Let's say that you are flipping a coin, and with heads you win a dollar, tails you lose a dollar. Then you'd expect that the average amount of money you'd win is zero.
Now, let's say it turns out the coin was weighted, so that heads came up half as often as tails. Then the amount you'd expect to win is:
$$\frac{2\cdot(-1)+1\cdot 1}{3} = -\frac{1}{3}$$
This is a weighted average of $-1$ and $1$, based on their frequencies.
The concept of expectation value or expected value may be understood from the following example. Let $X$ represent the outcome of a roll of an unbiased six-sided die. The possible values for $X$ are 1, 2, 3, 4, 5, and 6, each having the probability of occurrence of 1/6. The expectation value (or expected value) of $X$ is then given by
$(X)\text{expected} = 1(1/6)+2\cdot(1/6)+3\cdot(1/6)+4\cdot(1/6)+5\cdot(1/6)+6\cdot(1/6) = 21/6 = 3.5$
Suppose that in a sequence of ten rolls of the die, if the outcomes are 5, 2, 6, 2, 2, 1, 2, 3, 6, 1, then the average (arithmetic mean) of the results is given by
$(X)\text{average} = (5+2+6+2+2+1+2+3+6+1)/10 = 3.0$
We say that the average value is 3.0, with the distance of 0.5 from the expectation value of 3.5. If we roll the die $N$ times, where $N$ is very large, then the average will converge to the expected value, i.e.,$(X)\text{average}=(X)\text{expected}$. This is evidently because, when $N$ is very large each possible value of $X$ (i.e. 1 to 6) will occur with equal probability of 1/6, turning the average to the expectation value.
Best Answer
Maybe I didn't really understand your question, but average of the averages of these chunks is the same as the average of the whole list.
For example, suppose that your list is $[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]$
And you divide it into 5 chunks $[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15]$.
Average of each chunk : $[2],[5],[8],[11],[14]$
Average of average of chunks : $\frac{2+5+8+11+14}{5} = 8$
Average of the whole list : $\frac{1+2+...+15}{15} = 8$