Coloring $K_n$ via edge-weight sums This is a question inspired by 
and tangential to "A Question on 1, 2 ,3 Conjecture"—and certainly
much easier!
Suppose one assigns a random edge weight among $\{1,2,3,\ldots,k\}$ to each
edge of $K_n$, and then sums the incident edge weights to each vertex
and assigns those as vertex weights.
For example, here is $K_5$ with edge weights from $\{1,2,3,4,5\}$:
      
(The top vertex has weight $13=3+1+4+5$.)
In this case, the vertex weights do not form a proper coloring, because there
are two vertices assigned weight $14$.

What is the probability that $K_n$ will be properly colored when the
  edge weights are chosen uniformly from among $\{1,2,3,\ldots,k\}$?

Empirically the function is well-behaved. Here is an accounting over $1000$ trials
on $K_5$ for each $k$ from $3$ to $20$:
      

As requested, a plot of Lucia's $e^{-\frac{\sqrt{\frac{3}{\pi }} n^{3/2}}{2 k}}$ for $n=5$ (with $k$ on the horizontal axis):
   
 A: Here's a heuristic which suggests that the right scale for $k$ is on the order of $n^{3/2}$, and predicts the actual probability for large $n$.   More precisely, we conjecture the following:  Let $n$ be large, and suppose $k$ is close to $t n^{3/2}$ and we have in mind that $t$ is of roughly constant size.  Then the probability that one gets a proper coloring is (as $n$ goes to infinity) asymptotically $\exp(-\sqrt{3}/(2\sqrt{\pi}t))$.  As $t$ goes to zero this is exponentially small, and as $t$ goes to infinity it tends to $1$ at a linear rate.  
Here's the reasoning:  If $X_1$, $\ldots$, $X_n$ are chosen uniformly (with replacement) from $1$ to $k$, then the sum $X_1+\ldots+X_n$ has mean $n(k+1)/2$ and variance $n(k^2-1)/12$, and for large $n$ it is approximately normal with this mean and variance.  Therefore the sums for each of the $n$ vertices has mean $n(k+1)/2$ and wiggles around that on the scale of about $\sqrt{n}k$.  Let us imagine that different vertices have independent sums -- this is not quite true because they share an edge, but I would expect that effect to be small (this is a heuristic, not a proof after all!).  Then we have a situation like the birthday problem:  Pick $n$ numbers from an interval of length about $\sqrt{n}k$, what is the chance of a coincidence.  From the birthday problem we know that if $n^2$ is of size $\sqrt{n}k$ then there is a good chance of a coincidence.  (Note that for the proper coloring we want there not to be a coincidence of sums.)  This suggests that $k$ of size $n^{3/2}$ is the correct scaling. The actual formula was obtained by computing the analog of the birthday problem in this Gaussian setting 
(which is a nice exercise). 
The only thing preventing this from being a proof is the lack of independence for the sums at two vertices -- I think this can be fixed, but will need a better probabilist than me!   
