# Variance of the sum of the even numbers rolled

I tried to solve the following problem for my probability theory class :

A dice is rolled $$100$$ times. If $$X$$ is the sum of the even numbers rolled, find the expected value and the variance of $$X$$.

My attempt:

Let $$X_i$$ be the number of times we rolled $$i$$ $$\left(\text{for } i = \overline{1,6}\right)$$. For example, if after 100 tries, we rolled $$5$$ seven times, then $$X_5 = 7$$.

Then $$X=2X_2 + 4X_4 + 6X_6$$, so by the linearity of expectation, we have:

$$\mathbb{E}(X) = \mathbb{E}(2X_2 + 4X_4 + 6X_6) = 2\mathbb{E}(X_2)+4\mathbb{E}(X_4) + 6\mathbb{E}(X_6)$$

Also, $$\mathbb{P}(X_i = k) = {100 \choose k} \left( \dfrac{1}{6} \right)^k \left( \dfrac{5}{6} \right)^{100-k}$$ Therefore, $$X_i \sim \text{Bin}\left(100, \dfrac{1}{6} \right)$$, so we have $$\mathbb{E}(X_i) = \dfrac{100}{6}$$ $$\Rightarrow \mathbb{E}(X) = \dfrac{100}{6}(2+4+6) = 200.$$

To find the variance, I know that for two independent random variables $$A$$ and $$B$$, we have $$\text{Var}(A+B) = \text{Var}(A)+\text{Var}(B),$$ which would eventually solve my problem, but somehow, I can't figure out whether $$X_2, X_4, X_6$$ are independent or not... Or is this a wrong approach?

Thanks in advance for any help!

$$X_i$$ are NOT independent, because $$\sum\limits_{i=1}^6 X_i = 100$$ for example.

Consider $$Y_i\sim U\{1,2,3,4,5,6\}$$, represent each dice, and $$Z_i=1\{Y_i \text{ even}\}$$.

Now, you want to calculate expectation and variance of $$\sum\limits_{i=1}^{100}Y_iZ_i$$

Try to prove:

1. $$Y_iZ_i$$ and $$Y_jZ_j$$ are independent if $$i\neq j$$
2. $$E(Y_iZ_i)=2P(Y_i=2)+4P(Y_i=4)+6P(Y_i=6)$$

For variance, use that $$Var(X)=E(X^2)-E(X)^2$$ and try to analyze the variable $$(Y_iZ_i)^2$$. What are its possible values?

• Thank you. I see now that my random variables are not independent. I'll try to find the variance with the new approach. But I think my calculation of the expected value is correct, isn't it? Feb 24 at 13:18
• Yes! The expected value is correct Feb 25 at 14:35