When you divide the problem that way, you get the two cases which equal likelyhood:
in case mentioned boy is older:
B, B p = 1/3
B, G p = 2/3
in case mentioned boy is younger:
B, B p = 1/3
G, B p = 2/3
Why is it in both cases 1/3 vs. 2/3, and not, as you claim, 1/2. You have to look at the probability distribution of the underlying set. It is:
B, B = 1/4
B, G = 1/4
G, B = 1/4
G, G = 1/4
So all have the same likelyhood. In our breakdown into the two paths "younger and older" above notice how "B, B" is mentioned in both of them. Since "B, B" will occur on average just as often as "B, G" or "G, B", all the "B, B" cases will be split between the "boy younger" and "boy older" path, and it's probability of occurring in either path is thus halved when compared to it's alternative
When all Probabilitys are added up, the end result for the probability that you get B, B is the same.
p = 1/2 * p[path 1] + 1/2 * p[path 2]
p = 1/2 * 1/3 + 1/2 * 1/3 = 1/3