- boy - boy
- boy - girl
- girl - girl
So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.
To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability.
(This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)
Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.)
So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?
What the problem is really saying is this:
1) You have a large collection of families with two kids of varying genders.
2) You draw one of them at random. At this point, your only estimate of P(2 girls) is 0.25.
3) Someone tells you that the family you drew has at least one girl.
4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3.
The way you are interpreting it is this:
1) You have a large collection of families with two kids, at least one of whom is a girl.
2) Then the probability that the other child is a girl is clearly 50%.
As a reminder this is how the original post phrased the question:
Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls?
That’s not correct in general.
It’s only correct if you assume that “3) Someone tells you that the family you drew has at least one girl.” was equally likely to happen whether or not there were two girls.
That’s a quite strong assumption.
One can make different assumptions and get answers different from 1/3. For example, 1/2.
Ok, AIPedant.
I understand that you may find irritating that someone points out that the original problem is ill-posed and “the answer” depends on how we decide to “rephrase” it.
However, it doesn’t seem boring in the context of a discussion of how the problem is not well-posed and additional assumptions are required to get an answer.
So, in the original: "a family has two children. You're told at least one of them is a girl." What are the possible states? Well, assume first born is the girl, then you have 50% that the next is a girl. Then, assume that the first born was a boy, then there is no chance and the second born is the girl that you know of. So, at 50/50 on those chances, you have 50% chance of having a 50% chance, or a 50% chance of it being 0. I can't see how to combine those to get 1/3. :(
And the Monty Hall explicitly covers the case that a decision is made on which door is shown to you. I don't see any similar framing to this problem. Yes, the total states are GB, BG, GG, but only if you treat GG in such a way that either BG or GB was not a possible state. (That is, using G for girl that you know of, and g for unknown, then possible states are GB, Gg, gG, BG. There is no version of Bg or gB that is possible, so to treat those as equal strikes me as problematic.)
Specifically: it is not true that the firstborn has a 50-50 chance of being a girl, given you were told that the family has at least one girl. The firstborn has a 2/3rds chance of being a girl. This is the heart of your confusion.
In a broader sense there is an entire class of confusing conditional probability problems like this. Events which are causally independent in reality (e.g. gender of a child, which door Monty Hall hid the car behind) fail to be probabilistically independent when you have extra information. Yet these probability games are contrived in a way that our intuition takes over and we use our causal understanding even when a better probabilistic understanding gives you a better answer.
Consider, if you tell me that the Smiths have at least 1 girl, and I meet their daughter but you haven't met either, I have no way of knowing if I met the 1 girl or not. I could ask you, but you would just say, "I don't know, could be her. Could be the other kid. I just know they have at least 1 girl."
This is very different from the monty hall case, where the announcer knows what is behind all doors.
Similar. But different. I was using coins as an example elsewhere. If I flip a dime and a quarter, and tell you that one of them is heads, do you have increased chance of knowing if either particular one is heads? This is more liar's dice than it is monty hall.
This is not coherent as written:
If I flip a dime and a quarter, and tell you that one of them is heads, do you have increased chance of knowing if either particular one is heads? This is more liar's dice than it is monty hall.
The crux of my view on this problem is that I can make an "at least" or an "at most" statement only knowing what one of the coins is, specifically. So, we both flip a coin, I look at mine and say "at least one is heads." You do not get an increased odds of knowing your coin, obviously. (This is the general play of liar's dice. Just, with dice. :) )
That all said, I largely land on agreeing with one of the other commenters here. The assumptions you can bring to this scenario are rather large here. The assumption I would hold would be to mirror this with how it could happen "in the world." And most ways of selecting a family where "at least 1 is a girl" has you encountering a girl. That is, ordering the two events and revealing one at a time.
In modeling this, you could say that you have two coins, one marked with heads on each side and the other a standard fair coin. In this, I can easily see how your odds of seeing heads on the first coin is increased, but odds on them being both heads is now back to 1/2. I can see how this is not the same modeling you are using, but it is still one that I have a hard time shaking from my intuition on the problem.