Naive model. Let there be v vertices, and let p[n] be the probability that a
vertex has size n. Given a family of graphs satisfying these probabilities, do a flip on each. Then the probability that a vertex has size n goes from p[n] to the sum over k of ((p[n] + k/v) times the probability that the flip changes the number of n-faces by k). This net change should be zero. So we get: the sum over k of (k times the probability that the flip changes things
by k) should be zero.
vertex has size n. Given a family of graphs satisfying these probabilities, do a flip on each. Then the probability that a vertex has size n goes from p[n] to the sum over k of ((p[n] + k/v) times the probability that the flip changes the number of n-faces by k). This net change should be zero. So we get: the sum over k of (k times the probability that the flip changes things
by k) should be zero.
Solve[Expand@Simplify[Plus@@(((Plus@@(If[#===n,-1,1]&/@#))(Times@@(p/@#)))&/@Flatten[Outer[List,{n-1,n},{n,n+1},{n-1,n},{n,n+1}],3])]==0,p[n+1]]
{p[1+n]->-p[n]},p[1+n]->
2
p[n]
p[-1+n]
Since p[n] is not negative, the first solution is out. The second solution just says that p is exponential.
So p[n] = k r^n for some k and r. The Euler constraint for a torus says that
the sum over n of (6-n)p[n] must be zero. k is nonzero, so:
the sum over n of (6-n)p[n] must be zero. k is nonzero, so:
Solve[Sum[(6-n)r^n,{n,Infinity}]==0,r]
{r->0},r->
5
6
It must be the second one. Finally, the probabilities must sum to 1. So:
Sum[k(5/6)^n,{n,Infinity}]
5k
so k is 1/5. So the answer is:
p[n_]:=(5/6)^n/5
The answer could vary depending on boundary conditions, like: when is the flip moved allowed and when not? Should we allow faces with one edge? zero edges? etcetera.