This is precisely the first subtility you encounter when you learn probabilities : they are not identical. Paul is right that the formulation of the question doesn't refer to the Monty Hall problem at all, this isn't the same algorithm. But in this case, the probability turns out to be the same. That's the real confusion in Jeff's article.

I mean, it's not even my own deduction, it's what you are taught when you learn probabilities. It's a basic and core paradigm, and I'll digg it up from Wikipedia if somebody still doubts it :)

See my post above. You can take ordering into account if you'd like, but the result is the same, given the parameters of the problem.

Options 2 and 3 each still have the same probability weight as each of options 1 and 4.

No, they don't.

Again, ordering is irrelevant in this problem. We want to know only the probability that there will be a boy/girl pair, not the probability that the boy/girl pair was born in a particular order.

But the same result can be obtained when taking ordering into account -- the key observation is that if you subdivide options 2 and 3 to account for sibling ordering, then you also must subdivide options 1 and 4 to account for sibling order, resulting in 6 total possibilities, of which 2 are M/F sibling pairs, 2 are M/M pairs, and 2 are F/F pairs:

1) M/m

2) m/M

3) M/F

4) F/M

5) F/f

6) f/F

Given the knowledge that one sibling is female, you then exclude 2 of the 6 possibilities (the m/M and M/m pairs), to obtain 2/4 = 50% probability that the pair of siblings is of mixed sex.

The mistake you're making is that you're including ordering on the mixed-sex pairs, but not including ordering on the same-sex pairs.

If you include m and f in the universe of possibilities, then you must add the following combinations as well :

7) m/f

8) f/m

You are not allowed to skip arbitrarily some of the combinations of your universe (wich is now [M, F, m, f]). Probability : 2/3 =]

Don't be silly -- do you think I invented a couple of new sexes by adding lower-case letters? You're just getting thrown by the notation. I could have written the options as:

1) M/M

2) M/M

3) M/F

4) F/M

5) F/F

6) F/F

but I thought that was confusing, so I introduced a symbol to more clearly illustrate the differences between the ordering of the same-sex options.

"Somebody is wrong on the internet". I'm wasting my time and this is my last answer. If you haven't noticed, I only use strict mathematical arguments and I invite you to do the same if you intend to answer. Pure and clear maths please, no litteral arguments about the sexes or god knows, this is the only field where we can verify it.

Lacking elementary probabilities knowledge isn't as dramatic as refusing to learn it, please teach yourself now since nobody will look at this thread again.

Here is my last point, and you can ask any teacher of formal logic, probabilities or maths to verify it :

For universe [M,F], the table of possibilities is :

MM

MF

FM

FF

And that's it. I'm sorry but the table you just made up doesn't exist at all, please go ask one of your teacher about it. If you still believe you are right, and you can prove it, you just discovered a new field in mathematics and probabilities, congratulations.

Rather than insulting me, take a moment to think about the consequences of what you're saying: you're arguing that by announcing the sex of one child in a pair, the probability of the other child's sex being a particular value changes to 2/3. Does that make sense to you? Really?

Again, this has nothing to do with symbols or notation. There are two sexes, two symbols: M, F. The undergrad probability 101 mistake you're making is that options:

MF, FM

take order into account, while options:

MM, FF

do not. This is incorrect. If you take order into account for the mixed-sex case, you must take order into account for the same-sex cases. MM and FF encompass four options with ordering, not two.

I'm sorry if I sounded offensive, and I indeed was, I was tired when I wrote my last comment and my words didn't reflect my tought. I'd like to elucidate this problem once and for all, I really do believe we can both agree on a conclusion.

I'd like you to notice that your new table of probabilities imply that I have a chance of 1/6 to guess the gender setup of a family of 2 children. I don't think that makes sense either to you.

Let's make the experiment a bit clearer :

- We gather a number of families who have 2 childs.

- For each family, we announce the gender of one of the child, but we don't know wich one.

- We are then asked to guess the sex of the other child.

At this point, you believe that the probability to guess right is 1/2, and that the 2/3 probability doesn't make any sense. My claim is that you fall in the Monty Hall problem trap, wich is very counter-intuitive and doesn't seem to make sense at first.

But here is some clarification of the problem :

- What we are really asked is to guess the _gender_ setup of the family. So we need to establish the universe of possible family setups before answering. What are they ?

Even if we don't care about the order, we must acknowledge that there are 2 childs in the family, so there must be a first child, and a second child.

Setup 1 : both childs are boys : M/M

Setup 2 : both childs are girls : F/F

Setup 3 : the first child is a boy, and second child is a girl. M/F

Setup 4 : the first child is a girl, and the second child is a boy. F/M.

Why order matters in setup 3 and 4 ? Because M does not equal F, while M=M and F=F. We investigate not the individual itself, but the property of the individual (in this case, the gender). Therefore, M/F is not equal to F/M, and in the real world there must be a first and a second child.

If you are asked to write down all the possible setups of a family of two in the real world, you would write the same table. You'd say that :

Some families have 2 boys = 1 setup

Some families have 2 girls = 1 setup

Some families have one girl and one boy = 2 possible setups (1st one is a girl OR a boy).

Your argument of F/M = M/F implies that all families have _either_ one of the 2 setups, every first child is a boy, or a girl. But it doesn't work like this in real life. That is why order matters.

Conclusion : if we agree that there are 4 possible setups in a family of 2 childs, then we have a probability of 1/4 to guess the correct setup of the family given NO information. But if we are informed of the gender of one of the child, then one solution of the setup is removed (F/F or M/M), and we have a chance of 2/3 to guess right IF we chose the opposite gender (see Monty Hall problem).

And if we are informed of the gender of one specific child (1st one or 2nd one), then it leaves us with only 2 solutions ! And here, the probability becomes 1/2.