Combinatorics: Stochastics on a circular arrangement

1. Jul 15, 2011

I am dealing with the following interesting combinatorics problem, several reformulations of which haven't help me solve the problem:

Suppose we have a circular arrangement with M spots and we want to distribute N tokens over these spots such that there are token numbers $n_m$ that respect $\sum_{m}n_m = N$. Now we want to impose a discrete time stochastic dynamics such that in each time step every single token stays with probability $p$ and a "goes" with probability $q$ for jumps to the left and to the right (only to the neighbors). Therefore we have $p+2q=1$.
Now given the sets of token numbers $\{n_m\}$ and $\{n_{m}'\}$ of two consecutive time steps, what is the transition probability between these two sets of numbers?

The challenge in this question arises because the "macro" combinatorics has to be worked out starting from the "micro" combinatorics for each token, such that one has to consider several micro-configurations, each having a different probability, that give rise to the same macro-configuration.
With simple counting techniques, one gets pretty quickly stuck with this problem: on the one hand it has to be a local approach, taking into account 3 spots in a row (since tokens can stay or jump to the neighbors), on the other hand, all of the spots influence each other such also a macro-view has to be incorporated. I therefore think that a little more elaborated theory is needed.

Maybe one has seen an equivalent problem or has a fresh idea of reformulation.

I think it's a challenging, non-standard combinatorics problem, which makes it pretty interesting :-)

2. Jul 17, 2011

Stephen Tashi

The first insurmountable difficulty in the problem is understanding what you mean by a "token number". Is it something stamped on each token? Or is it stamped on the "spots"?

3. Jul 17, 2011

pmsrw3

He means the number of tokens at the spot.

4. Jul 18, 2011

Stephen Tashi

Then are the "sets" of token numbers actually vectors of token numbers?

5. Jul 18, 2011

pmsrw3

Umm, good point. Yeah, I guess they would have to be. Either that, or I have completely misunderstood the problem.

6. Jul 25, 2011