Further and deeper exploration of paradoxes and challenges of intuition and logic can be found in my recently published book, Probability, Choice and Reason.

You meet a man at a sales convention who mentions that he has two children, and you learn that one of them, at least, is a boy. What is the chance that his other child is a girl? It’s not a half, as most people think, but 2/3. What now if see someone on the street accompanied by a young lad whom he introduces as his son, and mentions that he has one other child. What is the chance that the other child is a girl? 2/3? No, this time it is ½. Puzzled? So was a Mr. John Francis, who challenged Marilyn Vos Savant, of Monty Hall fame, who had proposed the equivalent of these answers in response to a question to her ‘Ask Marilyn’ column in 1996. He wrote: “I have a BA from Harvard, an MBA from the University of Pennsylvania Wharton School, a math SAT score of 800 and a perfect score in the Glazer-Watson critical thinking test, but I’m willing to admit I make mistakes. I hope you will have the strength of character to review your answer to this problem and admit that even a math teacher and the person with the highest IQ in the world can make a mistake from time to time.”

Despite his SAT score, she was right, and he was wrong.

Here’s a way to look at it. Let’s imagine a red and a yellow door in a house which you have just entered. You are told that there is either a boy or a girl behind each of the doors. The choice of whether to place a boy or a girl behind each door is determined by the toss of a coin. Well, there are four possibilities:

• 1. Boy behind both doors.
• 2. Boy behind red door and girl behind yellow door.
• 3. Girl behind red door and boy behind yellow door.
• 4. Girl behind both doors.

So the probability that there is a girl behind one of the doors = ¾. The new information I now impart is that at least one of the doors has a boy behind it. So we can delete option 4. This leaves three equally likely options, two of which include a girl. So the probability that there is a girl behind one of the doors given this new evidence is 2/3.

This is directly comparable to the pairs of options available in a two-child family. The chance of two boys is 1/4 (indexed according to some discriminating factor, e.g. by age – older boy and younger boy), chance of two girls is 1/4 (older girl and younger girl), chance of boy and a girl is 1/2 (older boy and younger girl plus older girl and younger girl). The discriminating factor does not need to be age, so long as it is uniquely discriminating between each element of a pair (e.g. height, alphabetical order of first name, number of fingers on left hand, behind a particular door).

Now consider the case of the man you met at the Sales Convention, the one with two children, and one at least was a boy. From this information, what is the chance that his other child is a girl? Well, it could be a boy and it could be a girl. So surely it’s 50/50. Actually, it’s 2/3, and the two-door problem gives the reason. In telling you that one of his children was a boy, he was telling you the equivalent of saying that there was a boy behind one of the doors but nothing else, so you had excluded option 4 (girl behind both doors), leaving options 1, 2 and 3 alive, with equal probability. Among these options, two contain a girl and only one a boy. So the chance his other child is a girl is 2/3. That was your reasoning.

Now what if you meet the guy on the street. “Didn’t we meet at a Sales Convention? If you recall, you told me you had two children, and you mentioned a son. Pleased to meet you, sonny!” You had, as we’ve seen, worked out that the chance his other child was a girl as 2/3. “No, you’ve got the wrong guy,” he politely replies. “I do have two children, but that’s pure coincidence.” Armed now with this information, you instantly work out the chance that the other child is a girl is not 2/3, but ½. Why is the probability different this time? Does it really matter how you found out about the boy, whether you were told about him or saw him for yourself? In fact, it does.

If you know that the man at the convention has two children and at least one son, the chance his other child is a girl is 2 in 3. If you find out that the man in the street has two children, but you don’t know that he has a son until you see him, the chance his other child is a girl is 1 in 2.  The reason is straightforward. If you just happen to find out he has a son, without knowing in advance that he has at least one son, it is like opening a specific door (say the red door) and finding a boy behind it. The door represents any unique piece of discriminating information. Location, age, and so on would do as well. Now this rules out Option 4. It also rules out Option 3. The discriminating factor here in seeing a particular boy, not a generic boy. That leaves Option 1 (the remaining child is a boy) and Option 2 (the remaining child is a Girl). So what is the chance that the child behind the other door is a girl? If the probabilities are equal, they are both 1/2.

The problem can also be compared to asking the man at the Convention, before you know that he a boy, to toss a coin and produce one of his children at random. Or maybe draw a card from a deck. Black card means boy, red card means girl. In the case that he has a boy and girl, we might assume there is a 50-50 chance that he will produce a boy, and so that leaves a 50-50 chance that the remaining child is a girl. So, in any case where you find out that he has a boy, without any other information except that he has two children, a reasonable estimate of the chance that the remaining child is a girl is 50-50. But if you know that he has at least one boy, it is different information. Different information changes everything. It is like knowing that there is definitely a boy behind one of the doors but without knowing which. That rules out Option 4, but leaves the other three Options live, and there is a girl remaining in two of the Options (Options 2 and 3) and a boy in just one of them (Option 1).  So the chance the other child is a girl is 2/3. Put another way, you know that there is a 1 in 3 chance that he has two boys (BB, BG, GB) and so the chance the other child is a girl is 2 in 3.

