Easy(?) Probability

Originally posted by talzamir
I thought again.. yep. 50%

Correct! (simple proof?)
Many a textbook will give 1/3 with the above proof!

Originally posted by wolfgang59
Correct! (simple proof?)
Many a textbook will give 1/3 with the above proof!

Is the reasoning in my first post not adequate?

Its a slightly different question to the 1/3 one which was why I was in two minds and I now in one and agree 50%. Answering the phone makes the difference and makes a distinction between 1st and 2nd child. So probability phone answering child is a girl before the phonecall was 50% (100% after) and the other 50%. When she answers it makes no difference to the second child.

Originally posted by forkedknight
Is the reasoning in my first post not adequate?

I believe not.

Originally posted by wolfgang59
I posted a similar problem on here a couple of years ago, if you
remember that don't contribute, just sit back and enjoy the discussions!

I phone a work colleague up and his daughter answers the phone. He has
previously told me that he has two children at home, I had no idea he had
a daughter. What are the chances that the other child is also a girl?

hmm
He has
previously told me that he has two children at home

Ok, did do you know for sure:
The were both his children, they might have either or both been guests (his daughter might have been out that day)

his daughter answers the phone
Was she a child?

How about this,
His grown up daughter might be living with him, with her two baby sons.

Originally posted by wolfgang59
I believe not.

Ok, the probability of there being two girls in a family, I hope we can agree, is:
P(GG) = 1/4

The probability of calling a random family with two children, and having a girl answer is:
P(g) = 1/2
(this is assuming that both children are equally likely to answer the phone, which may not be true in real life, but we are assuming it's true here)

According to Bayes theorem:
P(A|B) = P(B|A)*P(A) / P(B)

P(g|GG) = 1 i.e. if we know the family has two girls, then the probability of a girl answering is 100%

therefore
P(GG|g) = 1 * P(GG) / P(g) = 1 * 0.25 / 0.5 = 0.5

You know there are two children, but you don't know their genders. So the probability that the child answering the phone is a girl (or a boy) is 50/50. If you call again and the other child answers, the probability of the second child being a boy or girl is also 50/50. I believe the coin toss example is correct, no matter how many times you toss the coin the probability of it coming up heads (or tails) will always be 50/50.

Edit: But now I'm thinking the probability of the second child being a boy might be slightly higher, if for no other reason than because many coin tosses tend to show close numbers for heads and tails.

Originally posted by lemon lime

Edit: But now I'm thinking the probability of the second child being a boy might be slightly higher, if for no other reason than because many coin tosses tend to show close numbers for heads and tails.

That's because the probability of heads/tails is 50/50, not because tails is more likely following a heads. This is the gambler's fallacy.

Originally posted by forkedknight
That's because the probability of heads/tails is 50/50, not because tails is more likely following a heads. This is the gambler's fallacy.

The probability of any one coin toss is always 50/50 no matter how many times the coin is flipped, but I was wondering about all coin tosses when viewed as a group.

With only two kids in a group the second flip is also 50/50, with nothing else to consider. But let's say for example there are 20 children in the household. Assuming the chances of the couple having either a boy or a girl is even with each birth, and the first 7 kids answering the phone are girls, is it still reasonable to assume the eighth call will be 50/50 (boy or girl)?

Prob of 2 girls given a girl took the call is , mathematically,
P(2 girls AND girl takes call)/P(girl takes call). The numerator = (2 girls) since, if that is the case, a girl must take the call.
So prob = P(2 girls)/P(girl takes call).
Num = 1/4
Denom =1/2
Result = 1/2
Could somebody who thinks the answer is 1/3 tell me what is wrong with this analysis?
Likewise if they want to give a comprehensive explanation of their answer I'll try and explain what I think is wrong with theirs.

Originally posted by 24en1KDexd29
Prob of 2 girls given a girl took the call is , mathematically,
P(2 girls AND girl takes call)/P(girl takes call). The numerator = (2 girls) since, if that is the case, a girl must take the call.
So prob = P(2 girls)/P(girl takes call).
Num = 1/4
Denom =1/2
Result = 1/2
Could somebody who thinks the answer is 1/3 tell me what is wrong with this a ...[text shortened]... omprehensive explanation of their answer I'll try and explain what I think is wrong with theirs.

If the question is phrased, "From all families with two children in which at least one of the children is a girl. What is the probability that the other child is also a girl?"

The answer is 1/3.

The apparent paradox is due to the "at least one child is a girl" statement. How do you determine that "at least" one child is a girl. There are not very many realistic scenarios where you could know that "at least" one child is a girl.

See the analysis on wikipedia for the difference between the two questions:
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Second_question

Originally posted by lemon lime
The probability of any one coin toss is always 50/50 no matter how many times the coin is flipped, but I was wondering about all coin tosses when viewed as a group.

With only two kids in a group the second flip is also 50/50, with nothing else to consider. But let's say for example there are 20 children in the household. Assuming the chances of ...[text shortened]... e phone are girls, is it still reasonable to assume the eighth call will be 50/50 (boy or girl)?

No, because that is a different question.

The question is what are the chances of the other child being a girl when there are only two children. With 20 children if you know 7 are girls the chance the rest of them are also girls is less than one percent.

The other difference is you aren't calling again to see if a girl answers the phone. In that case, with two children and you know one is a girl, the chances of a girl answering the phone on the call is 2/3. Edit: and that's assuming one of the children always answer and not the colleague himself.

Originally posted by forkedknight
If the question is phrased, "From all families with two children in which at least one of the children is a girl. What is the probability that the other child is also a girl?"

The answer is 1/3.

The apparent paradox is due to the "at least one child is a girl" statement. How do you determine that "at least" one child is a girl. There are not ve ...[text shortened]... nce between the two questions:
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Second_question

It is not a paradox. One child out of two being a girl is different than one child out of one being a girl.

We have sampled one child, so to speak and we know that child is a girl so that child is now taken out of the equation.

It's like two children are in a box and we take one child out and it is a girl what is the chance the other child is also a girl? Vs two children are in a box and one of them is a girl what are the chances of both being a girl?

There is no paradox unless you ask how we know that one child is a girl. It is not really a paradox because we don't have to know that one child is a girl, we are wondering what the probability both children are girls assuming one of them is a girl.

Originally posted by MISTER CHESS
It is not a paradox. One child out of two being a girl is different than one child out of one being a girl.

We have sampled one child, so to speak and we know that child is a girl so that child is now taken out of the equation.

It's like two children are in a box and we take one child out and it is a girl what is the chance the other child is also ...[text shortened]... wondering what the probability both children are girls assuming one of them is a girl.

Yes, which is why I said "apparent".

The posed questions do sound very similar:

"If there are two children, and we know at least one is a girl, what is the probability that the other is also a girl?"

"If there are two children, and we see that one of them is a girl, what is the probability that the other is also a girl?"