The definition of expected value and weighted average

Question

In the definition of expected value it is noted, that because the sum of all the probability coefficients of possible outcomes equal one, expected value can be viewed as weighted average. Why is it so, since the expected value is a sum and there is no division by the number of possible outcomes. I can't grasp the logic behind it. Can someone help?

Answer 1

When you take a weighted average, you don’t divide by the number of items. You divide by the sum of the weights. Here, the weights are the probabilities and their sum is $1$, so that division by the sum of the weights is implicit.

Answer 2

A weighted average is an average where each value included is multiplied by a "weight", summed and then divided by the sum of the weights. Note a "standard" average is a special case where each weight is $1$.

Consider that each "weight" is the probability of certain outcome with a numeric value. The expected value is the sum of the products of the "weights", i.e., probabilities, times the numeric values. Since the sum of all of the "weights", i.e., probabilities, is $1$, the expected value would be the same result if it was divided by the sum of the "weights", which would make it equivalent to the weighted average.