A fair $$n$$ ($$n > 1$$) faces die is rolled repeatedly until a number less than $$n$$ appears. If the mean of the number of tosses required is $$\frac{n}{9}$$, then $$n$$ is equal to _____.

A fair $$n$$-faced die ($$n > 1$$) is rolled repeatedly until a number less than $$n$$ appears. We need to find $$n$$ given that the mean number of tosses is $$\frac{n}{9}$$.

Set up the probability model.

On each roll, the probability of getting a number less than $$n$$ (i.e., rolling 1 through $$n-1$$) is:

$$p = \frac{n-1}{n}$$

The probability of rolling $$n$$ (not stopping) is:

$$q = 1 - p = \frac{1}{n}$$

Find the mean (expected value).

The number of tosses until the first success follows a geometric distribution. The mean of a geometric distribution with success probability $$p$$ is:

$$E[X] = \frac{1}{p} = \frac{1}{(n-1)/n} = \frac{n}{n-1}$$

Set up and solve the equation.

$$\frac{n}{n-1} = \frac{n}{9}$$

Since $$n > 1$$, we can divide both sides by $$n$$:

$$\frac{1}{n-1} = \frac{1}{9}$$

$$n - 1 = 9$$

$$n = 10$$

The value of $$n$$ is 10.