Well, I'm not entirely sure, but I do have an idea how it could be done.

What you do is fix the first ball of each pair, then "move" the other ones

(Drawing with O as black and X as white)

|O|O|O|O|O|X|X|X|X|X|X|

| | | | | | | | | | | |

Now, the leftover balls can be sorted in

ways.

only 1 of these is correct. (Namely with first all black balls then all white ones.)

Therefore the probability is

which comes down to

which seems about right to me. (Again, I'm not 100% positive this is the correct way, but the outcome seems to be right.)