Question

If a die is to be rolled until all sides have appeared at least once, find the expected number of times that outcome $1$appears.

Short answer

The expected number of times that outcome $1$appears is 2.45

Step-by-step solution

Given Information

If a die is to be rolled until all sides have appeared at least once,find the expected number of times that outcome $1$appears.

Explanation

It is known that if we roll a six-sided fair die, there are $6$possible outcomes, each one with a probability value $p=\frac{1}{6}$. Assume that die is rolled until all sides have appeared at least once, and in that case, let X be the number of rolls. Further, let random variable Xi represents the number of rolls needed to arrive at a new face on the die if i different faces have already appeared up. Then,

$X=1+X_1+X_2+X_3+X_4+X_5$

$p_2=\frac{4}{6}$, X$3$is a geometric random variable with parameters$p_3=\frac{3}{6}$, X$4$is a geometric random variable with parameters$p_4=\frac{2}{6}$and X5 is a geometric random variable with parameters$p_5=\frac{1}{6}$. Therefore,

$E\left[X\right]=1+\sum _{i=1}^{5}E\left[X_i\right]=1+\sum _{i=1}^5\frac{1}{p_i}=1+(\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1})=14.7.$

Explanation

Now, let's define indicator variables Ij, j=$1,2$, ...., X as:

$I_j=\{\begin{array}{ll}1,& \text{if}{E}_1\text{occurs}\\ 0,& \text{if}{E}_1\text{does not occur}\end{array}$

Whereby $E_1$denote the event :

$E_1="\text{the outcome}1\text{appears " ,}P\left\{E_1\right\}=p\text{.}$

Then, if Y represents the number of times that outcome $1$appears, we have that

$Y=\sum _{j=1}^X{I}_j$

and therefore the expected numberof time that outcome$1$appear is:

$E\left[Y\right]=E\left[\sum _{j=1}^X{I}_j\right]=E\left[E[\sum _{j=1}^X{I}_j\mid X]\right]=E\left[X\underset{=p}{\underbrace{P\left\{E_1\right\}}}\right]=pE\left[X\right]=\mathbf{2}\mathbf{.}\mathbf{45}.$

Final Answer

The expected number of times that outcome$1$ appears is 2.45