probabilities

[Tim]

Your permutations are wrong if you come up with a third - it puts equal weighting on finding a girls certificate for a boy/girl combination as a girl/girl combination.

Label all permutations by B&G, bold for the certificate of the child found. These are by the constraints applied in the OP of equal probability. There is no avoiding this.
1a)BB
1b)BB
2a)BG
2b)BG
3a)GB
3b)GB
4a)GG
4b)GG

Now we know that options 2b, 3a, 4a&b all fit the details. Second certificate to be found in 50% of the instances are for girls.

It is only if the selection of the first certificate is not random that you can get to the probabilty of one third, such as certificates are filed girls first, boys second then alphabetically by first name, and you take the first one in the file.

Consider it the odds of knowing that an heir will be male (assuming there is a male priority) given at least one child is female (this brings thirds into it)

Dammit many more answers presented. I'm with Tom. The one third answer is to a related question, but not the one posed.

[tom_e]

There was also no mention of the certificates being randomly selected, so we have:-

You said you found the birth certificate for one of the children.  If you meant you found both certificates and examined both of them before setting the question, that would be different.

a: Mm, mf, fm, ff
b: mM, mf, fm, ff
c: mm, Mf, fm, ff
d: mm, mF, fm, ff
e: mm, mf, Fm, ff
f: mm, mf, fM, ff
g: mm, mf, fm, Ff
h: mm, mf, fm, fF

g+h aren't two distinct outcomes, if both are girls I could pick either birth certificate to demonstrate one is a girl since I haven't said to whom it belongs, so we have:-

If you had both birth certificates to pick you would have said so.  Just no way that is consistent with your original statement.

And the fact they aren't visibly distinct is, AFAICT, not relevant to anything.

[Greenbank]

The sentence:-

"I found the birth certificate for one of the children and it shows she is female."

does not mean that did not find the other birth certificate and chose not to disclose what it said.

[tom_e]

Basic english comprehension.  The fact "I found the certificate for one of the children" means you are implying you didn't find the certificate for two of the children.

Note that we're not talking of a passive unseen other certificate.  For the probabilities to change to 1/3, you actually have to have [read both and] intended to deceive - can't happen by chance/probabilities alone.

[Tim]

Essentially the difference is between:
At least one is female (which is required to get to the answer of 1/3)
and
Found a birth certificate for one child and it was female (which gets you to 1/2)

Edit:
One maybe what you meant, the other is what you presented

[mrcharly-YHT]

Maybe I'm totally wrong, but I can only see TomE's answer as being correct if you rechoose the certificate.

I'm going round the twist.

There are only 3 possible combinations in the family; bb, bg, gg

Birth certificates in separate envelopes. You pick one envelope, open it and it is a girl.

So bb is ruled out.  Then the answer is 50% b or g.

But that's a different puzzle from the one originally posed, isn't it?

[tom_e]

There are only 3 possible combinations in the family; bb, bg, gg

No, there are four combinations.  You really can't combine re-ordered versions unless you count how many different ways they can occur first.  This is classical statistics, not quantum mechanics - the different states exist whether you can tell the difference between them or not.

[pcolbeck]

There are only three possible combinations of two children

bb
gg
bg

gb is a spurious fourth possibility as it's just a reordering of bg.

We have one birth certificate that show one child is a girl which removes the bb option leaving

gg
bg

So there are only two possible variations left so it 50% chance the other child is a girl.

If you allow gb and bg as options then it seems to imply its 1/3 but it isnt as if you allow ordering you have to also allow as second bb and a second gg ie two children of teh same sex but in the opsite order.

bb
bb
gg
gg
bg
gb

So six possibilities not four. Now since we know one is a girl we can remove all the bb options

gg
gg
bg
gb

we know one is a girl and two out of four of these remaining options allow a second girl so again 50% probability.

[pcolbeck]

There are only 3 possible combinations in the family; bb, bg, gg

No, there are four combinations.  You really can't combine re-ordered versions unless you count how many different ways they can occur first.  This is classical statistics, not quantum mechanics - the different states exist whether you can tell the difference between them or not.

That's where you are wrong there are 6 not 4 possible combinations if you allow reordering.

