Infinite Tree with Poisson Clocks

Question

Let $\mathcal{T}$ be the infinite countable $3$-regular tree graph. Pick a vertex in this graph, call it the root. Let the root carry the value $0$.

Next, assign $1$ to the neighbours of the root. Then, assign $0$ to the vertices at distance $2$ from the root. And so on, changing values layerwise.

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 neighbours.

What is the probability that the root changes value infinitely often?

Thank you.

(See a closely related question: Graph with Poisson Clock at each Vertex)

$\mathbf{EDIT:}$ A possible direction for a solution: Define a positive valued function $f$ on the vertices (or on the edges?), such that the sum of $f$'s values at all vertices is finite. Now, each time a vertex pops and changes value, $f$ is nullified there. If ticks at neighboring vertices cause that vertex to become unstable - not to carry the value of most of its neighbours - $f$'s original value at that vertex returns. If the sum of all $f$'s values at time $t$ is called $S_t$, then convergence at all vertices occurs iff $\lim_{t\to \infty}S_t$ exists.

Mind you, $S_t$ is a bounded stochastic process, and therefore has a bounded expected value. This gives rise to questions such as $E[S_1]=?$ $E[S_t]=?$ $\lim_{t\to \infty}E[S_t]=?$ Is the function $E[S_t]$ monotone decreasing?

(The set of times $t$ in the subscript $S_t$ is the set of ticking times. Because there are infinitely many clocks, every time interval contains infinitely many ticks (with probability $1$). In other words, the ticks are dense in $(0, \infty)$. By the Back-and-forth method, this set is isomorphic (in an sense of order) to the set of positive rational numbers.)

$\mathbf{EDIT:}$ Another possibly solving way: some use of the ergodic theory. Find some ergodic transformation (on the space where each element is the collection of ticking times) and apply an ergodic theorem. Maybe, even using the $S_t$ defined above.

Answer 1

The answer has to be 1. Here is a sketch proof (I don't have a full proof). Construct a graphical model by taking the product of $\mathcal T$ with $\mathbb R^+$. Mark on each copy of $\mathbb R^+$ the times when the clock will ring. At those points in the graphical model, put arrows from the vertex to its neighbors. This allows you, for each $t$ to partition $\mathcal T$ into regions with an ancestor: each region is coloured by following arrows from a unique vertex ("king"). By Borel-Cantelli, each king will be toppled sooner or later, so the root is a subject of a sequence of different kings. This can all be done before the colour of the kings is specified. Your rule is that the original colouring is specified by the parity of the distance from the origin, but it is clear that neither odd nor even is favoured in this model.