Four honest dice

The coefficient of $x^{14}$ in $(x+x^2+x^3+x^4+x^5+x^6)^4$ is the coefficient of $x^{10}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4 \; = \; (1-x)^{-4}(1-x^6)^4 \; = \; \big(1 - 4x^6 + \cdots\big)\sum_{j=0}^\infty \binom{j+3}{j}x^j$ and so is $\binom{13}{10} - 4\binom{7}{4} = 146$ making the probability $\tfrac{146}{6^4} = \tfrac{73}{648}$ , and so the answer is $73 + 648 = \boxed{721}$ .

There are 36^4 outcomes without the restriction of the sum of 14 points. With this restriction and with an 1 or a 6 on the red dice 2x21 outcomes. With this restriction and a 2 or a 5 on the red dice 2x25 outcomes. With this restriction and a 3 or a 4 on the red dice 2x27. Probability $\frac{146}{1296}$ = $\frac{73}{648}$ . a + b = 721.