Use Bayes rule to mechanically solve probability riddles (cloud.disroot.org)

> You're told that at least one of them is a girl.

> Likelihood of at least one girl

What the “mechanism” requires is “likelihood of being told that at least one of them is a girl”.

Use Bayes rule to correctly solve probability riddles:

https://news.ycombinator.com/item?id=45056790

    p(both are girls | you're told at least one is a girl)
       = p(you're told at least one is a girl | both are girls) * p(both are girls) / (
            p(you're told at least one is a girl | both are girls) * p(both are girls)
            +
            p(you're told at least one is a girl | they aren't both girls) * p(they aren't both girls)
        )

The solution there assumes that p(you're told at least one is a girl | both are girls) = p(you're told at least one is a girl | they aren't both girls).

Might be hug of death but the load times are horrifically slow.

This Monty Hall problem was asked to Marilyn vos Savant, a woman with an extremely high IQ, who solved it correctly, and many readers of her column, including PhD and mathematicians, declared her solution wrong.

Then careful analysis proved her correct.

https://en.wikipedia.org/wiki/Monty_Hall_problem#Savant_and_...

For the last one, why does the "born on a Tuesday" information change the result? I don't see how it isn't equivalent to "born on a day", since the day of the week has no connection to the rest of the scenario. I understand why "at least one boy" does matter.

Recent related discussion: https://news.ycombinator.com/item?id=45051798

I feel this the case because pretty much all probability riddles are bayes rules problems which is famously not intuitive

Just a tidbit for remembering bayes rule:

P(A|B)P(B) = P(A,B) = P(B|A)P(A)

The familiar forms

P(A|B) = P(A,B)/P(B) = P(B|A)P(A)/P(B)

immediately follow