Correct way to address the famous "brothers" conditional probability paradox OK, this question will probably stir some controversy. 

There is a well-known paradox (mentioned, among other sources, by Martin Gardner). 

A) You know that I have two children, and the oldest is a daughter. What is the probability that both are daughters? 
 (let's ignore the possibility of identical twins, which may muddle the question; and, for simplicity, assume that probability that a child is a daughter is exactly 1/2).
 This is very easy to answer.

B) Now, suppose you know that at least one child is a daughter (but you don't know if it's the oldest one or not).  What is the probability that both are daughters? 

The common way to answer is: you got extra information that the number of daughters, let's call it D, is at least one. 

Thus, $P(D=2 | D \ge 1) = P(D=2, D \ge 1)/P(D \ge 1) = (1/4)/(3/4) = 1/3$

The paradox is that the answer in (A) and (B) are different!

However, a student told me that this is a wrong way to look at it. 
How do you know that at least one child is a daughter? Perhaps you overheard me saying "my daughter ... (just had a birthday or something)". But in this case you were more likely to overhear in the first place, if I had had two daughters, and not one. This messes up the probabilities calculated above. 

To put it another way: let's generate the list of random 2-child families. Let's randomly "overhear" the gender of a child in each family on the list. Ignore suns and concentrate on daughters. Calculate which proportion of the families we picked had two daughters. Then your answer is 1/2 again. 

Which answer is the right one for (B)? 1/3 or 1/2?  

If you think it's 1/3, can you describe a realistic situation in which you can come up with the information "at least one girl"?
In calculating a conditional probability, the probability distribution of the given event is not under consideration.  It has already occurred.  To confuse this point by reintroducing some prior likelihood of having observed the given event is where your misunderstanding lies.

For instance, suppose I take a survey of people at random.  I ask them (1) how many children they have, and if they answer exactly two, then I ask them if both are boys.  If they say no, then I offer them $20 if they will come to a prearranged location at a specific time.

I then tell you to go to that prearranged location and time.  Then you will meet all these people, all of whom have exactly two children, of which at least one is female.  This situation meets your criteria:  you are given this sample.

Mathematically, you are conflating $\Pr[D = 2 \mid D \ge 1]$ with $$\Pr[D = 2 \cap D \ge 1] = \Pr[D = 2 \mid D \ge 1]\Pr[D \ge 1].$$  Once we are told $D \ge 1$, we are to treat the sample space as if all elementary outcomes for which the given condition is true have equal probability of occurrence.
Knowing that one child is a girl is different from hearing the gender of one child at random.

In the first case, say we ask a 2-child parent "Is one of your children a girl" and the answer is yes.  Then we are given nothing more than the fact that at least one child is a girl, and B applies.  Denote the event of a 'yes' answer as $X$.  Then $P(X) = 3/4$.

Now, if you "randomly overhear" that a child is a girl, call this event $Y$, you have $P(Y) = 1/2$, as it's the same as picking a random child from a random family, which is equally likely to be a boy or a girl.

If gg denotes the 2-girls event,
 you have $P(gg | Y) = \frac{P(gg, Y)}{Y} = \frac{1/4}{1/2} = 1/2$.