Question

Question: Suppose that a pair of fair dodecahedral dice is rolled.

a) What is the expected value of the sum of the numbers that come up?

b) What is the variance of the sum of the numbers that come up?

Short answer

Answer

a) The Expected value of the sum of the numbers that comes up is \(13\).

b) The variance of the sum of the numbers that comes up is \(\frac{{143}}{6}\).

Step-by-step solution

Given data

A pair of fair dodecahedral dice is rolled.

Concept of Probability

Probability is simply how likely something is to happen. Whenever we’re unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

Calculation for the expected value of the sum of the numbers

a)

Let us assume \({X_1}\;,\;{X_2}\) random variables \({X_1}(i,j) = i\;,\;{X_2}(i,j) = j\).

Here \({X_1}\) is the number appearing on the first die and \({X_2}\) is the number appearing on the second die.

Expected value is given as:

\(\begin{aligned}{}E({X_1}) &= \frac{1}{{12}}(1 + 2 + \ldots ..12)\\E({X_1}) &= \frac{{13}}{2}\end{aligned}\)

From the given data it is clear that:

\(E({X_1}) = E({X_2}) = \frac{{13}}{2}\)

Then, it gives

\(\begin{aligned}{}E({X_1} + {X_2}) &= E({X_1}) + E({X_2})\\E({X_1} + {X_2}) &= \frac{{13}}{2} + \frac{{13}}{2}\\E({X_1} + {X_2}) &= 13\end{aligned}\)

Therefore, the Expected value of the sum of the numbers that comes up is \(13\).

Calculation for the variance of the sum of the numbers

b)

Let us assume \({X_1}\;,\;{X_2}\) random variables \({X_1}(i,j) = i\;,\;{X_2}(i,j) = j\).

Here \({X_1}\) is the number appearing on the first die and \({X_2}\) is the number appearing on the second die.

\(\begin{aligned}{}E\left( {{X_1}^2} \right) &= \frac{1}{{12}}\left( {{1^2} + {2^2} + {3^2} + {4^2} + {5^2} + \ldots \ldots . + {{12}^2}} \right)\\E\left( {{X_1}^2} \right) &= \frac{{325}}{6}\end{aligned}\)

Variance can be determined as:

\(\begin{aligned}{}V({X_1}) &= E\left( {{X_1}^2} \right) - E\left( {\begin{aligned}{*{20}{c}}X\\1\end{aligned}} \right.\\V({X_1}) &= \frac{{325}}{6} - {()^{\frac{{13}}{2}}}\\V({X_1}) &= \frac{{325}}{6} - \frac{{13}}{2}\\V({X_1}) &= \frac{{143}}{{12}}\end{aligned}\)

Therefore, \(V(X1) = V(X2) = \frac{{143}}{{12}}\).

Sum of variances is given as:

\(\begin{aligned}{}V(X) &= V({X_1}) + V({X_2})\\V(X) &= \frac{{143}}{{12}} + \frac{{143}}{{12}}\\V(X) &= \frac{{143}}{6}\end{aligned}\)

Therefore, the variance of the sum of the numbers that comes up is \(\frac{{143}}{6}\).