Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to show it.
 A: No tedious algebraic manipulation required!
Consider the following experiment: say that we choose a sequence $x_1,x_2,x_3$ from $\{1,\ldots,n\}$, without replacement.  What is the probability that $x_1$ is the smallest of the three numbers?
On the one hand, this probability is $\frac{1}{3}$ by symmetry. Why?  Given any three distinct numbers $a,b,c\in\{1,\ldots,n\}$, the sequences $abc$, $acb$, $bac$, $bca$, $cab$, and $cba$ are all equally likely to occur.
On the other hand, we can split into cases based on what that smallest number is. We first choose which number $r$ we assign to $x_1$.  This assignment occurs with probability $\frac{1}{n}$.  Then, from the remaining $n-1$ elements, we choose any of the $n-r$ numbers bigger than $r$ for $x_2$ (probability $\frac{n-r}{n-1}$); finally, from the remaining $n-2$ elements, we choose any of the $n-r-1$ remaining elements that exceed $r$ (probability $\frac{n-r-1}{n-2}$).
So, all told, we find that the probability should be
$$
\sum_{r=1}^{n-2}\frac{1}{n}\cdot\frac{n-r}{n-1}\cdot\frac{n-r-1}{n-2},
$$
which differs from your sum only in limits of summation; and, you'll notice that the terms $r=n-1$ and $r=n$ that are missing here are both $0$ in your sum.
So, because these two different expressions (the sum and $\frac{1}{3}$) are both the probability of the same event, they must be equal.
Now, if you PREFER tedious algebraic manipulation, it actually isn't too bad here.  Note that changing your index of summation from $r$ to $k:=n-r$ yields
$$
\sum_{r=1}^{n}\frac{n-r-1}{n-2}\cdot\frac{n-r}{n-1}\cdot\frac{1}{n}=\frac{1}{n(n-1)(n-2)}\sum_{k=1}^{n-1}k(k-1).
$$ 
(Again, I've dropped indices of summation that contribute zero.) But, we can write
$$
\sum_{k=1}^{n-1}k(k-1)=\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k.
$$
Recall these identities:
$$
\sum_{i=1}^{m}i=\frac{m(m+1)}{2}\qquad\sum_{i=1}^{m}i^2=\frac{m(m+1)(2m+1)}{6}
$$
With these,
$$
\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k=\frac{(n-1)n(2n-1)}{6}-\frac{n(n-1)}{2},
$$
so that the original sum is
$$
\begin{align*}
\frac{(n-1)n(2n-1)}{6n(n-1)(n-2)}-\frac{n(n-1)}{2n(n-1)(n-2)}&=\frac{2n-1}{6(n-2)}-\frac{1}{2(n-2)}\\
&=\frac{2n-4}{6(n-2)}\\
&=\frac{1}{3}.
\end{align*}
$$
A: The sum can rewrite in this way:
$$\frac{1}{(n-2)(n-1)n}\sum_{r=1}^n(n^2-nr-n-nr+r^2+r)=$$
$$=\frac{1}{(n-2)(n-1)n}\left(n^3-n(\frac{n(n+1)}{2})-n^2-n(\frac{n(n+1)}{2})+(\frac{n(n+1)(2n+1)}{6})+(\frac{n(n+1)}{2})\right)=$$
$$=\frac{1}{6(n-2)(n-1)n}(6n^3-3n^3-3n^2-6n^2-3n^3-3n^2+2n^3+3n^2+n+3n^2+3n)=$$
$$=\frac{1}{6(n-2)(n-1)n}(2n^3-6n^2+4n)=\frac{1}{6(n-2)(n-1)}(2n^2-6n+4)=$$
$$=\frac{1}{3(n-2)(n-1)}(n^2-3n+2)=\frac{(n-2)(n-1)}{3(n-2)(n-1)}=\frac{1}{3}$$
