We roll five standard dice (sides numbered 1 to 6) and write down the sum of the top three i.e. of the 3 highest values.

What is the probability to get 15 ?

As mentioned by previous solution attempts, 3 dice must show 4,5,6 or 5,5,5 or 663. The rest of the dice have to be as follows :

For the case of 4,5,6 the 2 remaining results must be composed of 1,2,3,or 4's which allows 10 combinations, as follows :

1,1

1,2

1,3

1,4

2,2

2,3

2,4

3,3

3,4

4,4

__Each __of those combinations, together with the 3 results 4,5,6 **constitutes a complete 5 dice result**, which, __considering all 5^6 possibilities__, will occupy **5!** of those possibilities, therefore we get for the case of 4,5,6 (10*5!) satisfying results.

Similarly, for the case of 5,5,5 , the 2 remaining results must be composed of 1,2,3,4,5's, which allows 15 combinations, which will occupy (15*5!) results.

The 6,6,3 possibility requires the remaining 2 dice to be composed of 1,2,3's, allowing only 7 different dice outcomes ,and those again have to be multiplied by all possible 5! Perturbations , giving (7*5!) cases.

The final probability will therefore be :

(10+15+7)*5! / 5^6 = 3840/7776 = **0.493827**