Probability of rolling a specific number given different dice Given a set of dice types $d4$, $d6$, $d8$, $d10$, and $d12$ (see here for notation), is there a way to compute the probability of a given set of dice rolling greater than or equal certain number without having to come up with the entire sample space?
For example, is there a way to determine the probability of a $10$ or greater when rolling $2d4$ and $2d6$, without having to write out all possibilities and counting those which are greater than or equal to $10$?
 A: This is a partial answer, which will hopefully give you some insight on how one could approach this kind of problem.
Let the dice have sides $d_1,\dots,d_n$. Let $a_1,\dots,a_n$ be the rolls of the dice. We want to count the number of tuples $(a_1,\dots,a_n)$ such that $a_1 + \dots + a_n \ge m$ and $1 \le a_i \le d_i$.
Geometrically the region $1 \le a_i \le d_i$ forms a cuboid. We are thus looking at the number of lattice points inside the intersection of this cuboid and the region defined by $a_1 + \dots + a_n \ge m$.
I will now only consider the case where $n \le m < d_i + n$ for all $i$. In this case the region $a_1 + \dots + a_n < m$ is contained in the cuboid. Hence we can calculate the number of lattice points just by subtracting the number of points satisfying $a_1 + \dots + a_n < m$ from the number of points in the cuboid, which is just $d_1 \dots d_n$.
The number of points satisfying $a_1 + \dots + a_n < m$ is the sum of the number of solutions to $a_1 + \dots + a_n = k$ for $k = n, \dots, m-1$. This number is well-known to be ${k-1 \choose n-1}$. So we get
$$\sum_{k=n}^{m-1} {k-1 \choose n-1} = \frac{m-n}{n} {m-1 \choose n-1}.$$
(I calculated the sum using WolframAlpha.)
Thus the probability in this case would be
$$1 - \frac{\frac{m-n}{n}{m-1 \choose n-1}}{d_1 \dots d_n}.$$
Another approach suitable for computer calculation
Another way to look at the problem would be via generating functions. Let $f_i(x) = (x + x^2 + \dots + x^{d_i})/d_i$. Then the answer you are looking at is the sum of the coefficients of $x^{m},\dots,x^{d_1 + \dots + d_n}$ of the polynomial you get by expanding $f_1(x) \dots f_{n}(x)$. This is quite practical to calculate for small inputs.
