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?

[pixie373]

But here- having a boy or having a girl are equally likely...

[pixie373]

But I admit. I have 2 boys and 2 girls. If one of them was standing at the window, the chance of one of the others being the same is 1/3.

[Prudie]

How unbelievable that something so obvious has become so complicated.
I wouldn't even dream of disagreeing with Jim and FF who are undoubtedly right. I'll just live in my own happy world of knowing that it's (in the very precise situation of the OP) 1 in 2 thanks :-)

[Arksided]

Yes please I'm British... oh er... the chances are Equilibrium.

[jim360]

It's essentially just another example of the same problem, where people don't set up their approach correctly and so miss out on information. It's not difficult -- but it is important to be precise. People make it complex because they argue that BG and GB are, from the point of view of the problem, the same event (which is true) but then treat it as only one outcome, when in reality it's two separate outcomes that are counted in the same way. This is equivalent to setting 1=2, and of course any conclusion drawn is just wrong.

* * *

Just to fog the issue, the obvious extension is that if you met a family of 10 children, say, and were told that nine were girls, then the probability that the tenth is also a girl is 1/11 by the same argument. But this time it's a dodgy argument because you could well say that having so many children of the same gender is a rare enough event that it's possible this is a family where the female sperm are more likely to fertilise the egg for whatever reason. (In terms of coins, you could question whether or not the coin is fair).

[Prudie]

No, we'd say the probability that the 10th child is a girl 1 in 2 because it can only be one or another.

And clearly it is difficult.

[woofgang]

no, Dave because there are many more options between getting all six numbers or not and many different numbers to choose from.
To apply your lottery analogy to this question, I only have to get one number right to win and I only have a choice between two numbers.
Then surely i would have a 1 in 2 choice of winning the lottery?

[jim360]

Sigh... Prudie, it's dodgy thinking. The "tenth child" is ill-defined if you are told that nine are girls, because it could be GGGGGGGGGB or GGGGGGGGBG or GGGGGGGBGG or... the point is that there are evidently more permutations of nine girls and one boy than of 10 girls, but that each permutation of 9G/1B is euqally likely and as likely as 10G.

Hence 1/11, assuming 50/50 at each individual birth. All other thinking is flawed and misses this important detail that the order matters when counting how many relevant outcomes there are.

[sunny-dave]

I just keep going back to

"there are twice as many families with a boy and a girl, as there are with two boys"

- that is a complete and undeniable fact if we have 50/50 chances of boys and girls being born.

So, if we see a boy, there are two in three chances that the hidden child will be a girl.

Please explain why that is not true?

[Prudie]

Well as you're going to sigh - I'm off

[jim360]

Well I'm sorry for sighing, but having provided a fully-worked solution -- and not being the only one to -- and you still are ignoring it, not much else I can do.

In a family of n children, of which (n-1) are of the same gender, then the remaining child is going to have the same gender with probability 1/(n+1) assuming that boys and girls are equally likely. That's indisputable.

[jno]

we know one is a boy but we don't know which one. So he might be a boy with an older sister (GB), a boy with a younger sister (BG), or a boy with a brother (doesn't matter whether older and younger as both are BB). That would be 1 in 3. If I've got the reasoning right.

[Sipowicz]

I've got three children........one of each!

[Prudie]

Jim I'm not the only one to think it's 1 in 2 and presumably factor wouldn't have asked the question if it was so damned obvious to us thickos, nor would there be lots of artcles on the topic if it were so clear cut.

You can point out the error of my ways in a reasoned fashion but don't speak to me like I'm a child - and if you ever get that PhD I'd suggest you don't waste your talents on teaching.

[sunny-dave]

After a spot of lunch (well a tin of soup actually) - I've got one more comment.

I think the confusion is entirely about the difference between

1. the probability of a single yes/no event (which is 50/50)

and

2. the probability of such an event occurring as part of a series of events - which in this case is 1 in 3.



Prudie - I hope I haven't been patronising (I really don't want to be thought of like that). Can you have a look at my post at 13:38 and answer the question?

[-Talbot-]

You are stood with Mr Smith and his son waiting for Mrs Smith to come along with his other child. She will turn up with either a boy or a girl...50/50.

Please tell me how in this actual, physical situation can it be anything other than 50/50........ It doesn't matter if mrs smith turns up with a fifteen your old girl or a two year old boy.

Posters are just adding their own complications (IMO)

[sunny-dave]

-talbot-

Because (in the statistically correct model we are using) it is twice as likely that she will have had a girl as the other child, as a boy.

[jomifl]

How the question is asked doesn't alter the probability.

[jno]

///You are stood with Mr Smith and his son waiting for Mrs Smith to come along with his other child.///

Talbot - imagine you are stood with Mr Smith and a child but you can't see what sex it is waiting for Mrs Smith to come along with his other child.

Then what's your answer?

[New Judge]

Can we move on to the "Monty Hall 3 door problem", please? :-)