Further on this… I note that you state that the combinations of “BG” and “GB” are different (hence providing three possible combinations out of the original four after ascertaining that the final combination must contain at least one “G”). If that is the case, then you are stating that the order in which these children are placed (whether by age, or otherwise) is a factor here. In that case, should it not be four possible combinations out of an original six? Being “B1B2”, “B2B1”, “G1B1”, “B1G1”, “G1G2”, “G2G1” then reducing down to “G1B1”, “B1G1”, “G1G2”, “G2G1” or “BG”, “GB”, “GG”, “GG” and, once again, returning to the 50% chance of the other child being a “B”.