I found some answers to this question, but that was with only two dice, with 6 sides. What if I have 30 dice, with 10 sides each (from 1 to 10). I can't make a matrix and count as was suggested on the other post.
E.g. what is the probability of getting the sum of at least 100 with 30 dice that has 10 sides each?
On
Abraham de Moivre gave us the means to solve this, with $1800$ dice with $2$ sides, i.e. tossing a coin $1800$ times. In the first place, the mean and the variance of a sum of independent random variables are the sums of the separate means and variances. That is why standard deviations, i.e. square roots of variances, are used as measures of dispersion, rather than using the more obvious mean absolute deviation.
For a die numbered $1$ through $10$, the mean is $$ \frac{1+2+3+4+5+6+7+8+9+10}{10} = 5.5. $$ The mean square is $$ \frac{1^2 + 2^2 + 3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2}{10} = 38.5; $$ consequently the variance is \begin{align} & \frac{(1-5.5)^2 + (2-5.5)^2 + \cdots + (10-5.5)^2}{10} \\[5pt] = {} & \text{mean of squares minus square of mean} = 38.5 - 5.5^2 = 8.25. \end{align} Consequently for $30$ dice, the mean of the sum is $30\times 5.5 = 165$ and the variance is $30\times38.5 = 1155$. The standard deviation is $\sqrt{1155} \approx 33.98529$. Say you want the probability that the sum $X$ is at least $172$.
Note that $\text{“}X\ge172\text{''}$ is the same as $\text{“}X>171\text{''}$. When approximating the discrete distribution of the sum with a continuous distribution, one uses a continuity correction and finds the probability that the continuous random variable is greater that $171.5$. So we find that \begin{align} & \frac{171.5-\text{mean}}{\text{standard deviation}} = \frac{171.5 - 165}{33.98529} \\[10pt] \approx {} & 0.191259 \text{ standard deviations above the mean}. \end{align} So we seek the probability that a standard normal random variable is more than $0.191259$: $$ \int_{0.191259}^\infty \frac 1 {\sqrt{2\pi}} e^{-x^2/2}\,dx \approx 0.42416. $$ I got that last number from a software package.
This continuous approximation works well if the number of random variables being summed is "large". For an underlying distribution that's not too skewed, often a sum of $10$ random variables is enough.
What you want is called "the sum of independent random variables"
Let X and Y each be two random variables with discrete probability mass functions. Let Z denote a new random variable defined to be X + Y. Let x, y, z be possible values of X, Y, Z respectively.
$$Z = X + Y$$
$$P(Z = z) = \sum_{x, y \mid x+y=z} P(X = x, Y = y)$$
$$P(Z = z) = \sum_x P ( X = x, Y = z- x ) $$
The above process will work for two dice, of arbitrary number of sides.
Now you can think of the above random variables, as X being one die, Y being another, and Z being the sum of the two. You can then take the probability mass function of your new variable Z, and then add another dice to it. There is nothing stopping you from rinse and repeat to get the answer you are after.
Let $Q$ be yet another random variable, with possible values of q.
Let $K$ be yet another random variable which represents the sum of the values of three die, those values are $k$.
Note that:
$$K = Z + Q$$
$$P(K = k) = \sum_{q} P ( Z = z, Q = k - q ) $$
$$P(K = k) = \sum_{q} \sum_{x} P ( X = x, Y = z - x , Q = k - q ) $$
To get thirty die calculated you then need to repeat 27 more times.
$$P(\text{Sum} = 100) = \sum \text{die}_1, \sum \text{die}_2, \ldots, \sum \text{die}_{29} P( \text{die}_1 = \text{val}_1 \cdots )$$
It won't exactly be computationally easy, but it should work.
Here is a nice set of notes that works through some of the details.