1 in 15 is what I get.

a) 720 permutations total (6!)

b) it will only balance with a 1 2 3 on each side

c) call the balls 1a 1b 2a 2b 3a 3b then there are 6 permutations of any set of 3 balls on one side (3!)

d) there are 8 combinations of a balls or b balls taken in a group of 1 2 and 3.

e) 6*8=48

f) 48 in 720 = 1 in 15

now I hope I didn't make a mistake and my assumptions were correct.