1 in 733

It is soooo easy to make a mistake. I should've divided the total combinations of the fixtures by the permutations within themselves, i.e. AvB, CvD, EvF is the same combination as AvB, EvF, CvD.

So the total possible fixture combinations is ((20 choose 2) * (18 choose 2) *...*(4 choose 2))/10! = 654729075

The total possible fixture combinations featuring only teams playing in the same half is (((10 choose 2) * (8 choose 2) * (6 choose 2) * (4 choose 2))/5!)^2 = 893025

So 893025/654729075 = 0.001363961, i.e. 1 in 733

Posted By: yarmyyarmy, Nov 16, 16:30:41