## Wednesday, July 7, 2010

### Tuesday's Child

Here is a post at the corner concerning a probability question. It leads to a number of other posts and to Derbyshire's analysis here. If you follow the chain of posts back, it leads to other sites and considerable debate over the interpretation of this problem.

The analysis is tricky only if the problem is interpreted in something other than its straightforward, plain meaning. The statement of the problem is:

"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

Now the problems all come from that "born on a Tuesday" clause in the middle sentence. Take that out, and everyone agrees on the answer:

"I have two children. One is a boy. What is the probability I have two boys?"

This is the classic coin-flip enumeration problem. Having children is like flipping a coin, with heads = boys and tails = girls. The possible outcomes for two consecutive coin flips are:

Tails - Tails

or, in the boy girl terms:

Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl

Since we know that at least one of the children is a boy, the last case is ruled out and the probability that the speaker has two boys is one in three.

Returning to the original problem, the analysts all seem to think the phrase "born on a Tuesday" is very significant, but they can't agree on its significance. I don't think it adds anything to the problem at all. In the straightforward, obvious interpretation, it is only a statement after the fact of birth concerning the day of birth. It's like saying the boy was 8 lbs at birth, or was born with blue eyes. It doesn't say anything about the prior possibilities of weight or birthdays; it is only a statement about what in fact occurred. It doesn't say that one or both boys couldn't have been born on a Wednesday. If that had happened, the consequence would be that the problem would say:

"I have two children. One is a boy born on a Wednesday. What is the probability I have two boys?"

The answer to this question is the same as the answer to the Tuesday question and to the simpler question that does not refer to a day at all: 1/3.

Derbyshire calculates the probability as 13/27. He can only get there by interpreting the "Tuesday clause" as affecting the prior probabilities of birth. In other words, the case of two boys born on Wednesday need not be included in our enumeration of cases because it wasn't possible for both boys to be born on Wednesday, since we know one was born on Tuesday! I hope everyone can see the post facto fallacy in this reasoning. Anyone who would buy this line of reasoning is playing the role of the father in the following comic scenario:

One day you get a letter from the town correcting your son's birth certificate. He was born two seconds after midnight so he was actually born on a Wednesday rather than a Tuesday. With a heavy heart, you sit your son down and tell him the unfortunate news: "I'm sorry to tell you this son, but I'm not your real father. My son could only have been born on a Tuesday, and I've just learned you were born on a Wednesday."

By the way, this problem is not comparable to the Monty Hall problem. The Monty Hall problem is a genuinely counter-intuitive probability result. The Tuesday's Child problem is more like a riddle or joke that depends on deceptive or ambiguous language.

aletheist said...

If you flipped a coin twice, and you know that the first flip came up heads, then the probability that you got two heads is simply the probability that the second flip came up heads: 1/2. Likewise, if you have two children, and one is a boy, then the probability that you have two boys is also 1/2 (not 1/3), assuming that every child is equally likely to be a boy or a girl.

Despite Derbyshire's insistence to the contrary, the day of the boy's birth is just as irrelevant to the problem as the day of the coin's minting. The gender of the second child is statistically independent of any information whatsoever about the first child, just as the second coin flip is statistically independent of the first.

Of course, there is another ambiguity here, even if you leave out the day of the boy's birth. Which of the following is the correct interpretation of the second sentence?

1. At least one is a boy.
2. Exactly one is a boy.

My analysis above is based on the first case, so the answer is 1/2. In the second case, the answer is zero; the other child must be a girl. As Derbyshire points out, #1 is the mathematical-logical interpretation, while #2 is the everyday interpretation.

David T. said...

I agree with your coin flip analysis, because your condition is that "you know that the first flip came up heads." Yes, in that case, the probability of having two heads is 1/2. These are the possibilities:

Unfortunately, in the Tuesday's Child problem, we aren't told that the first child is a boy, only that one of them is. It might be the second one. So the possibilities are:

Boy - Boy
Boy - Girl
Girl - Boy

and the probability of having two boys is 1/3.

I agree with you about the ambiguity of the language. But it's a math problem, so the reader should be on his guard.

aletheist said...

No, I still disagree. The gender of the two children is statistically independent; it does not matter whether the boy that you know about was born first or second. The only two possibilities are (two boys) and (one boy and one girl), and the probability of each is 1/2.

Suppose that you know that a boy was born first; then the only two scenarios are boy-boy and boy-girl, and the probability of each is 1/2. Now suppose that you know that a boy was born second; then the only two scenarios are boy-boy and girl-boy, and the probability of each is (again) 1/2. How does the probability of two boys somehow change to 1/3 just because you do not know whether the boy that you know about was born first or second?

David T. said...

You're right that the birth of the children is statistically independent, but that doesn't mean you can ignore the conditional dependency. The first conditional is:

1) Given that a boy is born first, then the odds of having two boys are 1/2.

This is true. It corresponds to the two possibilities:

Boy-Boy
Boy-Girl

Since one of the two possibilities is boy-boy the probability is one-half.

The other conditional is:

2) Given that a boy is born second, the probability of having two boys is one-half.

This is true as well. It corresponds to the possibilities:

Boy-Boy
Girl-Boy.

But here's the rub. The unconditional problem is:

3) Given that at least one of two children is a boy, the probability of having two boys is.... what?

The solution to this problem is not found by just joining the two conditional problems. If you do that, you get the following possibilities:

Boy-Boy
Boy-Girl
Boy-Boy
Girl-Boy

and it appears the probability is one-half. But the Boy-Boy case has been counted twice, and it is really only one possibility in the unconditional problem. One of the two Boy-Boys has to be eliminated and you get a probability of one-half.

It's easy to test this empirically by doing a series of two coin flip experiments. Just repeat flipping two coins, eliminate any tail-tail outcomes, and I think you will find that the proportion of head-head results will be about one-third.