# Die puzzle with probability

I have basic understanding of probability but I couldn't get following question. Please help me to get it.

There are $n$ dice each with $m$ faces numbered from $1$ to $m$. Given a number $x$, find out the probability that the sum of all numbers obtained from the $n$ dice in a trial is greater than $x$.

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• Fact: Probability generating function (pgf) for the score of one die is $$\mathcal{P}(z) = \frac{1}{m} \left(z+z^2+\cdots+z^{m}\right) = \frac{z}{m} \frac{1-z^{m}}{1-z}$$
• Fact: The pgf of the total score (assuming independence of scores) $X$ is $\left(\mathcal{P}(z)\right)^n$.
• Fact: The generating function for the sequence $\mathbb{P}(X>k)$ is $$\sum_{k=0}^{n \cdot m} z^k \mathbb{P}(X>k) = \frac{1- \left(\mathcal{P}(z)\right)^n}{1-z}$$
Thus the answer to your question is $$\mathbb{P}(X>k) = [z^k] \frac{1-\left(\frac{z}{m} \frac{1-z^{m}}{1-z}\right)^n}{1-z}$$