Proof.
For two dice, the number of favorable cases is 5: (6, 2);(5, 3);(4, 4);(3, 5);(2, 6); so the
probability is 5/36 =
30/216
For three dice, the number of favorable cases is 21: (5, 2, 1), (4, 3, 1) counted with all permutations
(so 6 2 = 12), and (1, 1, 6), (2, 2, 4), (3, 3, 2), counted with the distinct permutations (so
3 3 = 9). The probability is 21/216
So the more likely case is for two dice.
