You already know the concept of expected values. It is an outcome multiplied by its probability, summed over all those outcomes. What is the expected value of an n sided die? Let us mathematically estimate the answer to this question.
![\\ \text{Let X be 1, 2, ...n sides of the die} \\ \\ p(X) = \frac{1}{n} \text{, assuming it is a fair die and each side is equally likely} \\ \\ E(X) = \sum\limits_{X=1}^n [p(X)*X] \\\\ E(X) = 1*\frac{1}{n} + 2*\frac{1}{n} = 3*\frac{1}{n} + ... + n*\frac{1}{n} \\ \\ = \frac{1}{n} [1 + 2 + 3 + ... n] = \frac{1}{n} \frac{n(n+1)}{2} = \frac{(n+1)}{2}](https://thoughtfulexaminations.com/wp-content/ql-cache/quicklatex.com-bfddc31a8f6a45367d579d105804f106_l3.png "Rendered by QuickLaTeX.com")
If you want to test: for a normal six-sided die, n = 6, E(X) = (6+1)/2 = 3.5. Similarly, the expected value of the sum of 2 dice. E(X + X) = E(X)+E(X) = 3.5 + 3.5 = 7.
It may be a bit worrying to some of you. How can the expected value of a fair die is one number, and that too, a fraction? You know, by now, that the expected value is more of a concept and is equal to the long term average, in this case, of several rolls of a die.
