Sex

[factor-fiction]

Following on from the entertaining Dice and Socks threads earlier this week I've now found the problem about the sex of children.

One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"

Thoughts please?

[scooping]

Don't think there is an answer to that.
Example what are the odds of heads if the person tossing has got 37 tails. The odds are always the same, that is to say that each toss has got an equal chance of being heads or tails. So, on the toss 38, the odds of heads are 2-1, the same they were on the first toss.

[woofgang]

Setting aside genetics and Mr Smith's family tendency to produce boys or girls, its 50/50 or 1 chance in 2. The information you are given regarding the gender of one child is irrelevant.

[Prudie]

Well there are 3 possible combinations of 2 children - BB,GG,BG (assuming order of issue is irelevant) so as we can ignore those with just girls it's 1/2.

Alternatively doing it via a tree method, if you start with B you can only have B or G next so still 1/2

[dzug2]

Well if you ignore the fact that slightly more boys are born than girls, then chances are 50:50 that the second child is a boy

But that's not the way the question is phrased. If you ignore previous knowledge then I'm not sure the answer is the same

[bert_h]

The answer, contrary to intuition, is: ONE-THIRD.

Considering all possible two-children families, the second child has a probability one-half of being the same sex as the first. Therefore half of all two-children families are the same sex (B+B or G+G), and half of them are one of each (B+G or G+B). Note especially that the three possibilities (GG, BB, and BG) are NOT equally likely.

Mr Smith's family are not (G+G), and the case in question (B+B) constitutes one-third of the remaining equally likely cases.

[factor-fiction]

scooping- by 2-1, if you mean 1/2, I agree for the coin toss.

When this was last on here, I argued that the probability was 1/2 (i.e. 50%) but the OP then insisted the answer 1/3 (33.33...%) and I have seen numerous articles which justify 1/3.

But whilst I can see the argument for 1/3 I feel uneasy as the question as I think the question is ambiguous and 1/2 can be justified too.

[factor-fiction]

Sorry bert-h, yours wasn't there when I started typing.
Despite all my typos I hope mine makes sense.
Yes, bert, the argument I have seen for 1/3 was based exactly on what you say.

But I still struggle with this. I will explain why shortly

[Prudie]

Firstly these children are already born so any stats about family trends or one sex being more likely than the other is totally irrelevant in this question.

And sorry so is your argument bert, we already know 100% that one child is a boy so GG is discounted at the start.

[sunny-dave]

It's a classic trick question - already nicely explained by bert_h.

In birth order you can have :

BG
BB
GG
GB

So (ignoring GG since we know that one is a boy) BB is a one in three chance.

(prudie's analysis misses the GB possibility)

[Prudie]

No I didn't, I said assuming order of issue is irrelevant therfore in this case BG abd GB are the same single event

[sunny-dave]

Nope - they are not the same event - and therefore twice as likely to happen.

[factor-fiction]

Sunny-dave/ bert-h-- yes, I understand that analysis but I am troubled by the wording of the question when it says "at least one is a boy". How can one know that 'at least one is a boy?
Suppose all I know is he has two children. I walk past his house and see a young boy standing at the window in his pyjamas. So now I know at least one of his children is a boy. But that tells me nothing about the sex of the other child- and isn't there pretty much a 50-50 chance of a child being a boy or a girl?

[factor-fiction]

There are lots of articles about this, by the way.
e.g.-
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

[ToraToraTora]

Evens, as you know one is a boy then it's even money that the other is also.

[sunny-dave]

I don't think so factor - you don't know whether he is the first or second born child, so the BG and GB events remain separate and so (in total) are twice as likely as the BB event.

It's very counter intuitive though (my brain is almost refusing to type this answer).

[ToraToraTora]

if he had 99 children and 98 of them are boys what is the chance that th eother one is a boy? also evens.

[factor-fiction]

Isn't that just complicating things, sunny-dave?
You know someone has exactly two children. You then see one of them and it's a boy. Doesn't matter if it's younger or older than the other child. Seeing the boy tells you nothing about the sex of the other

[sunny-dave]

How about rephrasing the question ...

Someone has tossed a coin twice.

You are told that the result wasn't two heads.

What are the odds that it is two tails?

It is (obviously) one in three.

How does that differ from this problem?

[factor-fiction]

I would say it depends- if they show me one tail (or I happen to spot that one is a tail) then I still know nothing about the other coin so I'd say it was 50-50 that the other was a tail

[sunny-dave]

Aha - so you think it depends on the way you are made aware of the information?

But the results of the toss (or sex of the children) has already been determined - so how can your subsequent acquisition of information affect the probability of the results?