SOLUTION: a fair six-sided die is rolled three times. what is the probability that the sum of the second and third rolls is less than the value obtained on the first roll?

A probability has a numerator and a denominator.  The denominator
is easy.  6 ways for the 1st roll times 6 ways for the 2nd roll 
times 6 ways the 3rd roll.  That's 6·6·6 = 6³ = 216

The numerator is not so easy.  The smallest sum the 2nd and 3rd rolls
can have is 2, and the largest sum the 2nd and 3rd rolls can have is 5,
when the 1st roll is 6.  So we make a chart:
    A               B           C             D            E           F 
               Possible       Number      Possible    Number of     Product 
 Sum of        2nd & 3rd      of ways     1st rolls    possible       of 
2nd & 3rd       rolls to      to get      for that     1st rolls    columns  
  rolls       get that sum    that sum     sum       for that sum    C & E 
---------------------------------------------------------------------------
    2              1+1          1         3,4,5,6          4           4
    3            1+2,2+1        2          4,5,6           3           6
    4          1+3,2+2,3+1      3           5,6            2           6
    5        1+4,2+3,3+2,4+1    4            6             1           4
---------------------------------------------------------------------------
                                                              Total = 20

So the numerator of the probability is 20.

Therefore the desired probability is 20 out of 216 or  which
reduces to 

Edwin