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

  

Posted by Dan Rosen on 2011-04-02 15:58:33