Appendix

Two solutions to the Boy-Girl Paradox using Bayes’ Theorem

Solution 1:

Consider a room containing two children. Assume equal probability that each child is a boy or a girl.

Let P (BB) = Probability of 2 boys; P (GG) = Probability of 2 girls; P (BG) = Probability of a boy and a girl.

So P (BB) = 1/4

P (GG) = 1/4

P (G and B) = 1/2

Now add the assumption that AT LEAST one is a boy. Represent this by B.

Let P (A I B) mean the probability of A Given B

So P (BB I B) = P (B I BB) x P (BB)/P(B) … by Bayes’ Theorem

P (B I BB) = probability of AT LEAST one boy given that both are boys = 1

It is clear that if both children are boys, at least one of the children must be a boy.

P (BB) = Probability of both being boys = 1/4, from the prior distribution

P (B) = probability of AT LEAST one being a boy, which includes BB and G.B = P (BB) + P (G.B) = 1/4 + 1/2 = 3/4

So P (BB I B) = 1/4 / 3/4 = 1/3, i.e. the probability that the other child is a boy is 1/3. The probability is 2/3 that the other child is a girl.

In other words, if at least one of the children is a boy, the chance that the other is a girl is 2/3.

Assume P (b) is the probability of drawing a boy from all possible cases (i.e. without the AT LEAST assumption).

Then, P (BB I b) = P (b I BB) x P (BB) / P (b)

= 1 x 1/4 / 1/2 = ½

Note that P (b), the probability of a boy in a random two-child family we can assume for present purposes to be the same as the probability of a girl, so the probability of each is ½.

P (b I BB) is the probability of a boy in a two-boy family, and is clearly 1.

P (BB) is the probability of both being boys = ¼ from the prior distribution.

So P (BB I b) = P (b I BB) x P (BB) / P (b) = 1 x 1/4 / 1/2 = 1/2, i.e. the probability that the other child is a boy is 1/2. The probability is 1/2 that the other child is a girl.

Solution 2:

What is the probability that someone has a girl, if he has two children, and you see him with his son?

Assuming that it was equally probable that you would have seen him with the boy or the girl, then the probability of seeing him with boy is ½, and with a girl is ½, if he has one of each. The correct calculation is now:

Let G2 be the probability he has a girl and P(B1) that he has a boy.

We want to find P(G2|B1).

P(G2|B1) = p(B1 and G2)/p(B1)

P(G2|B1) = 1/4/1/2 = ½

This is standard Bayesian analysis, but a complete solution requires an appeal to a “Possibility Theorem”.

“If something that might exist can’t be observed, it is more likely to exist than if it can be observed (with any positive probability) but isn’t observed.” (STRONG VERSION)

“If something might exist, it is more likely to do so than something which is more likely than it to be observed, but isn’t.” (WEAK VERSION)

This is critical in either formulation when assessing how the probability of a hypothesis being true might be affected by information which potentially exists and is relevant but is missing because it is for whatever reason unobserved or unobservable.

In this context, we must consider how likely is P(G2) to be true given the likelihood we would see the girl if she did exist.

If this is ½, the standard probability calculation is correct.

If there is a less than ½ chance that she would be observed even if she exists, however, then by appeal to this ‘Possibility Theorem’, the probability that she does exist is greater than half.

In particular, we require the more generalized rule:

P(G2IB1) = [P(G2 and B1) I (PB1Obs)/(PB1)] / [P(B1)IP(B1Obs)/P(B1)]

where P(B1Obs) = Probability that a boy would be observed

and so P(B1Obs)/P(B1) is the ratio of the probability that a boy would be observed to the actual probability of a boy existing.

In this case, this ratio could extend between 0 and 1, with 0.5 being a special case in which the probability that the other child is a girl is 0.5.

In the event that there is a zero probability that she would be observed even if she exists, P(G2IB1) = [P(G2 and B1) I (PB1Obs)/(PB1)] / [P(B1)IP(B1Obs)/P(B1)] =(1/4 x 2) / ¾, i.e.

the probability that she exists rises to 2/3.

In this context, knowing that someone has at least 1 boy is equivalent to observing the boy when the girl, should she exist, is unobservable.

P(G| at least 1 son) = p(G and at least 1 son)/p(at least 1 son)

P(G| at least 1 son) =1/2 / ¾ = 2/3

Without appeal to the ‘Possibility Theorem’, the solution to the paradox converges to 1/2 or 3/4. Appeal to the ‘Possibility Theorem’ shows that the actual probabilities can extend over a range between 1/2 and 3/4.

Exercise

1. You overhear a conversation in a bar in which a woman mentions her two children and also mentions her daughter. What is the probability that her other child is a boy?
2. The following week you see a woman in a park, and overhear her introducing her daughter, Barbara, to someone. “My other child’s at home”, you hear her say. What is the probability that her other child is a girl?