Step 1: Understanding the Question:
This problem follows a geometric distribution where we perform trials until a "success" occurs. A "success" is getting a number from $\{1, 2, \dots, n-1\}$.
Step 2: Key Formula or Approach:
In a geometric distribution, the mean number of trials required to get the first success is $1/p$, where $p$ is the probability of success in a single trial.
Step 3: Detailed Explanation:
A fair $n$-faced die has outcomes $\{1, 2, 3, \dots, n\}$.
Probability of success $p = P(\text{rolling a number } < n) = \frac{n-1}{n}$.
Mean of geometric distribution $= \frac{1}{p} = \frac{1}{\frac{n-1}{n}} = \frac{n}{n-1}$.
Given mean $= \frac{n}{9}$, we set up the equation:
$\frac{n}{n-1} = \frac{n}{9}$
Since $n \in \mathbb{N}$ and $n > 1$, we can divide by $n$:
$\frac{1}{n-1} = \frac{1}{9} \implies n-1 = 9 \implies n = 10$.
Step 4: Final Answer:
The value of $n$ is 10.