why the odds of the “two children problem” are 1/3?

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I was asked the question “a man states he has two children, and at least one of them are boys, what are the odds that the man has two boys?” I’ve been told the answer is 1/3, but I can’t wrap my head around it. Additionally, there is another version of the problem that states he has at least one boy born on a Tuesday. How does that change the odds? Why?

Edited to add (so people don’t have to sort through replies): the answer is 1/3 because “at least one boy” is accounting for B/G & G/B. The girl can be the first or second child. You can move the odds to 50/50 by rewording the question to “my first of two children is a boy, what are the odds the other child is a boy”

In: Mathematics

27 Answers

Anonymous 0 Comments

A fun way to think of this is a lottery, where each qualifying person can draw only once, and each person who draws has an equal chance for winning a prize for the family. There is always exactly one winner.

The families competing are two person families. The representatives are the two children, and their father, in this case.

The question is, once the qualifiers are given, what are the odds that a two son family wins?

If each father draws, it’s a 25%, obviously. One chance for each family to win.

If each eldest son draws, it’s 50%. Same for each youngest son. Half of the two-son’s competitors are disqualified, so the chances double.

If each daughter draws, it’s 0%. The two son families are disqualified, so how can they win?

If each son draws, it’s 50% again. While there are twice as many son-daughter pairs, and the two-daughter families are disqualified, the two-son families get two chances each, and even the odds. But what if they didn’t get that second ticket?

This is the two children problem. The lottery asks for a father to draw, but in order to do so, they must approach with exactly one son of their family. This is the father stating that they have two kids, and at least one is a boy. Even if they have two sons, their family only gets one ticket. And, remember, the two-daughter families are unable to draw, and the son-daughter families are still twice as numerous and still get their one ticket.

With their advantage lost, a two-son family will win 1/3 of the time, and a son-daughter family will win 2/3 of the time.

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