Family Has Two Children I know this has been answered a dozen times, and I know entirely how to get the 2 possible answers. My question revolves around the phrasing of this question, aren't (1) and (2) the exact same question?
A neighbor of yours has two children. Assuming that the gender of a child is like a coin flip, it is most likely that the neighbor has one boy and one girl, with probability of a half. The other possibilities are a quarter each (either two boys or two girls). 
(1) Suppose that you ask the neighbor whether she has any boys, and she said yes. What is the probability that one child is a girl? 
(2) Instead, suppose that you happened to see one of her children passing by, and it was a boy. What is the probability that the other child is a girl?
In both questions, it states that we have knowledge of the neighbor having a boy. The first question, she says yes (she has at least one boy). The second, we see at least one boy. Both asks the probability of the other child being a girl. Clearly, as I've read the other questions on this topic, as well as the wiki this is a play on words. So I guess I'm having trouble depicting which one is 2/3 and which is independent of each other and thus 1/2. 
 A: This problem demonstrates the difference between facts and events. And to understand it, you need to be painfully exact with procedure.


*

*Identify every independent outcome in your sample space, even those that may contradict what you will learn.

*Assign a probability to each outcome.

*An "event" is a set of outcomes defined by some common property. In these questions, the first event you need is the set of outcomes where you obtain the knowledge that there is at least one boy. It isn't necessarily the set where the fact is true, it is the set where you obtain the knowledge by the procedures described.

*The second event is the subset of the first where there is also a girl.

*The answer is found by adding up the probabilities for the outcomes in each set, and dividing the sum for the subset in #4 by the sum for the set in #3.


In Question #1, there are four outcomes: BB, BG, GB, and GG. Each has a probability of 1/4. The mother will answer "yes" for the set {BB, BG, GB}, and "no" otherwise. The subset {BG, GB} also has a girl. So the answer is (1/4+1/4)/(1/4+1/4+1/4)=2/3.
In Question #2, there is another factor needed to define the outcomes. Which child did you see? So there are eight outcomes: BB1, BB2, BG1, BG2, GB1, GB2, GG1, and GG2; where the number indicates which child you see. The probability of each is 1/8.
You saw a boy, meaning the set for #3 is {BB1, BB2, BG1, GB2}. Notice how some cases, where the fact "there is a boy" is true, that are not included. The subset where there is also a girl is {BG1, GB2}. The answer is {1/8+1/8)/(1/8+1/8+1/8+1/8)=1/2.
A: The second scenario is like you ask "Is your older child a boy?", since you are considering a fixed child. This is why the answer is $1/2$, instead of the $2/3$ you get if you ask "Do you have any boys?"
A: It is not a trick. There are four ways to have two children. They are, as you point out, $BG$, $BB$, $GB$ and $GG$. Each of these is equally likely. We are told there is at least one boy so our sample space is reduced to $BB$, $BG$ and $GB$. In two of these, there is a girl so there is a $\frac{2}{3}$ chance of having a girl.
This part is added to answer the edited question.
They are the same. Think in terms of a coin toss. If I see you toss a head, but I don't know whether it was the first toss or the second, the probability of having tossed a tail is $\frac{2}{3}$. It is only $\frac{1}{2}$ if I know which toss I am seeing. The same is true for the boys. If you knew her oldest was a boy then the probability of a girl would be $\frac{1}{2}$. 
There is no difference in this problem.
A: Even simple problems like this one are best worked out by using Bayes theorem. In part (2), it is important to realize that the data favors the hypothesis that there are 2 boys compared to the hypothesis that there is just one boy. In part (1) the data does not favor one hypothesis over the other.
For (1):
$$
P(1B1G|data) = \frac{P(data|1B1G)P(1B1G)}{P(data|1B1G)P(1B1G)+P(data|2B)P(2B)+P(data|2G)P(2G)}=\frac{1\cdot\frac{1}{2}}{1\cdot\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{4}}=\frac{2}{3}
$$
For (2):
$$
P(1B1G|data) = \frac{P(data|1B1G)P(1B1G)}{P(data|1B1G)P(1B1G)+P(data|2B)P(2B)+P(data|2G)P(2G)}=\frac{\frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{4}}=\frac{1}{2}
$$
A: 
Assuming that the gender of a child is like a coin flip
...
(2) Instead, suppose that you happened to see one of her children passing by, and it was a boy. What is the probability that the other child is a girl?

This is the easier one to answer: according to the assumption, the sex of the one child simply has no bearing on that of the other. If I flip heads, what's the probability that my next flip is tails? Or that my previous flip was tails?
For variety, here's another boy/girl puzzle that can be solved with the same observation. Consider a country where families keep having children until they have a son (so every family has exactly one son). What is the ratio of boys to girls?
This can be worked out through an infinite sum, but the easier approach is to see that the choice of whether or not to continue having children has no impact on the probability of any other child being a boy or girl; the ratio is therefore $1:1$.
Incidentally, the above country is a case where the answer to your question would in fact be different: since every family has exactly one son, then if we see a girl in a family with two children, the chance of the other child being a boy is of course $1$.
A: Think about the questions before clicking on the spoiler for the answers.  We can model two child families by coins in different ways.

There are four coins in a bag: two are fair, one is double-header, and one is a double-tailer.  I select one at random, noting which it is, and secretly toss it, covering it with a cup before you see the result. 
I assert that each side has a equal and independent probabilty for being a head; do you agree?
In senario A: I tell you that at least one side is a head.  What is the probability that they both are?

 There were three coins in the bag with at least one head, one of them has two heads, so $1/3$

In senario B: I lift the cup to show that the up-side is a head.  What is the probability that they both are?

 The upside has equal chance of being any of the four heads that were in the bag; two of these had a head on the other side of their coin.  $1/2$.


This time I have two fair coins and two cups (opaque).  I toss each and quickly cover each with a cup. 
I assert that each coin has a equal and independent probabilty for showing a head; do you agree?
Senario C: I peek under the cups and tell you at least one coin shows a head. What is the probability that they both do?

 $1/3$ as above $\tfrac 14/(\tfrac 14+\tfrac 24)=\tfrac 13$

Senario D: I shuffle the cups, and lift one to fairly reveal that it show a head. What is the probability that they both do?

 You know now that this coin shows a head, so the probability that the other does is $1/2$.

 But it is just happenstance that this coin was the one revealed, so let's check to be sure: They could have both been head with probability $1/4$ and when given that then you would have certainly veiwed a head, or the could have been a head and tail with probability $1/2$ and when given that you could have viewed the head with probability $1/2$.  So the conditional probability of them both being heads given that you viewed a head is $\tfrac 14/(\tfrac 14+\tfrac 12\tfrac 12)$ or again $\tfrac 12$.

A: Problem 1:
(A) Family has at least one boy. (B) Family has at least one girl.
$ Pr(B|A)=Pr(A\cap B)/Pr(A)=(1/2)/(1-1/4)=2/3 $
Problem 2:
(C) The child you saw is a boy. (D) the other child is a girl.
$ Pr(D|C)=Pr(C\cap D)/Pr(C)=Pr(C)*Pr(D)/Pr(C)=1/2$
, because C and D are independent.
