Variant of boy or girl paradox Here is an interesting question my friend brought up:

You are invited to a family party. A Boy opens the door for you. There
  are two children there. What is the probability that a boy opens door for
  you next time?

This is my solution:
$$P(\text{boy second time | boy first time}) = \frac{P(\text{boy both time)}}{P(\text{boy first time)}}$$
$$=\frac{1/3+1/3*1/2*1/2+1/3*1/2*1/2}{1/3+1/3*1/2+1/3*1/2}=3/4$$
The $1/3$ comes from the fact that given we have one boy already so the combination can only be {boy, girl}, {girl, boy}, {boy, boy}. The 1/2 comes from for each time there is a 1/2 probability that a boy will open the door if there is one boy one girl.
Is my calculation correct? I feel not confident about the 1/3 argument.
Edited: @Rolf proposed another point that actually the probability of two boys and one boy/one girl should be 1/2, not 1/3 each.
Edited again: Actually there are two approaches, as shown below. The key is to whether to assume the prior or not in the calculation.
 A: A boy opened the door the first time around, but we are told there are two children (and we are assuming the boy is one of the two). Assuming there is an equal chance the other child is a boy or girl, that means there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood. Hence, a boy opening the door at any time is 
$$\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$$
So ... @James ... you got the right answer ... but the wrong method. In fact, you made two mistakes.  The first mistake was the same I originally made (see link to my original answer below) that with a boy opening the door there would be a $\frac{1}{3}$ chance of there being two boys and a $\frac{2}{3}$ chance of there being one boy. No, that is really just $\frac{1}{2}$ and $\frac{1}{2}$. The second mistake is that you used that new sample space, that is based on the fact that a boy opened the door the first time, to calculate the probability of a boy opening the door the first time! No, if you calculate it this way, you should make no assumptions at all and thus you have a sample space of BB, BG, GB, and GG. .. this is what Robert did in his answer.
Original (incorrect) answer
A: This is a classic underspecified problem.

You are invited to a family party. A Boy opens the door for you. There are two children there. What is the probability that a boy opens door for you next time?

We can make many sets of assumptions about facts not mentioned, and get different responses.
As an example, what if we assume we live on an island where there are no girls.  Then the probability next time it is opened by a boy is clearly 100%.
What if we assume children always alternate turns opening doors, and each child has a 50% chance of being a boy or a girl.  Then the probabilty is 50%.
If we assume a random child opens the door, that children are randomly boys or girls, then the probability is 75%.
If we assume that, if there are two boys, neither opens the door and instead an adult does, and if there is a boy and a girl they take turns, then the probability is 0%.
So unspecified criteria about how the door is chosen to be opened and the distribution of children and how they interact permits the probability to be anywhere from 0% to 100%.
This is the same issue that the classic Monte Hall problem has: there are reasonable assumptions that can set the answer anywhere.  People pick some set they consider reasonable and draw a conclusion with them.
Now, on a test, you'd just answer 75% and state your assumptions directly, then draw conclusions from those assumptions.  But that is a test taking strategy, not a mathematical truth.
Assume each child is randomly of either gender.
Assume a random child always opens the door.
      BB   BG   GB   GG
BB   100%  0%   0%   0%
BG    25%  25%  25%  25%
GB    25%  25%  25%  25%
GG    0%   0%   0%  100%

on the left is the gender of the two children.  Each of the rows has equal probability.  We could simply divide every probability by 4, but a later step makes that not required as we are going to normalize it anyhow.
On the top is who opens the door 1st and second time.  In the middle is the chance that this happens.
We now eliminate every case where a boy doesn't open it the first time:
      BB   BG  
BB   100%  0%  
BG    25%  25% 
GB    25%  25% 

Now add up percentages:
      BB   BG
     150%  50%

and normalize:
      BB   BG
      75%  25%

75% chance that a boy opens the door the 2nd time.

