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There is a red and a yellow door in the puzzle 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).

The following week, you see a man walking along the street, accompanied by a young boy. You think you recognise him as the guy you met at a Sales Convention the previous year. He had mentioned his two children and his son, and nothing further about them, and you had deduced the chance his other child was a girl was 2/3. It is like the 2-Doors problem. He had effectively told you there was a boy behind one of the coloured doors but nothing else, so you had excluded option 4 but left 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.

Back to the guy on the street. “Didn’t we meet last year 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 man on the street has a daughter is 1/2! Why is the probability different this time? Does it matter how you found out about the boy? 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 probably closer to 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 the boy is not age or height or alphabetical order of first name, but location. That leaves Option1 (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 him the man 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, that 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.

If you see the boy, however, and you know there was a 50-50 chance that you would do so if he had a boy and girl, that is different information to simply being told that he has at least one son. It is like identifying one of the coloured doors (say the red door) and peeking behind it. Say it’s a boy. That leaves an equal chance there is a boy or a girl behind the yellow door. The different information sets can be compared to tossing coins. The possible outcomes are HH, HT, TH, TT. If you already know there is ‘at least’ one head, that leaves HH, HT, TH. The probability that the remaining coin is a Tail is 2 in 3. If, on the other hand, you don’t have any information other than that a coin has been tossed twice, the possibilities are HH, HT, TH, TT. Equal chance of a Head or a Tail if you uncover one coin randomly. So there is a 50-50 chance you will see a Head if you uncover one of the coins randomly, and there is a 50-50 chance that the coin you haven’t uncovered is a Tail. What goes for coins goes also for people. Actually, it’s not quite that simple. This assumes that there was a 50-50 chance that the man in the street would choose a boy or a girl (if he had one of each) to accompany him on his walk! The more likely it is that he would choose to take a boy with him than a girl (if he had a girl), the more likely it is that the other child is a girl. It is like tossing two coins, but only ever choosing to show a Head if there is a Head and a Tail. In that case, the likelihood that the other coin is a Tail is 2/3. It is the equivalent of announcing that you have at least one Head (or one boy) whenever you do have one, but never allowing the possibility of a Tail (or a girl) to be observed. It can be stated this way. “If something that might exist is unobservable, it is more likely to exist than if it can be observed but isn’t observed.” This is the strong version of what I term the ‘Possibility Theorem.’ Bottom line is that when we only display the boy when he exists, but never display the girl when she exists, it is more likely that the unobserved child is a girl. If we display both with equal probability, and they exist with equal probability, then when we observe one, the probability that the other exists is the same, unless some other information exists. It’s the difference between knowing that a boy is behind the red or yellow door, or perhaps both, on the one hand, and opening the red door and finding a boy behind it, on the other hand, when we know that boys and girls have been assigned to the doors by the toss of coin. In the first case, the probability of a girl behind a door in the puzzle house is 2 in 3; in the case where we open the red door and reveal the boy, the probability of a girl behind the other door is 1 in 2. We are assuming, of course, that there is actually a 50-50 chance of a boy and a girl, a Head and a Tail. If we change the assumptions or change the information set, we change the answer. Paradox? Or common sense? You decide.

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’ (let’s call it the Vaughan Williams Possibility Theorem) shows that the actual probabilities can extend over a range between 1/2 and 3/4.

Further Reading and Links

https://selectnetworks.net

From → Probability, Puzzles

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