Question 1098975
Let's look at the throws when the second die is a 1.
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1) 
This would lead to x=5 out of 6.
Similarly for the second die equal to 2.
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2) 
This gives x=4 out of 6.
All the way to the second die of 6.
(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
which has a x=0.
Summing them all,
{{{x=5+4+3+2+1+0=15}}}
So 15 out of 36 throws have the first dice exceeding the second. 
Similarly 15 of 36 will have the second dice exceeding the first.
Finally 6 of 36 will have the first equal to the second. 
{{{15+15+6=36}}}
{{{36=36}}}
There are no other possibilities.