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Buzzie | 00:35 Thu 22nd Jan 2009 | Riddles
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A couple have two children (and they are a normal couple with a 50:50 chance of having a girl or a boy). You are intorduced to one of their children who is a girl. What is the probability that the other child is also a girl? There is a QI claxon waiting for the unwary.....
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Slightly in favour of the other one being a girl, as there is a small chance they could be identical twins in which case they would both have to be girls.
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No twins involved!
Still 50/50 surely? The fact that the other child is a girl is neither here nor there.
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Oh there goes the klaxon!!!

Think about the different ways in which it is possible to have 2 children
boy/boy
boy/girl
girl/boy
girl/girl

At the risk of anohte klaxon, I don't see it.
another*
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So you now know that it can't be B/B.

In 2 of the other 3 cases the second child is a boy so the probability of the other child being a girl is 1/3
Ah of course. Excellent! People assume that it you are referring to the first child.

I like that one.
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Yep, that was the mistake I made when I 1st heard it!
It's also a load of rubbish - if you believe that kind of logic, then join the flat earth society.
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Big Dave,

The question does not refer to the probability of giving birth to a girl or a boy which, I agree is as near 50/50 as makes no difference.
The question asks IF you know someone has 2 children & you meet one, what is the probability of the other being the same sex. It's a difficult concept to grasp but if you look at how another-view has laid it out then it becomes clear. As AV says it is irrelevant whether you have met the 1st born or not.
I repeat what I said - the logic is seriously flawed. Look at my first post for the correct answer.

By saying that you have met one of them then you have fixed whether that is the older or younger. Which it is does not matter. You can't then vacillate between the two in the course of the logic.

If you told this at the end of the evening in the pub, you might be able to persuade those present that you had a case.
For what it is worth, and I do have a maths degree so I know something about probability, the answer of 1/3 is correct.

You can read an explanation here:

http://www.jimworthey.com/puzzle.html
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Thankyou Rollo!! I was about to go into a lengthy explanation about probability theory but I don't need to now!!
I've seen the answer but I disagree with it. What does it matter whether the 2nd child is older or younger? That doesn't make a difference to the odds. Ultimately, they can only have either a boy or girl. BigDave - I agree with you.
Just to throw a spanner in the works...

What if the second child hasn't been born yet - it's in the womb and you don't know its sex. Isn't the probability of the foetus being a girl 50-50 ? The sex of the existing child isn't relevant is it?
If it takes 6 men 4 hours to dig a 20 foot long 3 foot deep trench how long does it take wee Johhny to get his granny's messages on a Friday?
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I wish I'd never posted the damn Q!!!

For the last time here we go:

You know that there are 2 children.
You know that one of them is a girl.

The question asks about the probability of the other child also being a girl, NOT the probability of giving birth to a girl or boy.
SO: there are 4 possible combinations of having 2 children, B/B, G/G, G/B & B/G

Knowing that one is a girl means you have eliminated B/B from the possibilities, leaving G/G, G/B & B/G as the options.
Therefore, in only one out of those 3 options is the other child a girl, hence the probability of 1/3
Hi Buzzie - what about the answer to my question?
The answer of 1/3 is obviously correct.

If the combinations are :

b/b
b/g
g/b
g/g

Then out of the 3 combinations that have a girl, 2 are boys.

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