[Lecture Mode]
There are three basic rules of probability that are in play here. Now these rules apply to events that are independent, ie the outcome of one tells you nothing about the outcome of the other. Every seems to agree that the sexes of siblings are independent, so these two rules would apply in this case.
1. The sum of the probabilities of all possible events must equal 1.00 (100%), no more, no less. (One of the events must happen, therefore there is a 100% that at least one of them will happen.)
2. To find the probability of any two events, A and B, happening at the same time, multiply the probabilities together. Using standard probability notation, this is P(A and B) = P(A) * P(B).
3. To find the probability of one of several events happening, add the probabilities together. Again, using standard probability notation, P(A or B) = P(A) + P(B).
If you don't understand these 3 rules, I'll wait. Don't go on until you understand and agree with these three rules. If you understand them, but don't agree with them, don't read any further. Instead, go take a basic probability course. I can't tell you anything that will make you see the reality of this situation.
Now that you understand and argree with the three rules above, let's start. It seems most everybody agrees that P(boy)=.50 and P(girl)=.50. True, experimental probability shows this not to be the case, but it does simplify matters. (Girls are slightly more than 50% of the world's population).
What is the probability of two boys?
#2 says P(boy AND boy) = P(boy) * P(boy) = .5 * .5 = .25
By a similar calculation, I can show that P(girl AND girl) = .25. There seems to be no dispute about these two facts. So let's move onto another question.
What is the probability of a boy and a girl?
There some that argue that this boy-girl and girl-boy should be consider the same outcome. So by #1, the two probabilities above, plus this one must have a sum of 1.00 (unless you think there is another possible outcome). What does that tell us about P(one of each)? That is must be .50 (1.00 - .25 - .25 = .50)
The answer here shows you that two boys, two girls and one of each are NOT equally likely. This is what has tripped all of the 50%'er up. If you see these as the only possibilities, that is fine. Just remember that these three are not equally likely and therefore cannot be assigned the same probabilities.
But wait you may say, I thought one of each would be .5 * .5 = .25! Well, that's right, but that's not what you have here. Here, you're saying the first event can be either a boy or a girl. By rule #3 P(boy or girl) = P(boy) + P(girl) = .5 * .5 = 1.00. Then, the second event must be different, ie, if A=boy, it must be that B=girl to have one of each. So P(second is different) = .5 (this works even if A=girl. Go ahead, try it, I'll wait). So now P(one of each) = P(either sex) * P(other sex) = 1.00 * .5 = .5
Let me put it another way. P(boy AND girl) = P(boy) * P(girl) = .5 * .5 = .25, and also P(girl AND boy) = P(girl) * P(boy) = .5 * .5 = .25. So this gives P(one of each) = P(boy AND girl) + P(girl AND boy) = .25 + .25 = .50 No matter how you address it, getting one of each sex is TWICE as likely as getting two boys or two girls. This is VERY crucial to understanding how the answer is arrived at in this problem. If you think there are three possibilities [bb, gg, 1 of each], you must agree that they are not equally likely. If you want all equally likely possibilities, you must use [bb, gg, bg, gb] as this is the only way to get the sum to be 1.00.
If you know that one of two children is a boy, what is the probability that the other is a girl?
Let's look at the three outcome situation first. P(boy AND boy) = .25, P(girl AND girl) = .25, P(one of each) = .50 and we want P(girl if we know one is boy). Only the first has a boy, and P(boy AND boy) + P(one of each) = .75, and only the third has a girl, that is .50 out of .75, which is .50/.75 = 2/3 = .66 so there is a 66.6666.....% chance that if you know that one of two children is a boy, the other is girl.
If you prefer the four outcome situation, P(boy AND boy) = .25, P(girl AND girl) = .25, P(boy AND girl) = .25, P(girl AND boy) = .25 and the only the first, third and fourth outcomes come into play. Since these are equally likely, (ie, same probability), and 2 of the 3 have a girl, the probabilities is 2/3 = .6666... or 66.6.....%.
The lesson is you must know if your outcomes are equally likely or not. Once the probabilities of your chosen outcomes are known, be they equally or unequally likely, the answer will be the same.
[/lecture mode]
Oh, and by the way, I have two daughters, and rarely when talking to other do I identify them as "my older daughter" or "my younger daughter". I usually just say "my daughter", not really caring whether the listeners knows I have another one or not, so that semantic argument holds no water what so ever.
ETA: Try this one on for size. You and your two friends, Tom and Harry, are captured in Lithistania by Geristal extremists. The terrorist will execute two of you at dawn, just to prove they are a real threat to the Lithistanian government. What is the probability you will be executed at dawn? (This part is easy) But later, you overhear the guards arguing about who has to take Harry to the execution room to be beheaded at dawn. Now, what is the probability that you will be executed at dawn?