When intuition and math probably look wrong
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When intuition and math probably look wrong
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Re: When intuition and math probably look wrong
#2It then goes on to acknowledge that the 33% answer would be correct if we specifically choose a two child family that fits the parameters of the problem beforehand.
It's all about why the information was selected.
Re: When intuition and math probably look wrong
#3Re: When intuition and math probably look wrong
#4conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) = 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)
Re: When intuition and math probably look wrong
#5The statement "I have two children, one of whom is a son born on a Tuesday" is semantically ambiguous. It can mean (1) "I have two children, and the quantity of them who are males born on a Tuesday is exactly one", (2) "I have two children, at least one of whom is a male born on a Tuesday", or even (3) "I have two children, and the maximum quantity of males born on the same Tuesday is one".
Re: When intuition and math probably look wrong
#6Ok, I was about to rage about yet another article going on about the Two Children problem and getting it wrong by leaving out a whole host of children (i.e. children in families of more or less than two children). Then it surprised me by not only acknowledging it, but acknowledging that the the 50% answer is correct when we are selecting from an arbitrary family (as the original problem is usually presented). It then…
Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order.
So given 2 children, there are only 3 possibilities ...
boy, boy
girl, girl
boy, girl
So if you're told one of them is a boy ...
boy, girl
boy, boy
And that's it ... a 50% probability that both children are boys.Re: When intuition and math probably look wrong
#7Ok, I was about to rage about yet another article going on about the Two Children problem and getting it wrong by leaving out a whole host of children (i.e. children in families of more or less than two children). Then it surprised me by not only acknowledging it, but acknowledging that the the 50% answer is correct when we are selecting from an arbitrary family (as the original problem is usually presented). It then…
> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order. So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl So if y…
Older child boy, younger child boy
Older child boy, younger child girl
Older child girl, younger child boy
Older child girl, younger child girl
Of those only one is precluded by saying (at least) one child is a boy.Re: When intuition and math probably look wrong
#8Earlier quoted context omitted.
> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order. So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl So if y…
You can't mean that. There are four possibilities: Older child boy, younger child boy Older child boy, younger child girl Older child girl, younger child boy Older child girl, younger child girl Of those only one is precluded by saying (at least) one child is a boy.
Oh well, I guess this is what the article is talking about :)
Re: When intuition and math probably look wrong
#9isn't this the same kind of "intuitive" reasoning that fails at simpson's paradox? http://en.wikipedia.org/wiki/Simpsons_paradox conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) = 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)
p(boy = 2 | boy >= 1) = p(any child = boy)
And this was the original problem that led them to the 33% probability.Re: When intuition and math probably look wrong
#10Firstly, as presented it is clearly ambiguous. It is intended to be ambiguous, but in such a way that people who are familiar with the original version will get suckered into believing that it's well formed.
Secondly, if presented precisely, the answer usually given is either 13/27 or 1/2, depending on which version.
Finally, this is like the Monty Hall problem all over again. There are people arguing vehemently and without listening at all, demonstrating clearly that they are excellent at missing the point.
In case you're wondering, here's one statement and answer.
Suppose on knock on people's doors and ask - Do you have exactly two children? If they answer no, I move on. If they answer yes I then ask - Is at least one of them a boy born on a Tuesday? If they say no, I move on.
If they look surprised and say "Yes," what is the probability that they have two boys?
Answer: 13/27.
Yes, it really is.
If you replace the second question with "Is at least one of them a boy with red hair, left-handed, plays piano, was born on Tuesday, and has a cracked left upper incisor" then if the answer is "Yes" then the probability of both children being boys is almost exactly 50%.
If, instead, you replace the second question with "Is at least one a boy" then the probability of two boys is 1/3.
Finally, suppose you see a parent that you know has two children in the park with a boy. Now the probability of two boys is 50%, because, assuming uniform probabilities, having two boys makes it more likely you see them with a boy.
tl;dr: It's hard, and depends precisely on the assumptions you make.