If you prefer more math, we can use Bayes's theorem.  Here are the initial probabilities (same chart as above, except divided by 4):
      BB   BG   GB   GG
BB    1/4   0    0    0
BG    1/16 1/16 1/16 1/16
GB    1/16 1/16 1/16 1/16
GG    0    0    0    1/4 

now we want the probability of boy boy given boy first, or:
P(BB|BX) = P(BX|BB) * P(BB) / P(BX)

P(BB|BX) = 100% * (3/8) / (1/2)
= 75%

Alternatively, eliminate the cases where a boy doesn't open the door first:
      BB   BG  
BB    1/4   0  
BG    1/16 1/16
GB    1/16 1/16
GG    0    0   
---------------
      3/8  1/8  = 1/2

Giving is P(XB|BX) = (3/8)/(1/2) = 3/4
A: There's definitely one boy, so half of the possibilities have a probability of one:
$$\frac{1}{2}\times 1=\frac{1}{2}$$
Assuming an equal chance of either boy or girl, the other half will have an equal chance of being a boy: 
$$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$
Add them:
$$\frac{1}{2}+ \frac{1}{4}=\mathbf{\frac{3}{4}}$$
Done.
A: I would think about it differently. Next time, the chance that second kid (no matter it's sex) opens door is $P(second) = \frac{1}{2}$. 
Chance of second kid being girl is $P(girl) = \frac{1}{2}$.
Therefore, you have chance $P(second) * P(girl) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ girl opens the door. Therefore $P = 1 - P(second) * P(girl) = \frac{3}{4}$
A: Let's suppose we know that there will be two children at the party, and we make the rather doubtful assumptions that each is equally likely to be boy or girl, and each time someone opens the door it is equally likely to be each of the two children, independent of who opens the door on other occasions.  
With probability $1/4$ both children are boys, and the door is opened by a boy both times. 
With probability $1/4$ both are girls, and the door is opened by a girl both times.
With probability $1/2$ one is a boy and one is a girl, and the door openers are equally likely to be (boy,boy), (boy,girl), (girl,boy), (girl,girl).
Then 
$$ \frac{P(\text{boy both times})}{P(\text{boy first time})} = \frac{(1/4) \cdot 1 + (1/2) \cdot (1/4) + (1/4) \cdot 0}{(1/4) \cdot 1 + (1/2) \cdot (1/2) + (1/4) \cdot 0} = \frac{3}{4}$$
A: While everyone arrives at the same answer, I have a problem with some of the reasoning:
"There's definitely one boy, so half of the possibilities have a probability of one:"
I think that there are three possibilities, with probabilities of 1, 1/2 and 1/2.
"there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood." I think it is twice as likely to be mixed-sex as not. Here is my solution:
The possibilities for the children are BB, BG, and GB, since they are distinguishable by age. By Bayes theorem, given that the door was opened by a boy the probabilities for these combinations are 1, 1/2 and 1/2, each divided by their sum, or 1/2, 1/4 and 1/4. Thus, there is a probability of the door being opened by a boy the second time of 1 in the first case and 1/2 in each of the other two cases. Multiplying these probabilities by the probabilities of the cases and adding gives 1/2 + 1/8 + 1/8 = 3/4.
We assume the children take turns opening the door or that they respond at random.
A: There are either 1 or 2 boys in the family; call that count $X$.  
You want the probability that a boy opens the door next time, given one did this time.  Call those events $B_N, B_T$ respectively. Bayes' Rule gives us that:
$$\begin{align}\mathsf P(B_N\mid B_T)&=\dfrac{\mathsf P(X{=}1, B_T, B_N)+\mathsf P(X{=}2, B_T, B_N)}{\mathsf P(X{=}1, B_T)+\mathsf P(X{=}2, B_T)}\\[2ex]&=\dfrac{\tfrac 24\cdot\tfrac 12\cdot\tfrac 12+\tfrac 14\cdot\tfrac 11\cdot\tfrac 11}{\tfrac 24\cdot\tfrac 12+\tfrac 14\cdot\tfrac 11}\\[2ex]&=\dfrac 34\end{align}$$

Alternatively, as has been proposed.  A boy will open the door next time (given one did this time) if it is the same boy, or its the other child which turns out to be a boy$$\mathsf P(B_N\mid B_T)=\tfrac 12+\tfrac 12\cdot\tfrac 12=\tfrac 34$$

So both approaches give the answer of $~\dfrac 34~$.

To test that we are not just accidentally getting the same answer, consider if it was that there were three children in the family.   The first approach gives us $\tfrac{\tfrac 38\tfrac 19+\tfrac 38\tfrac 49+\tfrac 18\tfrac 99}{\tfrac 38\tfrac 13+\tfrac 38\tfrac 23+\tfrac 18\tfrac 33}=\tfrac 23$, while the second approach gives us $\tfrac 13+\tfrac 23\cdot\tfrac12=\tfrac 23$.
