Let a, b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $$1, 2, 3, 4$$. If the probability that $$ax^2 + bx + c = 0$$ has all real roots is $$\frac{m}{n}$$, $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to ________

Fair tetrahedral die with faces 1,2,3,4. For $$ax^2+bx+c=0$$ to have all real roots, we need $$b^2 - 4ac \geq 0$$.

Total outcomes = $$4^3 = 64$$.

We need $$b^2 \geq 4ac$$. Enumerate favorable cases:

b=1: $$1 \geq 4ac$$, impossible for $$a,c \geq 1$$.

b=2: $$4 \geq 4ac \implies ac \leq 1 \implies a=c=1$$. 1 case.

b=3: $$9 \geq 4ac \implies ac \leq 2$$. Cases: $$(1,1),(1,2),(2,1)$$. 3 cases.

b=4: $$16 \geq 4ac \implies ac \leq 4$$. Cases: $$(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(4,1)$$. 8 cases.

Total favorable = $$0+1+3+8 = 12$$.

Probability = $$\frac{12}{64} = \frac{3}{16}$$.

$$\gcd(3,16) = 1$$, so $$m = 3, n = 16$$.

$$m + n = 19$$.

The answer is $$\boxed{19}$$.