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The Surprisingly Paradoxical Gender Problem

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The Surprisingly Paradoxical Gender Problem

The recently deceased Martin Gardner was known for, among other things, his mathematical puzzles. Here’s one of them.

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

The alleged answer may surprise you. Knowing the child’s gender and day of birth does actually affect the proability. Allegedly. The answer is hidden below.

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(I’m not sure how this spoiler plugin is going to interact with the RSS feed, so just to be on the safe side I’m throwing a warning here that the answer is coming up and you should stop reading now if you don’t want it spoiled!) The chance of me having two boys is 13/27

I’ll also say that the answer is wrong , and in fact your initial intuition is right.Let’s step back and simplify the problem for a moment. Get rid of the day of the week. I have two children, one of which is a boy. What is the probability that I have two boys?

The answer to this simpler problem is supposed to be 2/3. The reason you don’t know whether it’s my first or second child which I’m referring to. Thus the possibilities are Boy Boy, Boy Girl, and Girl Boy. Further, each of these outcomes is equally likely.

Let’s turn the problem into a game. I’ll flip two coins and tell you the outcome of one. If you guess what the other is, you win. Following the logic above, the optimal strategy would seem to be to choose the opposite of whatever I say. If I choose heads, say tails. If I choose tails, say heads. You should win 2/3 of the time, right?

Wrong. Consider how the strategy works. You will always win if I flip two different faces, and you will always lose if I flip the same face twice. So there’s a 1/2 chance you’ll win (I flip HT or TH) and a 1/2 chance you’ll lose (I flip TT or HH).

So let’s play a round. I’ve just flipped two coins. I pick one at random and announce that it is heads. As I’ve just proven, you have a 50% chance of winning with your optimal strategy. Yet, the logic above still holds true! The possibilities for my flip are HH, HT and TH, and each of those three possibilities is equally likely. So you should win 2/3 of the time.

What gives?

This is something that’s been bugging me for a couple days, and I’ve finally figured out the answer: there’s a hidden factor not being accounted for. Specifically, the likelihood that I’ll announce heads!

Consider those three results again. With HH, I’ll always announce heads. Since there’s a 1/4 chance of my flipping HH, and this is the only way for you to lose, the probability I’ll announce heads and you’ll lose is 1/4*1=1/4.

If I flip HT or TH, there’s only a 1/2 chance I’ll announce heads. Thus the chance that I’ll flip HT and announce heads is 1/4*1/2=1/8, and the chance I’ll flip TH and announce heads is also 1/8. Add them up and the chance that I’ll announce heads and you’ll win is 1/4.

Following the same logic, the chance of me flipping tails and you winning is 1/4, and the chance of me flipping tails and you losing is 1/4. Thus, we’ve finally got your 50/50 chance of winning despite what seems like should be an optimal strategy.

This game is exactly analogous to the simplified gender problem above, and so I’ve proven that the chance of me having two boys is 50/50.

Now let’s bring it back to the original problem.

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

Let’s get some terminology and numbers down.

TB=Tuesday Boy (a boy born on a Tuesday). Base chance = 1/7 (Tuesday) * 1/2 (Boy) = 1/14

NTB=Non-Tuesday Boy (a boy born on any other day). Base chance = 6/7 (not Tuesday) * 1/2 (Boy) = 3/7

G=Girl. Base chance = 1/2.

The possibilities for my children are:

Child 1C1 BaseChild 2C2 BaseChance I’d say “Tuesday Boy”Overall chance
TB1/14TB1/1411/14 * 1/14 * 1 = 2/392
TB1/14NTB3/71/21/14 * 3/7 * 1/2 = 6/392
TB1/14G1/21/21/14 * 1/2 * 1/2 = 7/392
NTB3/7TB1/141/23/7 * 1/14 * 1/2 = 6/392
G1/2TB1/141/21/2 * 1/14 * 1/2 = 7/392

So the chance of two boys is 14/392, and the chance of a boy and a girl is 14/392. In other words, the two possibilities are equally likely.

It’s worth emphasizing that this is all due to the completely random nature of things, and specifically the fact that, since they’re “my” children, it’s safe to assume that I didn’t decide beforehand that I wanted at least one boy born on a Tuesday and take extreme measures to ensure that I got one.

If, on the other hand, we rephrased the question as

If I go out searching for a woman who has two children, one of which is a boy born on a Tuesday, then what is the probability that she has two sons?

Then the original answer, 13/27, is correct. The difference here is that I’ve decided to search out a subset of possibilities, rather than taking a specific case and just announcing the gender and day of birth of one random child.

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