This has been killing me all week. It’s a probability problem known as the “Tuesday boy” problem. I’ll simplify the problem by reducing the possibility space.
Alice has two children. What is the probability that she has two boys given that:
a) at least one of her children is a boy?
b) at least one of her children is a boy; and at least one of her children is left handed?
c) at least one of her children is a boy, and he is left handed?
Assume that left/right handedness are equally likely.
Let’s look at a). A naive first answer might reason as follows: we know that one child is a boy. The probability of the other child being a boy is a half. Thus the probability of two boys is a half. And this is wrong, because it doesn’t count the number of possibilities correctly. There are not two relevant possibilities: boy and girl. There are, in fact, three when you take the order of the children into account: girl-boy, boy-girl, boy-boy. So the answer to a) is one third.
Our first reaction to b) might be that, as we are being asked about gender, the left-handedness of one of the children is irrelevant. And, in this case, we are correct. To see why, let’s write out the possibilities: let B/G = boy/girl, L/R = left/right handed. There are 9 ways to have one boy, and one left-hander:
(BL)(BL) ; (BL)(BR) ; (BR)(BL) ; (GL)(BL) ; (GL)(BR) ; (GR)(BL) ; (BL)(GL) ; (BL)(GR) ; (BR)(GL).
Of these, the first three have two boys, giving a probability of a third, as before. This is a general feature of probabilities: adding useless information doesn’t change your probabilities.
It seems that the exact same reasoning would apply to c). Knowing that the boy is left-handed shouldn’t change anything. But let’s have a look at those possibilities …
(BL)(BL) ; (BL)(BR) ; (BR)(BL) ; (GL)(BL) ; (GL)(BR) ; (GR)(BL) ; (BL)(GL) ; (BL)(GR) ; (BR)(GL). (1)
Thus the correct answer to c) is 3/7.
Huh?! We can reverse our previously given axiom: if the probabilities have changed, then we must have been given useful information. But how is the fact that the boy is left-handed relevant to the probability that the other child is a boy? In fact, the more information given about the one child that is a boy, the more likely it is that the other child is a boy. For example, if we’re told that “one child is a boy, born on Tuesday”, then the probability of two boys is 13/27. If we’re told he was born on the 1st of October, then it’s 729/1459 = 0.4996. The probability approaches a half.
This is bonkers, isn’t it? Suppose Alice is making you guess the gender of her other child. A bully, unbeknownst to Alice, is making you guess “boy”. You have a 1/3 chance of being correct. You have a stroke of genius. “Is your boy left-handed?”, you ask. It doesn’t matter what the answer is – once you know, your chances of being correct rise to 3/7. The more questions you ask, the better your chances. But how? How is this information relevant?
The resolution of this paradox comes by looking more closely at the possibilities left behind in (1) above. Let’s get together a huge group of two-child families. Send “girl-girl” (GG) families home – one quarter depart. The BB families make up a third of the remainder. Now, send home any family that doesn’t have a left-hander. Once again, one quarter leave. BB still makes up a third. Now send home any family that doesn’t have a left-handed boy. All the BB families are safe. But the some of the GB and BG families must leave – those where the boy is right handed.
The lesson is as follows: the information that the boy is left-handed is useful information because it removes the scenarios where Alice has one right-handed boy and one left-handed girl, but doesn’t remove the possibility that Alice has one right handed boy and one left-handed boy. The more restrictions are placed on the boy, the less likely that a family with only one boy will be able to fulfill those restrictions.
The more pressing issue is that our intuitions are often mistaken when it comes to probability.