Children, coins, and Monty Hall
June 30, 2010 at 6:18 AM by Dr. Drang
You may have seen links to this Science News article floating around the internets this week. It describes a new variant on a old probability problem and a dispute about its solution. I think the article does a good job describing both the old and new problems and their conventional solutions, but not such a good job with the dispute. I want to discuss the old problem, the difficulties that can arise in interpreting it, and its relationship to another famous probability problem.
The original problem is sometimes called the Two Child Problem and can be stated like this:
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
The naive answer is 1/2: Boys and girls are equally likely and the sex of the child we’re told about has no effect on the sex of the unknown child.
The more sophisticated answer is 1/3: There are four equally likely permutations of two children. If we order the children by age, those permutations are girlgirl, girlboy, boygirl, and boyboy. Since we are told that there’s at least one boy, girlgirl is eliminated and we have three equally likely possibilities, only one of which has two boys.
The even more sophisticated answer is “Wait a minute—how did you decide which child to tell me about?”
Because there are some extraneous issues that can confuse the Two Child Problem (boys and girls really aren’t equally likely, and what about identical twins?), and because it doesn’t lend itself to thinking about repeated trials (if Mr. Smith keeps having two children again and again, Mrs. Smith will get pissed), I want to use a probabilistically equivalent problem:
I’m in my office. You’re in another room where you can’t see, hear, feel, smell, or taste what I’m doing. I flip two fair coins onto my desk, take note of how they landed, and cover each of them with an upsidedown cup. I call you into my office and tell you (honestly) what I’ve done. I then reveal one of the coins to be a head and ask you the probability that the other coin is also a head. What should you answer?
As with the Two Child Problem, there’s the naive answer of 1/2, the more sophisticated answer of 1/3 (based now on permutations of HH, HT, TH, and TT), and the even more sophisticated answer that we’re going to explore.
Strangely, what you should answer depends not on any immutable laws of probability, but on my behavior, which hasn’t been fully defined in the problem statement.
Is the problem set up so that I must always reveal a head? Do you know that before you enter my office?
Twentyfive percent of the time I’ll flip two tails; what should I do then? Leave you to rot in the other room? Flip again until I get at least one head? Change the problem and reveal a tail instead?
If you know ahead of time that I won’t ask you into the room unless I’ve flipped at least one head, and if you know that I must reveal a head, then you’re justified in saying that the probability that the other coin is a head is 1/3. But without those constraints on my behavior, you don’t have that justification.
Suppose the problem were stated this way:
I flip two coins onto my desk, cover them up and call you in. When you enter I reveal one of them and ask you the probability that the other is the same. What should you answer?
Clearly, this time you should answer 1/2, because the coins are the same in two of the four cases (HH, HT, TH, TT).
If my behavior isn’t constrained—or you don’t know that my behavior is constrained—you wouldn’t know the difference between the earlier coinflipping game and this one. So maybe the naive answer of 1/2 isn’t so naive, after all. This is the point made by Yuval Peres in the aforelinked Science News article.
We can look at the Two Child Problem the same way: are we asking this question only of men that have at least one son, or would we change the question to be about girls if Mr. Smith had two daughters? Unless you know the constraints, you don’t know what the best answer is.
If you’re familiar with the Monty Hall Problem, you may see some similarities between it and the Two Child Problem (or our coinflipping equivalent). In both cases, there’s an element of chance and you’re given information that may or may not influence your answer. Here’s the Monty Hall Problem, as given on its Wikipedia page:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host [Monty Hall^{1}], who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
I won’t give you the answer; if you don’t already know it, you can follow the Wikipedia link. The important point is that there is an answer, because Monty’s behavior—unlike mine in the coinflipping problem—is well defined:
 There is always a car behind one door and a goat behind each of the other two, and Monty knows which is which.
 After you make your first choice, there will always be another door with a goat behind it.
 Monty always reveals a door with a goat behind it.
Because of these constraints, and because you know these constraints going in, you can develop a strategy. In the Two Child Problem and our original statement of the coinflipping equivalent, you can’t develop a strategy until you know more.
If you’re having trouble following the logic, don’t despair—you’re in good company. According to the Science News article, Martin Gardner didn’t understand the subtleties of the Two Child Problem the first time he wrote about it.
Update 6/30/10
There’s a bit more about the Two Child Problem here

This setup is similar to the “Big Deal of the Day,” the last contest on each episode of the old TV game show Let’s Make a Deal. The host of Let’s Make a Deal was Monty Hall, hence the name of this problem.) ↩