Expectation ${S\over{n(2n-1)}}$ with cancelling odd and even pairs Here's a question from my probability textbook:

A bag contains $2n$ counters, of which half are marked with odd numbers and half with even numbers, the sum of all the numbers being $S$. A man is to draw two counters. If the sum of the numbers drawn be an odd number, he is to receive that number of shillings; if an even number, he is to pay that number of shillings. Show that his expectation is worth (in shillings)$${S\over{n(2n-1)}}.$$

Alright, here's what I did. We are calculating$${{\text{sum of all the odd pairs}-\text{sum of all the even pairs}}\over{\binom{2n}{2}}} = {{\left(\sum_{k = 1}^n e_k\right)\left(\sum_{k = 1}^n o_k\right) - \left(\sum_{k = 1}^n o_k\right)(n - 1) - \left(\sum_{k = 1}^n e_k\right)(n - 1)}\over{\binom{2n}{2}}} = {{\left(\sum_{k = 1}^n e_k\right)\left(\sum_{k = 1}^n o_k\right) - S(n - 1)}\over{n(2n-1)}}.$$ So to conclude, we want to show that$$\left(\sum_{k = 1}^n e_k\right)\left(\sum_{k = 1}^n o_k\right) = Sn.$$ This is where I'm stuck, I tried messing around for a while but didn't get anywhere. Can anyone help me out?
EDIT: Peanut helpfully writes as a comment the following:

You can't it show because it's not true. Take $n=2$, $e_1=2$, $e_2=4$, $o_1=1$, $o_2=3$. Then $(2+4)(1+3)\neq(2+4+1+3)\cdot2$. There's a mistake in your reasoning.

Then where did I go wrong? How can I salvage my approach? In my approach I would prefer to not count any intermediate averages but rather just sum up everything explicitly and divide everything by $n(2n-1)$.
 A: Points to note:

*

*Suppose the average odd coin is $\bar o$ and the average even coin is $\bar e$


*so $S=n\bar o + n \bar e$ and the average coin has value $\dfrac{\bar o + \bar e}{2}$


*so the average pair has value $\bar o + \bar e$,


*as does the average pair which has an odd sum,


*and therefore also the average pair which has an even sum.


*There are $n^2$ equally likely odd pairs and $n(n-1)$ even pairs, and $n(2n-1)$ in total


*so the expected gain is $\dfrac{n^2 (\bar o + \bar e) - n(n-1)(\bar o + \bar e)}{n(2n-1)} = \dfrac{n (\bar o + \bar e) }{n(2n-1)}=\dfrac{S}{n(2n-1)}$
A: I find this reasoning more intuitive:
there are $\binom{2n}{2} = n(2n-1)$ possible pairs (denote such a pair by $(x, y))$. Each has a probability $1/[n(2n-1)]$ of being chosen. So the expected gain is$$=\frac{1}{n(2n-1)}[\sum_{(x,y) \text{ s.t.}\\ x+y \text{ odd}}(x+y)-\sum_{(x,y) \text{ s.t.}\\ x+y \text{ even}}(x+y)]$$
Now consider any pair $(x, y)$ s.t. $x+y$ is odd. Fix $x$. How many $y$ are such that $x+y$ is odd? If $x$ is even there are $n$ such values. If $x$ is odd there are again $n$ such values. So in the end, each number will figure $n$ times in the first sum. So its value is $nS$. Similarly for the second sum you can find that its value is $(n-1)S$. Thus the difference between the squared parentheses is $S$.
A: The denominator $\text{(sum of odd pairs)} - \text{(sum of even pairs)}$ simply doesn't simplify as you do it on the next line.
It is correct to take the sum of the even pairs and write it as $(n-1)\sum_{k=1}^n e_k  + (n-1) \sum_{k=1}^n o_k$, the logic being that each number (even or odd) contributes to $n-1$ different even pairs. You are also right that this simplifies to $(n-1)S$.
However, it is not correct to take the sum of the odd pairs and write it as $\left(\sum_{k=1}^n e_k\left)\right( \sum_{k=1}^n o_k\right)$. This expands out as a sum of terms $e_i o_j$ for $1 \le i,j \le n$. Each pair $\{e_i, o_j\}$ is a pair with odd sum, but instead of using the sum $e_i + o_j$ in the expectation, you are using the product $e_i o_j$.
Instead, argue as for the even pairs: each number (even or odd) contributes to $n$ different pairs with even sum. So the sum of the odd pairs should be $n \sum_{k=1}^n e_k + n \sum_{k=1}^n o_k$, which simplifies to $nS$. The final answer is
$$
   \frac{nS - (n-1)S}{\binom{2n}{2}} = \frac{S}{\binom{2n}{2}} = \frac{S}{n(2n-1)}.
$$
