This is the answer
This was a famous question Pascal asked Femat in the 17th century.
A sum of 11 is obtained with the following 6 combinations:
(6, 4, 1) (6, 3, 2) (5, 5, 1) (5, 4, 2) (5, 3, 3) (4, 4, 3).
A sum of 12 is obtained with the following 6 combinations:
(6, 5, 1) (6, 4, 2) (6, 3, 3) (5, 5, 2) (5, 4, 3) (4, 4, 4).
Each combination of 3 distinct numbers corresponds to 6 permutations, while each
combination of 3 numbers, two of which are equal, corresponds to 3 permutations.
Counting the number of permutations in the 6 combinations corresponding to a sum
of 11, we obtain 6 + 6 + 3 + 6 + 3 + 3 = 27 permutations. Counting the number of
permutations in the 6 combinations corresponding to a sum of 12, we obtain 6 + 6 +
3+3+6+1 = 25 permutations. Since all permutations are equally likely, a sum of 11
is more likely than a sum of 12.
Note also that the sample space has 63 = 216 elements, so we have P(11) =
27/216, P(12) = 25/216.:D
