Variant of boy or girl paradox

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.

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 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 The first mistake was the same I originally made (see link to my original answer below) that with a boy opening the door there would be a $\frac{1}{3}$ chance of there being two boys and a $\frac{2}{3}$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