Let $ X $ Be the Number of Faces that Never Showed Up in $ n $ Dice Rolls - What's $ \mathbb{E} \left[ X \right] $?

Question

This is a example from a book. Suppose a fair die is rolled n times, and let X be the number of faces that never show up in these n rolls. $E(X)=?$

The method as suggested in the book is:-

Define $A_i=$ $i^{th}$ face is missing . $\therefore X= \sum_{i=1}^{6}A_i$.

Here is my problem , According to the method suggested above,$ X$ can take values from {0,1,2...6}.

Then the sample space of $ X={0,1,2,3,4,5}$, since at least 1 face has to show up in the n rolls.

Where am I going wrong?

Answer 1

By way of enrichment here is how to solve it using EGFs. Supposing that the die has $q$ faces and is rolled $n$ times we have from first principles for the expectation

$$\mathrm{E}[X] = \frac{1}{q^n} \sum_{p=0}^q p {q\choose q-p} n! [z^n] (\exp(z)-1)^{q-p} \\ = n! [z^n] (\exp(z)-1)^q \frac{1}{q^n} \sum_{p=0}^q p {q\choose p} (\exp(z)-1)^{-p} \\ = n! [z^n] (\exp(z)-1)^q \frac{q}{q^n} \sum_{p=1}^q {q-1\choose p-1} (\exp(z)-1)^{-p} \\ = n! [z^n] (\exp(z)-1)^{q-1} \frac{q}{q^n} \left(1+\frac{1}{\exp(z)-1}\right)^{q-1} \\ = n! [z^n] \frac{q}{q^n} \exp((q-1)z) = \frac{q}{q^{n}} (q-1)^n = q\left(1-\frac{1}{q}\right)^n.$$

What we see here confirms the result from linearity of expectation with $q$ indicator variables for each possible value and $(1-1/q)^n$ the probability of that value not appearing.

Observe that this technique will produce higher factorial moments and hence the variance, e.g. we get

$$\mathrm{E}[X(X-1)] = n! [z^n] (\exp(z)-1)^q \frac{q(q-1)}{q^n} \sum_{p=2}^q {q-2\choose p-2} (\exp(z)-1)^{-p} \\ = n! [z^n] (\exp(z)-1)^{q-2} \frac{q(q-1)}{q^n} \left(1+\frac{1}{\exp(z)-1}\right)^{q-2} \\ = n! [z^n] \frac{q(q-1)}{q^n} \exp((q-2)z) = \frac{q(q-1)}{q^{n}} (q-2)^n = q(q-1)\left(1-\frac{2}{q}\right)^n.$$

Recall that

$$\mathrm{Var}[X] = \mathrm{E}[X(X-1)] + \mathrm{E}[X] - \mathrm{E}[X]^2$$

so that we obtain

$$\mathrm{Var}[X] = q(q-1)\left(1-\frac{2}{q}\right)^n + q\left(1-\frac{1}{q}\right)^n - q^2\left(1-\frac{1}{q}\right)^{2n}.$$

Answer 2

The trick is to use the following property of the Indicator Function (Called Indicator Function - Mean, Variance and Covariance):

$$ \boldsymbol{1}_{A} \left( x \right) = \begin{cases} 1 & \text{ if } x \in A \\ 0 & \text{ if } x \notin A \end{cases} \Rightarrow \mathbb{E} \left[ \boldsymbol{1}_{A} \left( x \right) \right] = P \left( A \right) $$

Defining $ {A}_{i} $ as the event of the $ i $ -th face doesn't appear.

The, as indicator function of the set:

$$ {\boldsymbol{1}}_{{A}_{i}} = \begin{cases} 1 & \text{ if } \text{The $ i $ -th face doesn't appear} \\ 0 & \text{ if } \text{The $ i $ -th face does appear} \end{cases} $$

This gives the following equality:

$$ X = \sum_{i = 1}^{6} {\boldsymbol{1}}_{{A}_{i}} $$

By the Linearity Property of the Expectation Operator:

$$ X = \sum_{i = 1}^{6} {\boldsymbol{1}}_{{A}_{i}} \Rightarrow \mathbb{E} \left[ X \right] = \mathbb{E} \left[ \sum_{i = 1}^{6} {\boldsymbol{1}}_{{A}_{i}} \right] = \sum_{i = 1}^{6} \mathbb{E} \left[ {\boldsymbol{1}}_{{A}_{i}} \right] $$

Utilizing the property of the Indicator Function:

$$ \mathbb{E} \left[ X \right] = \sum_{i = 1}^{6} \mathbb{E} \left[ {\boldsymbol{1}}_{{A}_{i}} \right] = \sum_{i = 1}^{6} P \left( {A}_{i} \right) = \sum_{i = 1}^{6} {\left( \frac{5}{6} \right)}^{n} = 6 {\left( \frac{5}{6} \right)}^{n} $$

To Do

• Add simulation.