[Rich753]

[

https://secure.wikimedia.org/wikipedia/en/wiki/Boy_or_Girl_paradox

It depends how the question was phrased:-

(Note that this refers to the question as number of boys...)

"
    * From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
    * From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.[3][4]
"

In my phrasing the family containing at least one girl has been preselected.
[/spoiler]

I don't think so - your phrasing was "a friend of mine has two children" - that surely preselects families with two children, thereafter you selected one child at random and inspected the birth certificate, plainly the second case.

[Tim]

There are only three possible combinations of two children

bb
gg
bg

Yes, but you then must consider the relative probabilities of the combinations:
two boys - 25%
one boy and one girl - 50%
two girls - 25%

The advantage of listing everything out (at equal chance) means you can stop worrying about relative probabilities.

[mrcharly-YHT]

There are only 3 possible combinations in the family; bb, bg, gg

No, there are four combinations.  You really can't combine re-ordered versions unless you count how many different ways they can occur first.  This is classical statistics, not quantum mechanics - the different states exist whether you can tell the difference between them or not.

Ah, thank you that make sense.

First child Second child
b                 g
b                 b
g                 b
g                 g

those are the four possible states, right?

Now, if I understand you, you are saying that when you choose the cert, you are doubling the possible states:

Choose first child's bc
b                 g
b                 b
g                 b
g                 g

Choose 2nd child's bc

b                 g
b                 b
g                 b
g                 g


Leading to 8 possible states

b                 g
b                 b
g                 b
g                 g
b                 g
b                 b
g                 b
g                 g

Which gives us 2 states where the 2nd cert if for a girl, and 4 where it is for a boy. So 4/6 for a boy and 2/6 for a girl.

[Tom B]

How similar is this to the Monty Hall problem?

[tom_e]

mrcharly - in your last list of eight, you forgot that you'd chosen the certificate for the first column in the first four rows, and the 2nd column in the last four rows.  So you need to identify rows in the first four with a g on the left, and rows in the last four with a g on the right.

Scratch that, I think I misunderstood you.

You just need to not silently rule out two of the eight rows, without saying why!
There were eight states.  Go looking through those eight for the girl certificates.

[Hello, I am Bruce]

I agree with tom_e that "I found the birth certificate for one of the children" means you did not find the certificate for the other child, and the answer to the original problem is 1/2, but it can be rephrased in a way that yields 1/3.

Here's how to compute the answer using Bayes' law...

A friend of mine has two children. I found the birth certificate for one of the children and it shows she is female. What is the probability that the other one is also a girl?

We want to know the probability of two girls , given that one (and only one) arbitrarily chosen certificate shows a girl.  Denote this P(GG | cert-G).  By Bayes' law

P(GG | cert-G) = P(cert-G | GG) * P(GG) / P(cert-G)

Our prior probability of two girls P(GG) = 1/4 (because 1/4 families have 2 girls), our prior probability of finding a girl's certificate P(cert-G) = 1/2 (because half of all certificates are for girls).  Our conditional probability of finding a girls certificate given that both children are girls P(cert-G | GG) = 1.

P(GG | cert-G) = P(cert-G | GG) * P(GG) / P(cert-G)
   = 1 * (1/4) / (1/2)
   = 1/2

The alternative question is

A friend of mine has two children. I asked him if (at least) one of them is a girl and he said yes. What is the probability that the other one is also a girl?

In this case we calculate P(GG | yes).

P(GG | yes) = P(yes | GG) * P(GG) / P(yes)

The prior probability of answering yes P(yes) = 3/4 (because 3/4 families have at least one girl).

P(GG | yes) = 1 * (1/4) /(3/4)
 = 1/3

[Kathy]

Out of curiosity, do either of the children have a goat or a sports car?

[tom_e]

At least one of them has either.

[mrcharly-YHT]

I agree with tom_e that "I found the birth certificate for one of the children" means you did not find the certificate for the other child, and the answer to the original problem is 1/2, but it can be rephrased in a way that yields 1/3.

Here's how to compute the answer using Bayes' law...

