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.