked at random from a large set of two-child families, is a boy, what is the probability that both children in this family are boys?’

When such a puzzle is made interesting by being framed in ordinary speech — as Gary Foshee apparently did — then commonly no answer is unequivocally correct.

The clearest way to see where this equivocation arises is by using the Bayes formulation; to calculate the probability unequivocally, one would need to know the probability that the speaker will say ‘I have at least one boy’ (rather than ‘I have at least one girl’) in the case that s/he has one of each. ]]>

Seems to me that everyone’s overlooking the key factor here, which isn’t so much about the question being asked and whether or not it’s vague, but the mechanism proposed for producing the answer. That is, the probability hinges not on which father you picked, but on what exactly you’ve established that the father is going to say, when he announces his result.

That is, if the father will announce “My first child is a son, my second child is a son” then the odds of him making that statement (everything being equally probably) is 1/4. Because there are only four distinguishable outcomes. If in that scenario I was to tell you, before the answer was revealed, that one of the children was born on a Tuesday it wouldn’t change anything – there are still 4 distinguishable outcomes, thus the odds are still 1/4. If I told you one of the children was a boy, then now there are only 3 possible outcomes, and the odds are 1/3. Telling me that one child is a boy born on Tuesday doesn’t provide any additional information, no more so than telling you one child is a son and I had a salad for lunch would – because the number of distinguishable outcomes hasn’t changed.

Note that if the father was to announce “My other child is a son,” then the odds of that child being a son or daughter are 1/2 (and it doesn’t matter what the first child is, or on what day it was born). Why? Because there are only two possibilities, and the problem didn’t establish any conditions or dependencies.

Finally, it’d be similar if the father were to announce without specifying the order, e.g., “two boys,” “two girls,” or “one of each,” then each outcome is 1/3, and if you stipulate one is a boy it’s now 1/2 that he has two boys.

Given the way the original puzzle was worded, I think it’s reasonable to conclude that Foshee intended the first case, that he would announce “the son born on tuesday is my first child, the second child is also a boy” in which case there are 3 possible outcomes given the stipulation and thus the odds are 1/3.

–s

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