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 ?

(In reply to Solution contradicting all previous ones by Dan Rosen)

First, your notation 5^6=7776 is incorrect. That's 6^5 that's equal to 7776, which is indeed the correct overall number of possibilities.

Now to the actual calculational errors:

You say that each of the ten pairs you list, together with, say, 4,5,6, accounts for 5! of the overall 7776 possibilities. Well that is the case for 1,2; 1,3 and 2,3, but not for the others, as those involve duplicates. I'll take the 4,4 case as that combines two redundancies, one internal and one in a sharing with the 4,5,6 that count.  There are only 20, not 5!=120 cases:

44456
44465
44546
44564
44645
44654
45446
45464
45644
46445
46454
46544
54446
54464
54644
56444
64445
64454
64544
65444

Those with only the internal redundancy, such as 2,2 with the 4,5,6 allow for 60 possibilities, which is still less than the 120 you count:

22456 22465 22546 22564 22645 22654 24256 24265 24526 24562 24625 24652 25246 25264 25426 25462 25624 25642 26245 26254 26425 26452 26524 26542 42256 42265 42526 42562 42625 42652 45226 45262 45622 46225 46252 46522 52246 52264 52426 52462 52624 52642 54226 54262 54622 56224 56242 56422 62245 62254 62425 62452 62524 62542 64225 64252 64522 65224 65242 65422

This occurs with all the 5,5,5 results. This is especially true of 5,5 with 5,5,5. That's only one of the 7776 possbilities, not 120 of them.

BTW, 5! gives you the number of permutations of 5 distinct objects, not perturbations.

Edited on April 2, 2011, 5:18 pm

Posted by Charlie on 2011-04-02 17:17:18