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The Boy or Girl Paradox

November 27, 2014

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 red door and boy behind yellow door.
  • 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. The following week, you see a man walking along the street, accompanied by a young boy. You think you recognise him and ask him if you know him. “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!” “No, you’ve got the wrong guy,” he 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 at the sales convention has a daughter and the chance that this random guy on the street has one as well. It is just like the Two Doors problem. It’s 2 in 3 in both cases. Or is it? Does it matter how you found out about the boy? 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, but not definitely. 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 could be compared to asking him to toss a coin and produce one of his children at random. 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. You now 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, 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. The different information sets can be compared to tossing a coin twice. 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 Vaughan Williams ‘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.

Further Reading and Links

https://selectnetworks.net

From → Probability, Puzzles

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