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Rolling to score (Posted on 2010-11-15) Difficulty: 4 of 5
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 ?

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.3333 (3 votes)

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Solution re (4) contradicting the contradiction ! Comment 17 of 17 |

As I mentioned before my previous solution was mistaken - see Charlie's comments, so I give here my corrected version which is in principle identical to the results others have reached, though to my taste more straight forward :

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 results, which, considering all 6^5 possibilities, will occupy the number of possible perturbations for this result. The results not containing identical digits will occupy 5! perturbations each. Results containing n identical digits will occupy only 5!/n! possibilities, and results containing both n identical digits as well as additional m identical ones will occupy only 5!/(n!*m!) possibilities.  Therefore, considering the 10 above combinations, we get for the case of 4,5,6 the following possibilities : (3*5! + 6*5!/2! + 1*5!/3!) = 740<o:p></o:p>

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. These 15 combinations will occupy the following possibilities : (6*5!/3! + 4*5!/((2!*3!) + 1*5!/5! + 4*5!/4!) = 181.

The 6,6,3 possibility requires the remaining 2 dice to be composed of 1,2,3's, allowing only 6 different dice outcomes ,and those again have to be multiplied by all possible  perturbations as follows : ( 4*5!/(2!*2!) + 1*5!/(3!*2!) + 1*5!/(2!) ) = 190.

The final probability will therefore be :

<o:p> </o:p>

              (740+181+190)/7776 = 1111/7776 = 0.1428755 <o:p></o:p>

 


  Posted by Dan Rosen on 2011-04-03 05:18:46
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