Drunkard's Walk - Leonard Mlodinow [27]
The next step, according to Cardano, is to sort through the outcomes, cataloguing the number of heads we can harvest from each. Only 1 of the 4 outcomes—(heads, heads)—yields 2 heads. Similarly, only (tails, tails) yields 0 heads. But if we desire 1 head, then 2 of the outcomes are favorable: (heads, tails) and (tails, heads). And so Cardano’s method shows that Alembert was wrong: the chances are 25 percent for 0 or 2 heads but 50 percent for 1 head. Had Cardano laid his cash on 1 head at 3 to 1, he would have lost only half the time but tripled his money the other half, a great opportunity for a sixteenth-century kid trying to save up money for college—and still a great opportunity today if you can find anyone offering it.
A related problem often taught in elementary probability courses is the two-daughter problem, which is similar to one of the questions I quoted from the “Ask Marilyn” column. Suppose a mother is carrying fraternal twins and wants to know the odds of having two girls, a boy and a girl, and so on. Then the sample space consists of all the possible lists of the sexes of the children in their birth order: (girl, girl), (girl, boy), (boy, girl), and (boy, boy). It is the same as the space for the coin-toss problem except for the way we name the points: heads becomes girl, and tails becomes boy. Mathematicians have a fancy name for the situation in which one problem is another in disguise: they call it an isomorphism. When you find an isomorphism, it often means you’ve saved yourself a lot of work. In this case it means we can figure the chances that both children will be girls in exactly the same way we figured the chances of both tosses coming up heads in the coin-toss problem. And so without even doing the analysis, we know that the answer is the same: 25 percent. We can now answer the question asked in Marilyn’s column: the chance that at least one of the babies will be a girl is the chance that both will be girls plus the chance that just one will be a girl—that is, 25 percent plus 50 percent, which is 75 percent.
In the two-daughter problem, an additional question is usually asked: What are the chances, given that one of the children is a girl, that both children will be girls? One might reason this way: since it is given that one of the children is a girl, there is only one child left to look at. The chance of that child’s being a girl is 50 percent, so the probability that both children are girls is 50 percent.
That is not correct. Why? Although the statement of the problem says that one child is a girl, it doesn’t say which one, and that changes things. If that sounds confusing, that’s okay, because it provides a good illustration of the power of Cardano’s method, which makes the reasoning clear.
The new information—one of the children is a girl—means that we are eliminating from consideration the possibility that both children are boys. And so, employing Cardano’s approach, we eliminate the possible outcome (boy, boy) from the sample space. That leaves only 3 outcomes in the sample space: (girl, boy), (boy, girl), and (girl, girl). Of these, only (girl, girl) is the favorable outcome—that is, both children are daughters—so the chances that both children are girls is 1 in 3, or 33 percent. Now we can see why it matters that the statement of the problem didn’t specify which child was a daughter. For instance, if the problem had asked for the chances of both children being girls given that the first child is a girl, then we would have eliminated both (boy, boy) and (boy, girl) from the