I've been studying power sets and thought this was interesting. $\frac{x+y+\frac{x+y}{2}}{3} = \frac{x+y}{2}$
This is if we have a set of 2 elements, x and y, and we get it's powerset, find the averages of each subset then the average of those averages. It's the same as the average of the 2 elements. Does anyone know why this is? Or is there a proof?
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Think about it intuitively. If we have three numbers $x,y$ and $z$ with $z$ exactly halfway between $x$ and $y$, then the average of $x,y$ and $z$ should just be $z$. But what is the number halfway between $x$ and $y$? It is just the average.
Of course, you can give a proper proof rather easily: $$\frac{x+y+\frac{x+y}2}{3}=\frac{2(x+y)+(x+y)}6=\frac{3(x+y)}6=\frac{x+y}2$$ but you were probably able to do this already - the intuition is perhaps just as important here.
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This is indeed the case in general: if we take the average of each subset of $\{1,2,\ldots,n\}$ and then average those averages, we find the average $\frac{n+1}{2}$ of the numbers $1$, $2$, ..., $n$.
In fact, something stronger is true: the average already equals $\frac{n+1}{2}$ if we only consider the $k$-element subsets of $\{1,2,\ldots,n\}$ for some $1 \leq k \leq n$. There are $\binom{n}{k}$ such sets and each $1 \leq i \leq n$ occurs in exactly $\binom{n-1}{k-1}$ of them (there are that many ways to choose the remaining elements). This means that the sum of the elements of each $k$-element set, summed over all sets, equals $\binom{n-1}{k-1}(1+2+\cdots+n)$. The sum of all averages equals $\frac{1}{k} \binom{n-1}{k-1}(1+2+\cdots+n)$ and this is equal to $\binom{n}{k} \frac{n+1}{2}$ by the property $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$. Therefore, the sum of all averages equals $\frac{n+1}{2}$ times the total number of sets under consideration, meaning that the average of the averages equals $\frac{n+1}{2}$
This is a consequence of the following fact: if $X$ is any nonempty set of numbers$^*$, and $n\in\mathbb{N}$, the average of the averages of the $n$-element subsets of $X$ is equal to the average of $X$. For instance, if $n=2$ and $X=\{x_1, x_2, x_3\}$, then the average of the averages of the $n$-element subsets of $X$ is $${{x_1+x_2\over 2}+{x_1+x_3\over 2}+{x_2+x_3\over 2}\over 3}={2(x_1+x_2+x_3)\over 2\times 3}={x_1+x_2+x_3\over 3};$$ generalizing this to arbitrary $X$ and $n$ is a good exercise.
$^*$ What kind of number? Well, really, any kind of number works - integers, real numbers, even complex numbers - in general, $X$ can be any subset of a divisible abelian group.