A friend of mine has two children. I found the birth certificate for one of the children and it shows she is female. What is the probability that the other one is also a girl?

We want to know the probability of two girls , given that one (and only one) arbitrarily chosen certificate shows a girl.  Denote this P(GG | cert-G).  By Bayes' law

P(GG | cert-G) = P(cert-G | GG) * P(GG) / P(cert-G)

Our prior probability of two girls P(GG) = 1/4 (because 1/4 families have 2 girls), our prior probability of finding a girl's certificate P(cert-G) = 1/2 (because half of all certificates are for girls).  Our conditional probability of finding a girls certificate given that both children are girls P(cert-G | GG) = 1.

P(GG | cert-G) = P(cert-G | GG) * P(GG) / P(cert-G)
   = 1 * (1/4) / (1/2)
   = 1/2

The alternative question is

A friend of mine has two children. I asked him if (at least) one of them is a girl and he said yes. What is the probability that the other one is also a girl?

In this case we calculate P(GG | yes).

P(GG | yes) = P(yes | GG) * P(GG) / P(yes)

The prior probability of answering yes P(yes) = 3/4 (because 3/4 families have at least one girl).

P(GG | yes) = 1 * (1/4) /(3/4)
 = 1/3

My brain hath spun. I get 1/3 girl and 2/3 boy - so is my listing of states correct?
(I think I know why I failed math at uni)

[Hello, I am Bruce]

How similar is this to the Monty Hall problem?

In the Monty Hall problem, the host has full information and is trying to help you to win by eliminating a goat.  So it's a bit like the family problem where the person asking the question knows the sex of both children and chooses which child to tell you about.

[tom_e]

My brain hath spun. I get 1/3 girl and 2/3 boy - so is my listing of states correct?

Your listing of eight states was good, you just then needed to separate out the ones in the first half with g on the left, and the ones in the second half with g on the right (and not spuriously draw a line through two states).

The second part of your deleted post looked exactly right to me.  Once you'd got over my confusingness.

[tom_e]

The alternative question is

A friend of mine has two children. I asked him if (at least) one of them is a girl and he said yes. What is the probability that the other one is also a girl?

Dang, I'd been repeatedly reading that one of yours and hadn't spotted the difference.  I see it now. 

Arguably it still ends up as english comprehension, because if the friend has two girls and you ask "Is one of them a girl?" then there's at least a 60% chance they won't say "yes", or "no", they'll say "both of them are girls". 

[Attitudeless Badger]

"
A friend of mine has two children. I found the birth certificate for one of the children and it shows she is female. What is the probability that the other one is also a girl?
"

Slightly confused here, as substituting names for each, I get "I found the birth certificate for Child A and it shows she(Child A) is a girl.  What is the probability that Child B is a girl?"

Why can you not eliminate that way (answers in plain english, please)

[mrcharly-YHT]

My brain hath spun. I get 1/3 girl and 2/3 boy - so is my listing of states correct?

Your listing of eight states was good, you just then needed to separate out the ones in the first half with g on the left, and the ones in the second half with g on the right (and not spuriously draw a line through two states).

The second part of your deleted post looked exactly right to me.  Once you'd got over my confusingness.

Ah, so I should do the elimination of the bb states, remove the lines where the chosen cert is for b, ending up with
b                 g
b                 b
g                 b
g                 g

b                 g
b                 b
g                 b
g                 g

50/50 girl boy.

[tom_e]

No reason to separately eliminate bb states first. 
Just remove the lines where the chosen cert is for b.  It is sufficient.

[The Seldom Killer]

How similar is this to the Monty Hall problem?

In the Monty Hall problem, the host has full information and is trying to help you to win by eliminating a goat.  So it's a bit like the family problem where the person asking the question knows the sex of both children and chooses which child to tell you about.

Also the Monty Hall problem is based on a pattern of behaviour that eliminates an option. Therefore the response is based on the contenders knowledge of the show. If ignorant then the transfer of probability applies, if not then the effective decision is practically delayed until after the elimination and so the transfer doesn't take place. If a 50/50 split in familiarity is assumed then I guess you could argue that only half of the probability is transferred so the answer remains the same but the odds aren't as good.