Comment author:nono
17 February 2011 10:48:59AM
2 points
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"Renormalizing leaves us with a 1/3 probability of two boys, and a 2/3 probability of one boy one girl."
help me with this one, i'm n00b.
If one of the kids is known to be a boy (given information), then doesn't the other one has 50/50 chances to be either a boy or a girl? And then having 50/50 chances for the couple of kids to be either a pair of boys or one boy one girl?

Comment author:Morendil
17 February 2011 11:54:51AM
1 point
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one of the kids is known to be a boy

That's not the given; it is that "at least one of the two is a boy". Different meaning.

For me, the best way to get to understand this kind of exercise intuitively is to make a table of all the possibilities. So two kids (first+second) could be: B+B, B+G, G+B, G+G. Each of those is equiprobable, so since there are four, each has 1/4 of the probability.

Now you remove G+G from the table since "at least one of the two is a boy". You're left with three: B+B, B+G, G+B. Each of those three is still equiprobable, so since there are three each has 1/3 of the total.

Comment author:matteri
06 May 2011 05:37:24PM
0 points
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And in hope of clarifying for those still confused over why the answer to the other question - "is your eldest/youngest child a boy" - is different: if you get a 'yes' to this question you eliminate the fact that having a boy and a girl could mean both that the boy was born first (B+G) and that the girl was born first (G+B). Only one of those will remain, together with B+B.

Comment author:prase
17 February 2011 01:20:41PM
*
0 points
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This sort of problem is often easier to understand when modified to make the probabilities more different. E.g. suppose ten children and information that at least nine of them are boys. The incorrect reasoning leads to 1/2 probability of ten boys, while actually the probability is only 1/11. You can even write a program which generates a sequence of ten binary values, 0 for a boy and 1 for a girl, and then prompts you whenever it encounters at least nine zeros and compare the relative frequencies. If the generated binary numbers are converted to decimals, it means that you generate an integer between 0 and 1023, and get prompted whenever the number is a power of 2, which correspond to 9 boys and 1 girl (10 possible cases), or zero, which corresponds to 10 boys (1 case only).

Such modification works well as an intuition pump in case of the Monty Hall problem, maybe is not so illustrative here. But Monty Hall is isomorphic to this one.

## Comments (190)

Old"Renormalizing leaves us with a 1/3 probability of two boys, and a 2/3 probability of one boy one girl." help me with this one, i'm n00b. If one of the kids is known to be a boy (given information), then doesn't the other one has 50/50 chances to be either a boy or a girl? And then having 50/50 chances for the couple of kids to be either a pair of boys or one boy one girl?

That's not the given; it is that "at least one of the two is a boy". Different meaning.

For me, the best way to get to understand this kind of exercise intuitively is to make a table of all the possibilities. So two kids (first+second) could be: B+B, B+G, G+B, G+G. Each of those is equiprobable, so since there are four, each has 1/4 of the probability.

Now you remove G+G from the table since "at least one of the two is a boy". You're left with three: B+B, B+G, G+B. Each of those three is

stillequiprobable, so since there are three each has 1/3 of the total.And in hope of clarifying for those still confused over why the answer to the other question - "is your eldest/youngest child a boy" - is different: if you get a 'yes' to this question you eliminate the fact that having a boy and a girl could mean both that the boy was born first (B+G) and that the girl was born first (G+B). Only one of those will remain, together with B+B.

*0 points [-]This sort of problem is often easier to understand when modified to make the probabilities more different. E.g. suppose ten children and information that at least nine of them are boys. The incorrect reasoning leads to 1/2 probability of ten boys, while actually the probability is only 1/11. You can even write a program which generates a sequence of ten binary values, 0 for a boy and 1 for a girl, and then prompts you whenever it encounters at least nine zeros and compare the relative frequencies. If the generated binary numbers are converted to decimals, it means that you generate an integer between 0 and 1023, and get prompted whenever the number is a power of 2, which correspond to 9 boys and 1 girl (10 possible cases), or zero, which corresponds to 10 boys (1 case only).

Such modification works well as an intuition pump in case of the Monty Hall problem, maybe is not so illustrative here. But Monty Hall is isomorphic to this one.