Dice Probability question?

Let R be the number on the red die and W be the number on the white die. Then we have the three disjoint cases:

(1) R > W

(2) R < W

(3) R = W

We therefore know that

P( R>W ) + P( R<W ) + P( R=W ) = 1.

By symmetry, we get that

P( R>W ) = P( R<W ).

Also, the probability that they show the same number is fairly easy to compute. Once you roll the red die, regardless of the value on the red die, the probability that the white die shows the same number is 1/6. That is,

P( R=W ) = 1/6.

Putting this together, we get

2 * P( R > W ) + 1/6 = 1,

so

P( R > W ) = 5/12.

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Just because I think it's interesting: If instead we had an (n+1)-sided die (ignore the practical problems of doing this), then comparing my answer with Mike T's, you get

(1 + 2 + 3 + ... + n) / (n+1)^2 = n/( 2(n+1) )

and then by multiplying both sides by (n+1)^2:

1 + 2 + 3 + ... + n = n(n+1)/2.

36 total possible rolls

W  R

1  2-6: 5 rolls

2  3-6: 4 rolls

3  4-6: 3 rolls

...

(5+4+3+2+1+0) / 36