Probability the other child is also a boy: a linear algebra approach?

Question

The classic conditional probability problem states: you learn that somebody has two children and that one of them is a boy; what is the probability both children are boys? As a prior we have two independent discrete uniform random variables: $X_{1}$ -- sex of the first child and $X_{2}$ -- sex of the second child: $$p(X_{1}) = \begin{cases}0.5 \; | \; x_{1} = B \\ 0.5 \; | \; x_{1}=G\end{cases}$$ $$p(X_{2}) = \begin{cases}0.5 \; | \; x_{2} = B \\ 0.5 \; | \; x_{2}=G\end{cases}$$ We use the multiplication rule to get the joint uniform discrete pdf (I know, I should call it pmf), which is: $$p(X_{1},X_{2}) = \begin{cases} 0.25 \; | \; x_{1} = B, x_{2}=G \\ 0.25 \; | \; x_{1} = G, x_{2}=B \\ 0.25 \; | \; x_{1} = B, x_{2}=B \\ 0.25 \; | \; x_{1} = G, x_{2}=G \\ \end{cases}$$ Now we are given additional information (evidence) that one the children is a boy, so we can condition the distribution, i.e. slice and scale it to arrive at. $$p(X_{1},X_{2} | X_{1} \cup X_{2} = B ) = \begin{cases} 0.33 \; | \; x_{1} = B, x_{2}=G \\ 0.33 \; | \; x_{1} = G, x_{2}=B \\ 0.33 \; | \; x_{1} = B, x_{2}=B \\ \end{cases}$$ Now this case writeout is all very ugly and manual (but at least manageable due to low 2-dimensionality). Is there a nice elegant way with linear algebra to get this conditional distribution? There becomes even more of a need for this when we start adding additional independed random variables like $X_{3}$ -- the first child is born on a particular day of the week, $X_{4}$ -- the second child is born on a particular day of the week; and condition say on Tuesdays.