A plane Poisson process with uniform intensity scatters "sites" about the plane. If I'm not mistaken, in a sense the "average" Voronoi diagram of that set of sites is a honeycomb. I know it's been proved that the average number of edges of the cells is $6$, and I've read (but not in anything published very recently) that the probabilities that the number of edges is equal to $n$ for $n\in\{3,4,5,\ldots\}$ is known only numerically.


My question is whether there is any sense in which the average shape could be said to be a honeycomb or the number of edges in an average Voronoi diagram can be shown to be $6$ without any sort of probability distribution on the set of sites?


This is a vague hunch which I suspect was inspired in part by the time I read in Seymour Geisser's book on predictive inference a way of deriving Student's small-sample confidence intervals without using probability. Although probability was not mentioned, the squaring function as an objective function was relied on.



"Average" Voronoi diagrams without probability?
Here is an argument that Béla Bollobás showed me once. (this was motivated by a physics paper where a simulation was done showing that the average number of edges per face
was 5.997$\pm$ 0.005). 


Take a large number of seeds (i.e. points generating the Voronoi diagram) and make the assumption that there are no *multiple points*: points of that are simultaneously closest to four or more seeds. (This is certainly the case with probability 1 if the seeds come from a Poisson Processes, but of course it's much more general than that). 


Then use Euler's formula. Each vertex of the Voronoi diagram is nearest to exactly three seeds. Each of the $\binom 32$ pairs of seeds gives an edge in the Voronoi diagram, so that each vertex has three edges emanating from it. Let $v$ be the number of vertices, $e$ be the number of edges and $f$ the number of faces. Then $f-e+v=2$ by Euler's formula. 
Also $2e=3v$, so that $f=2+\frac e3$. Let $\rho$ be the average number of edges per face. Then $e=\rho f/2$, so that $f=2+\rho f/6$. Hence when the number of faces becomes large, the number of edges per face approaches 6. 


