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

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Anonymous 0 Comments

The “math class” answer is 1/3 for reasons that everyone else is explaining. In real life the answer is “it depends”. The difference is part of a subject of math called *Bayesian Inference*.

The *method* in which you find out information is just as important to the math problem as the actual information you learn.

The answer is 1/3 when those three combinations are equally likely. You might run into this situation during a meeting of the “has two kids” club and you ask a fellow parent “do you have at least one boy?” The question is usually constructed so that this is the most natural interpretation.

The answer is 1/2 in other situations. Say for example you are canvassing houses in the “has two kids” clubs, and one of the children answer, and it’s a boy. Assuming that boys and girls are equally likely to answer the door, the answer is 1/2 because BB is twice as likely to result in a boy answering a door than either of BG or GB. IMO this is generally a more common real-life situation.

The pieces of information that you learn are the same – but in probability it matters *how* the information is revealed to you.

This is also the key insight in the Monty Hall problem.

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