Today's I&I segment comes via my younger brother, who directed me to this thread on the Something Awful Forums, which after some time devolves into arguing about how probability works with respect to the particular problem. My aim here is to point out the problem, and elucidate the proper solution. So without further ado, I present to you the Boy or Girl Paradox:
You are walking around town when you come upon a mother and her son. She makes chit chat with you, and mentions that she went to the park on a picnic with her two children the other day.
Assume for a moment that the odds a girl or boy is born are equal. Now, what is the probability the other child is a girl?
There are two answers that crop up continually throughout the thread: Either the probability of the unobserved child being a girl is 1/2, or the probability of same is 2/3. Now, there are all kinds of ways you can organize the information when you work this through, but the two ways it's done most in the thread is age, and whether or not the child is observed. The 2/3 answer comes up as a result of a mistake in the interpretation; either there's a missed piece of information, a misinterpretation of what the mother is actually saying, or it's mistaken as a Monty Hall Problem (which I won't be getting into here; that's a whole different can of worms.).
The major misinterpretation of this problem is as follows: Some people read this and think the question is asking what the probability of a child being a girl will be given that at least one of the two is a boy. In that case, the probability IS 2/3; There are four possible configurations, BB, BG, GB, GG, and based on what you're told, only one of them, the GG case, is invalid. The problem with this? This isn't what the question is asking.
One of the issues with ordering the children by age is that you're not given any information about the ages, which can lead to this false interpretation, as outlined by HHHH:
Here's how you get 2/3:
There are four configurations of sexes, each with equal probability:
BB, BG, GB, GG
Obviously the last one is impossible since at least one is a boy:
BB, BG, GB
So this means that there are only three possibilities. Since a girl shows up in two of the three possibilities, and we've already established they're equal, the probability of the other child being a girl is 2/3!
Yes, I know it's wrong, but the more challenging question is what is the fallacy with this logic?
The fallacy here is that although you know one of the children is a boy, you also don't know which child she brought, either child A or child B. Because of that, there are two possible cases for the BB arrangement. So, labeling this by age, and first-born with a subscript 1, second-born with a subscript 2, we get the following cases for the boy being brought to the park:
| Brought | Left |
| B1 | B2 |
| B2 | B1 |
| B1 | G2 |
| B2 | G1 |
The crowd that are insisting that the probability is 2/3 are assuming in the BB case that the woman is bringing the first son, and forgetting that it's possible that the second son could be with her, and the first at home. They're arranging it by age, but forgetting that there are two possible arrangements for the two-boy case, because there are two distinct boys in that case. As such, the probability of the other child being a girl is in fact 1/2.
Now, let's look at the problem another way, also equally valid: Observed or unobserved. The table here goes as follows, first child is observed, second is the unobserved:
BB, BG, GB, GG.
Again, four possibilities here, and we can scrub the last two, since the observed child is not a girl, thus leaving BB and BG as the two remaining possibilities, and the probability of the unobserved child being a girl is once again 1/2.
In truth, this problem is trivial; the question gives us the answer straight up. The probability of a child being either a boy or a girl is equal. Since the gender of the observed child has no bearing on that of the second, children being independent trials, we can say straight off that the probability of the second child being a girl is 1/2.
This problem was even attacked by one poster using Bayes' Theorem, and not only did the answer turn out to be 1/2, but when he used the probabilities of 2/3 for a girl and 1/3 for a boy, Bayes' theorem yielded a probability sum of 4/3, which is impossible; by definition, the sum of all probabilities in any given scenario must be 1.
Of course, given that the only two possibilities for a chromosome pair (trisomies notwithstanding) are XX (female) and XY (male), biology on its own bears out that the probability must be 1/2. Why do you think there's a roughly even split of men and women in the world? SA user semicolon said it best:
It's 50% because biology doesn't give two shits about your tables and theorycrafting.
No, it most certainly does not.
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