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 ?

There are 6^5 possible combinations which comes to 7776.  I created a matrix of five columns, 7776 rows deep which contained all these combinations.  then in five subsequent columns, I assigned a value of 10^x for the corresponding cell column in the first matrix.  For example, if a 6 appeared in the first column of a row, then the value 100000 was calculated for the first column in the subsequent matrix.  I then added the values of these cells and derived a number in which each digit represented the quantity of values for that position.  For example, a roll of 42326 would appear as 101120.  (Note, the digits of these 6-digit numbers will always equal 5).

I then truncated these numbers to get rid of the last two digits, leaving four digit numbers.  I then used search options to find the combinations needed for a value of 15.  These are as follows:

03xx, 04xx, 0500 - 555 
(Three or more rolls of 5, and no rolls of 6) = 181 combinations
111x,112x,1130 - 654 
(One roll of 6, one roll of 5, and at least 1 roll of 4) = 740 combinations

2001, 2002, 2003 - 663
(Two rolls of 6, no rolls of 5, no rolls of 4, and at least one roll of 3) = 190 combinations

Total combinations of 15 = 1111

Probability = 1111/7776 ≈ 1/7  

Posted by hoodat on 2010-11-16 16:15:02