There are 36 possible ordered pairs of dice rolls.
Let $X_i$ be 1 if the sum of the values of the two dice is 12 and 0 otherwise.
The probability of getting a sum of 12 is $ \frac{1}{36} $ because the only way to get a sum of 12 is to roll a 6 and a 6.
$ Pr(X=1) = \frac{1}{36} $
To find the expected value, we divide 1 by the probability of getting a sum of 12.
$ E(X) = \frac{1}{ \frac{1}{36}} $
$ E(X) = 36 $