# Find the variance of the following dice

Let be a dice with the sides $$\{0,0,6,0,3,3\}$$ which will be rolled $$n$$ times since $$n\in\mathbb{N}$$. Now I'm modelling the following experiment: Let $$X_i$$ be the random variable of the $$i$$-th throw and $$S:=\sum_{i=1}^n X_i$$ the sum of the diced values. The exercise is to determine the variance of $$S$$.

I've calculated the mean of $$S$$: Since the random variables are independent, then

$$E(X_1) = 0\cdot P(X_1=0)+3\cdot P(X_1=3)+6\cdot P(X_1=6) = 0 + \frac{3}{3}+\frac{6}{6} = 1+1=2,$$ so

$$E(S) = E(X_1+\ldots+X_n) = n\cdot E(X_1) = 2n.$$

Then, the second moment of $$X_1$$ is

$$E(X_1^2) = 0 + \frac{1}{3}\cdot 9+\frac{1}{6}\cdot 6^2 = 3+6=9,$$

and the variance of $$X_1$$ is $$Var(X_1) = E(X_1^2)-E(X_1)^2 = 9 - 4 = 5.$$

For the sum $$S$$, I determine by the aid of the independece of $$X_i$$:

\begin{align*} Var(S) &= Var(X_1+\ldots + X_n) = n\cdot Var(X_1) = 5n.\\ \end{align*}

Are my calculations and reasons correctly?

• Isn't $P(X_1=3)=1/2$? Sep 3 '19 at 10:43
• I've corrected the numbers of the dice. It has to be $\{0,0,6,0,3,3\}$. Sep 3 '19 at 10:46

All is fine but $$\operatorname{Var}(X_1+\ldots + X_n) = \operatorname{Var}(n\cdot X_1).$$

You have to take the variance summandwise in case of independence, i.e. $$\operatorname{Var}(S)= \operatorname{Var}(X_1+\ldots + X_n) =n\cdot \operatorname{Var}(X_1).$$

Illustration: Let $$X=\text{weight of a pack of sugar in kg}$$. In that case $$X_1+\dots +X_n$$ is the sum of the weights of onethousand packages in kg whereas $$1000 X$$ is the weight of one package, measured in grams.

• Could you please elaborate Why first one is incorrect ?? Sep 3 '19 at 10:58
• @tourism For showing that you only need a counterexample. Sep 3 '19 at 11:37
• Because you deal with a sum and not with changing units. Sep 3 '19 at 12:43
• I understood... Got confused with $Var(X_1+X_2+...X_n) = Var(n \times X_1)$ which is not true Sep 3 '19 at 20:32

You took a nice route when you went for analyzing $$S$$ as the sum $$X_1+\cdots+X_n$$.

However an essential mistake is made by calculation of variance.

We do not have: $$\mathsf{Var}(X_1+\cdots+X_n)=\mathsf{Var}(nX_1)$$.

What we do have is:$$\mathsf{Var}(X_1+\cdots+X_n)=\mathsf{Var}(X_1)+\cdots+\mathsf{Var}(X_n)$$which can be exploited nicely.

This on base of independence of the $$X_i$$.

• Thanks for the answer! I've corrected the formula of the variance and I hope that the result should be now correctly. Sep 3 '19 at 11:01
• You are welcome. Everything seems fine to me now. Sep 3 '19 at 11:05