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Four houses, each with two children in them.

House A has two boys.
House B has an older boy, and a younger girl.
House C has an older girl, and a younger boy.
House D has two girls.

You walk into a house at random and see one girl. What are the chances that you are in house D?

Had immense fun with moonbeam's question (and the Monty Hall problem) last night. My fiancée prefers the more obvious answers to both problems!

Re: the boy/girl problem. The bit that melts my head is that if you ask the girl if she was the younger child, the chance becomes 50/50, whatever she answers.

This means that maths is broken, and we should never have come down from the trees.

Part-timer - it's 50/50 isn't it? This is the same problem in reverse.

Mr_Grue wrote:Re: the boy/girl problem. The bit that melts my head is that if you ask the girl if she was the younger child, the chance becomes 50/50, whatever she answers.

Yes. Fun, isn't it? It's a bit like that bloke's cat... (And also kind of not like it at all.)

The key is that, until you have an answer, she could be younger or older. Once you know the answer, you have eliminated either house B or house C, as well as house A.

Part-timer - it's 50/50 isn't it? This is the same problem in reverse.

It's my attempt to illustrate how the maths works. It's the same problem.

Or, if you have a 50% chance of taking a blue sock from an infinite sock drawer, and a 50% chance of getting a pink sock, and I take two socks at random from the drawer, what's the chance that I have a pair of pink socks? Now if I show you that I have at least one pink sock, what is the chance that I have a pair of them?

Just to clarify - the answer is 1/3.

Simple (ish) explanation:

You are on an island .... and on this island live 4 million couples - each having 2 children. Each couple is normal, insofar as they have a 50/50 chance of producing a boy or a girl (let's ignore twins, etc for the sake of this exercise).

Statistically speaking the couples will be split thus:
a. 1 million couples will have an older boy and a younger girl.
b. 1 million couples will have 2 girls.
c. 1 million couples will have 2 boys.
d. 1 million couples will have an older girl and a younger boy.

Agreed so far ??

You now meet one of the couples at random and are told that they have at least one girl. You know that this couple cannot be from group c.

You are now left with 1 million couples who have 2 girls .... and 2 million couples that have a boy and a girl.
what are the odds that the other child is a girl ..... well it's 1 million in 3 million i.e. 1 in 3.

I knew that this riddle would kick up a fuss (mission accomplished ).

For all you non-believers:

http://www.theanswerbank.co.uk/Quizzes- ... 94606.html

http://www.jimworthey.com/puzzle.html

http://forums.xkcd.com/viewtopic.php?f=3&t=311

http://forums.xkcd.com/viewtopic.php?f=3&t=33809

http://forums.xkcd.com/viewtopic.php?f=17&t=32659


There's loads of proof in them links that the answer is 1 in 3 and NOT 50/50.

la la la la la - not listening!

He's right you know. 1/3 it is. Hard logic beats "common sense".

Part-Timer wrote:Four houses, each with two children in them.

House A has two boys.
House B has an older boy, and a younger girl.
House C has an older girl, and a younger boy.
House D has two girls.

You walk into a house at random and see one girl. What are the chances that you are in house D?

Had immense fun with moonbeam's question (and the Monty Hall problem) last night. My fiancée prefers the more obvious answers to both problems!

If you only see one girl in the house, then the chance that you are in house D is 0%, because it has 2 girls, and the other one you see is a boy?

It doesnt state you only see one child, who is a girl. It states that you only see one girl

Blapsing_Beard wrote:It doesnt state you only see one child, who is a girl. It states that you only see one girl

This would suggest that the children are housebound and each house consists of a single room.

A village has two hospitals, one at the top of the hill, and the other at the bottom.

The hospital at the top has a maternity ward that averages 45 births a day. The hospital at the bottom of the hill has a maternity ward that averages 15 births a day. Over a year, approximately as many boys are born as girls. A statistician decided to establish how many days each hospital experienced a 60:40 split between boys and girls. He knows what he's doing. Does he expect to find:

a) considerably more 60:40 days in the larger hospital than the smaller hospital,
b) considerably more 60:40 days in the smaller hospital than the larger hospital, or
c) an almost equal number of 60:40 days in both hospitals,

and why?

But.... The maths is broken!
Can we, perhaps, go back to ice related riddles and have a new thread for maths riddles?

Edit:

Mr_Grue wrote:A village has two hospitals, one at the top of the hill, and the other at the bottom.

Woah Woah Woah...

It's a very fecund village.

Kippers?

I'd have said that they'd be the same?

Nerp. (though I thought that too)

Mr Grue tosses a coin, and it lands heads up.

Mr_Grue wrote:A village has two hospitals, one at the top of the hill, and the other at the bottom.

The hospital at the top has a maternity ward that averages 45 births a day. The hospital at the bottom of the hill has a maternity ward that averages 15 births a day. Over a year, approximately as many boys are born as girls. A statistician decided to establish how many days each hospital experienced a 60:40 split between boys and girls. He knows what he's doing. Does he expect to find:

a) considerably more 60:40 days in the larger hospital than the smaller hospital,
b) considerably more 60:40 days in the smaller hospital than the larger hospital, or
c) an almost equal number of 60:40 days in both hospitals,

and why?

None of the above - he decides to get a life instead!