Blue pair of dice outscoring the red pair

Four fair six-sided dice are thrown: two are red and two are blue. What is the probability that the total on the blue dice is strictly greater than the total on the red dice?

$575/1296$. The blue and red totals are identically distributed, so blue-higher and red-higher are equally likely and the answer is $\tfrac12(1-P(\text{tie}))$. A tie has probability $\sum_s c_s^2/1296 = 146/1296$, where $c_s$ counts the ways two dice make $s$ (1,2,…,6,…,2,1). Hence $\tfrac12\cdot\tfrac{1150}{1296}=\tfrac{575}{1296}$.