One way is to use generating functions.
If you have $n$ dice, with sides $a_1,\dots,a_n$, then the coefficient on $x^k$ of the polynomial
$$
\prod_{i=1}^n \left( \frac{1}{a_i} \sum_{j=1}^{a_i} x^i \right)
$$
yields the probability of a sum of $k$ with these dice.
Summing the appropriate values will give you the probability of a sum above a given value.
For example, if you have a three-sided, a four-sided, and a six-sided die, then we want to look at
$$
\left(\frac{1}{3}\sum_{i=1}^3 x^i\right)\left(\frac{1}{4}\sum_{i=1}^4 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{72} x^{13}
+ \frac{1}{24} x^{12}
+ \frac{1}{12} x^{11}
+ \frac{1}{8} x^{10}
+ \frac{11}{72} x^9
+ \frac{1}{6} x^8
+ \frac{11}{72} x^7
+ \frac{1}{8} x^6
+ \frac{1}{12} x^5
+ \frac{1}{24} x^4
+ \frac{1}{72} x^3.
$$
Then, for example, the probability that the sum is greater than $10$ is $\frac{1}{12}+\frac{1}{24}+\frac{1}{72}=\frac{5}{36}$.
Another example. For the probability of throwing 9 or greater with two six-sided dice, we look at
$$
\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{36} x^{12}
+ \frac{1}{18} x^{11}
+ \frac{1}{12} x^{10}
+ \frac{1}{9} x^9
+ \frac{5}{36} x^8
+ \frac{1}{6} x^7
+ \frac{5}{36} x^6
+ \frac{1}{9} x^5
+ \frac{1}{12} x^4
+ \frac{1}{18} x^3
+ \frac{1}{36} x^2
$$
Summing the coefficients on $x^9$, $x^{10}$, $x^{11}$, and $x^{12}$, we find the probability of throwing $9$ or higher is $\frac{5}{18}$.
