Question 1064597
There are {{{6^3}}} or {{{216}}}  possible outcomes. 
Look at the possible rolls that have 1 as the first roll,
(1,1,1)=3
(1,1,2)=4
(1,1,3)=5
(1,1,4)=6
(1,2,1)=4
(1,2,2)=5
(1,2,3)=6
(1,3,1)=5
(1,3,2)=6
(1,4,1)=6
When the first roll changes to 2, 4 of them are removed,
(2,1,1)=4
(2,1,2)=5
(2,1,3)=6
(2,2,1)=5
(2,2,2)=6
(2,3,1)=6
When the first roll changes to 3, 3 more are removed,
(3,1,1)=5
(3,1,2)=6
(3,2,1)=6
When the first roll changes to 4, 2 more are removed,
(4,1,1)=6
There are no rolls that begin with 5 or 6 in the solution set.
So out of 216 possible outcomes, 20 have a sum of 6 or less.
{{{P=20/216}}}
{{{P=5/54}}}