Expected value of the sum of two dice. Came across this question:
We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.

*

*Find the expected value of $X$.


*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.

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:

*

*$\mathbb{E}(X_1+X_2)=\mathbb{E}(X_1)+\mathbb{E}(X_2)$ for any two variables $X_1,X_2$;

*$\mathrm{Var}(X_1+X_2)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)$ provided $X_1,X_2$ are independent.

