Like tempestn noted in another comment, enumeration is straightforward for small families, and makes it crystal clear that for any family size N, boys and girls

Like tempestn noted in another comment, enumeration is straightforward for small families, and makes it crystal clear that for any family size N, boys and girls have the same average number of brothers: Gender: B G # brothers: 0 0 B avg # brothers: 0 G avg # brothers: 0 Gender: B,B B,G G,B G,G # brothers: 1,1 0,1 1,0 0,0 B avg # brothers: (2 + 0 + 0) / (2 + 2) = .5 G avg # brothers: (0 + 2 + 0) / (2 + 2) = .5 Gender: B,B,B B,B,G B,G,B G,B,B B,G,G G,B,G G,G,B G,G,G # brothers: 2,2,2 1,1,2 1,2,1 2,1,1 0,1,1 1,0,1 1,1,0 0,0,0 B avg # brothers: (6 + 6 + 0 + 0) / (3 + 6 + 3) = 1 G avg # brothers: (0 + 6 + 6 + 0) / (3 + 6 + 3) = 1 Gender: B,B,B,B B,B,B,G (x4) B,B,G,G (x6) B,G,G,G (x4) G,G,G,G (I collapsed the different permutations here) # brothers: 3,3,3,3 2,2,2,3 1,1,2,2 0,1,1,1 0,0,0,0 B avg # brothers: (12 + 24 + 12 + 0 + 0) / (4 + 12 + 12 + 4) = 48/32 G avg # brothers: (0 + 12 + 24 + 12 + 0) / (4 + 12 + 12 + 4) = 48/32 In a family of N children, both boys and girls will have, on average, (N-1)/2 brothers. Arguments for unequal results are slipping in assumptions that make the observed distribution different from the enumerated one. It's an exercise for the reader to prove that this equality holds for higher N, it's quite easy (the way I formatted the arithmetic shows how to match up terms).