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
The Monty Hall problem doesn’t apply in this way. The order of birth has absolutely nothing to do with this situation. People are assigning values to this situation that don’t actually exist.
If we wanted to Monty Hall this situation, it would be that person has three children and we know one is a girl. We guess which one, in order, and then the parent tells us one of other two is a boy. It would benefit us to change our guess to the remaining child of those two, as there is a 2/3 chance that is a girl, as opposed to our initial 1/3 chance.
Alternatively, we could Bertrand’s box it. In that case there are three parents, each with two children. One has two boys, one two girls, and one a boy and a girl. If one of those parents revealed the sex of one child at random, telling us that the child is a girl, there is a 2/3 chance the other is also a girl.
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