How to argue why one dice is more rigged than the other?

Question

Let $\omega$ be a finite set and $P : \Omega \rightarrow \mathbb{R}$ be a probability measure.

You are given a set of three dices $\{A, B, C\}$. The following table describes the outcome of six rollouts for these dices, where each column shows the outcome of the respective dice. (Note: assume the dices are standard six-sided dices with values between 1-6) $$ \begin{array}{|c|cccccc}{A} & {4} & {4} & {2} & {4} & {1} & {1} \\ \hline B & {3} & {6} & {3} & {3} & {4} & {3} \\ \hline C & {5} & {5} & {2} & {1} & {1} & {1}\end{array} $$

(1) Estimate the expectation and the variance for each dice using unbiased estimators. (Show your computations).

(2) According to the data, which of them is the “most rigged”? Why?

So for (1) I calculated: \begin{equation*} \begin{aligned} \overline{x}_A &= \frac{8}{3} &, \qquad s_A^2 &= \frac{34}{15} \\ \overline{x}_B &= \frac{11}{3} &, \qquad s_B^2 &= \frac{22}{15} \\ \overline{x}_A &= \frac{5}{2} &, \qquad s_C^2 &= \frac{51}{10} \\ \end{aligned} \end{equation*}

In order to figure out which one is the most "rigged" one I calculated the expectation and variance of a "perfect" dice:

\begin{equation*} \begin{aligned} \mathbb{E}[X] = \frac{7}{2} \qquad \mathbb{V}[X] = \frac{35}{12} \end{aligned} \end{equation*}

As the dices differ in expectation and variance from a "perfect" dice I am trying to figure out how I could argue which one is the most "rigged" one?

The professor mentioned the Kullback–Leibler divergence in another setting very briefly and said that the KL divergence is a measure for the "difference" of distributions. Is this the tool he wants us to use here?

Answer 1

Your professor might want you to calculate the Kullback-Leibler-divergence for each dice with respect to a perfect dice. Whichever dice has the highest divergence is the one that is most rigged.

Since we are dealing with discrete random variables, we can use eq. (1) from Wikipedia (here, $Q$ and $P$ denote the probability distributions of discrete random variables $X,Y:\Omega\to\mathcal{X}$ such that $Q\{x\}=0\implies P\{x\}=0$ for all $x\in\mathcal X$. Whenever $P\{x\}=0$, we will consider the sum term below to equal $0$ aswell.)

\begin{equation}D(P \parallel Q) = \sum_{x\in\mathcal{X}} P\{x\} \log\left(\frac{P\{x\}}{Q\{x\}}\right).\end{equation}

For instance, based on the observations (which are indeed fairly scarce), we approximate the dice A by ($\Bbb P_A$ denotes the probability measure of the random variable $A$):

• $\Bbb P_A\{3,5,6\}=0$,
• $\Bbb P_A\{1\}=\frac13$,
• $\Bbb P_A\{4\}=\frac12$,
• $\Bbb P_A\{2\}=\frac16$.

Then the Kullback-Leibler-divergence of $A$ with respect to $X$ is

\begin{equation} D(\mathbb P_A\parallel \mathbb P_X)=\frac13\log\left(\frac63\right) + \frac12\log\left(\frac62\right)+\frac16\log\left(\frac66\right) =\frac{\log2}3+\frac{\log3}2\approx0.780. \end{equation}

I will leave the calculations for $B$ and $C$ to you.