Expected value of the sum of two dice.

Question

Came across this question:

We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.

1. Find the expected value of $X$.

2. Find the variance of $X$.

I'm not sure how to do either, but this was my thinking for part 1:

$$E(X) = 2((1/6)^2) + \\ 3(2(1/6)^2) + \\ 4(2(1/6)^2 + (1/6)^2) + \\ 5(2(1/6)^2 + 2(1/6)^2) + \\ 6(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\ 7(2(1/6)^2 + 2(1/6)^2 + 2(1/6)^2) + \\ 8(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\ 9(2(1/6)^2 + 2(1/6)^2) + \\ 10(2(1/6)^2 + (1/6)^2) + \\ 11(2(1/6)^2) + \\ 12((1/6)^2)$$

The reason I multiplied some by 2 is because it could possibly switch up or permute. So, for example, for 4, the two sums that could give us 4 are (3,1) and (2,2), so I multiplied one of the probabilities by 2 because (3,1) could come as either (3,1) or (1,3) whereas (2,2) can only come in one form.

Answer 1

$$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)=3.5+3.5=7$$

can you calculate $V(X)$?

$$V(X_1)=E(X_1^2)-E^2(X_1)\approx 2.917$$

thus

$$V(X)=2\times 2.917=5.8\overline{3}$$

Answer 2

What you have is correct so far.

However, a simpler approach is to write $X=X_1+X_2$, where $X_1,X_2$ are the values on the individual dice, and use the facts that:

1. $\mathbb{E}(X_1+X_2)=\mathbb{E}(X_1)+\mathbb{E}(X_2)$ for any two variables $X_1,X_2$;
2. $\mathrm{Var}(X_1+X_2)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)$ provided $X_1,X_2$ are independent.