N-Urns N-Color ball modelling as Markov Chain I am trying to model a system which can, mostly, be simplified to elements of different groups changing groups among themselves. I want to understand how frequently the elements change group and how frequently they stay in the same group (after a large number of interactions). My initial idea was to treat it as a Ehrenfest urn problem with N-Urns and N-Colors (the initial group). A ball after being selected can change urn or stay in the same.
For example (with N=2), all the red balls start in urn 1 and all the blue balls in urn 2 (the number of blue and red balls is not necessary the same). Select randomly a ball, for example a red one from urn 1 and it can move to urn 2 with probability p (or stay with 1-p).
In my problem, I have the final state of the urns (after enough time for mixing) and try to find the transition matrix(ces) based on that.
However, I can't seem to find any source with anything that looks like my problem. All the cases I see are, in general, much simpler and I can't find a way to apply them to my problem. Is an Ehrenfest urn what I am looking for or is there any other any name? Am I missing something and is there any other way to model this problem?
Thanks!
 A: The study of the CNO-cycle in stellar nuclear processes has some similarity with this inverse problem (using observed ratio's of C, N und O to calculate the reaction rates), although the multiple urns would make it more difficult. 
Basically you are looking for matrices such that observed final state is an equilibrium state (eigenvector with eigenvalue 1). 

simple example
With $x$ blue balls in urn 1 and $y$ blue balls in urn 2 you have an equilibrium if the ratio of fluxes is  $a/x$ from urn 1 to urn 2 and $a/y$ from urn 2 to urn 1. 
Indeed observe the eigenvector and eigenvalue of the matrix $$ \begin{pmatrix} 1-\frac{a}{x} & \frac{a}{y} \\ \frac{a}{x} & 1- \frac{a}{y} \end{pmatrix} \begin{pmatrix} x \\y \end{pmatrix}  = 1 \cdot \begin{pmatrix} x \\ y\end{pmatrix}$$ 
General
For more urns you are looking for $M v = v$ or $(M-I) v = 0$. Thus any matrix $M$ such that $(M-I)$ is made up of rows that are perpendicular to $v$ will do. So let $P$ be a n-1 x n matrix with rows perpendicular to $v$, and let $N$ be any n x n-1 matrix with at least a non-zero entry in every row. then $(N P)+I$ is a matrix with $v$ as eigenvector with eigenvalue 1. 
In our example this is 
$$ M-I = \begin{pmatrix} b \\ c  \end{pmatrix} \begin{pmatrix} -\frac{1}{x} & \frac{1}{y} \end{pmatrix} =  \begin{pmatrix} -\frac{b}{x} & \frac{b}{y} \\  -\frac{c}{x} & \frac{c}{y} \end{pmatrix}$$
where we impose in addition $b=-c$ according to the equation of continuity (for the solution to remain physically relevant the flow out of an urn is equal to what is inside the cell urn itself, ie the sum of values in a single column equals to 1)
