Graph with Poisson Clock at each Vertex Let $G$ be a connected, undirected graph, with countably infinite set of vertices and countably infinite set of edges. Assume that the degree of each vertex is finite, and moreover, the degrees of all vertices are uniformly bounded.
Let each vertex carry one of two values: $1$ or $-1$.
Now, equip each vertex with a Poisson clock ($\lambda=1$), all clocks are independent. At each time the clock of a vertex ticks, the vertex updates its value to be the value of most of its neighbors (in case of a draw $-$ the value of the vertex remains unchanged).
Does there exist such a graph as described, with certain initial values at vertices, such that with $\mathbf{positive}$ probability, there will be a vertex where the value is $not$ eventually constant?
Thank you.
$\mathbf{EDIT:}$ If you wish, for a beginning, analyze the example given by domotorp in his comment (which could be a solution): take the $3$-regular tree with initial values as follows: pick one vertex, it will be $1$. The vertices around it will be $-1$. The vertices at distance $2$ from the initial vertex are again $1$. And so on, changing the value layerwise. In this graph and initial values, will there be a vertex, that with positive probability, will not converge? 
(Even if the answer for this example is NO, the fact that every vertex in the example almost surely converges is also nontrivial, and a proof of this will also be upvoted.)
 A: Yes, my example is easy to modify after some thought, just take a thick enough layer for each level.
More precisely, let $f$ be a sufficiently fast growing function, and define the initial value on any vertex at distance $f(n)\le d< f(n+1)$ from the root as $(-1)^n$.
Given $f(n)$, one can also pick a large enough $f(n+1)$ such that independently of the later values the root will get $(-1)^n$ in some time that depends only on $f(n+1)$ with at least $50\%$ probability.
This guarantees that with $1$ probability it will switch values infinitely often (just like every other vertex).
In more details, suppose first that $f(n+1)=\infty$, i.e., all but a finite number of vertices have the value $(-1)^n$.
Lemma. If all vertices at distance $>r$ have value $(-1)^n$, then with probability $1$ after a while all vertices will have value $(-1)^n$.
Proof. Vertices further than $r$ can never change value. If a vertex at distance $r$ gets value $(-1)^n$, it will never change again. So eventually all vertices at distance $r$ will also obtain value $(-1)^n$ and we can use induction on $r$.
From this it follows that the root will obtain value $(-1)^n$ in some $T(n)$ time with $90\%$ probability.
If $f(n+1)$ is not $\infty$, but just large enough compared to $T(n)$, then the probability of any vertex at distance $r+1$ from the root changing value is less than $10\%$.
Therefore, this won't affect the probability of the root changing value.
A: Doesn't quite answer the question, but in this paper --
http://people.hss.caltech.edu/~tamuz/papers/retention.pdf
-- it is stated that Tessler and Louidor showed that in the infinite d-regular tree with d even (and random tie-breaking), for uniformly random initial spins, almost surely there are vertices that change their spin infinitely often.
I couldn't actually find the paper being referred to, and it doesn't clearly answer the question because of the random tie-breaking.
