A fair n(n 1) faces die is rolled repeatedly until a number less than n appears.

Question

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 \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

In this case, the success is defined as getting a number less than $n$ when rolling an $n$-sided die. The probability of success, $p$, in each roll is then $(n-1)/n$, and the probability of failure, $q=1-p$, is $1/n$.

$=\sum _{i=1}^{\infty}{p}_{i}{x}_{i}=1\cdot \frac{n-1}{n}+\frac{2}{n}\cdot (\frac{n-1}{n})+\frac{3}{{n}^{2}}(\frac{n-1}{n})+\ldots$

$\frac{n}{9}=(1-\frac{1}{n})S$ ......(1)

where

$\begin{gathered} S=1+\frac{2}{n}+\frac{3}{{n}^{2}}+\frac{4}{{n}^{3}}+\ldots \ \frac{1}{n}S=\frac{1}{n}+\frac{2}{{n}^{2}}+\frac{3}{{n}^{3}}+\ldots \ ---------------- \ (1-\frac{1}{n})S=1+\frac{1}{n}+\frac{1}{{n}^{2}}+\frac{1}{{n}^{3}}+\ldots \end{gathered}$

$\begin{gathered} \Rightarrow (1-\frac{1}{n})S=\frac{1}{1-\frac{1}{n}} \ \Rightarrow \frac{n}{9}=(1-\frac{1}{n})	imes \frac{1}{{(1-\frac{1}{n})}^{2}}=\frac{n}{n-1} \end{gathered}$

$\Rightarrow$ n = 10

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