TITLE: Charged particles
QUESTION [8 upvotes]: We are creating a circular hub consisting of charged 0 and 1 particles next to each other, beginning with four of them: 0, 1, -0 and -1 in this order. Every 1 sec we randomly select one of each kind and add it next to the last one. Whenever a 0 particle finds itself next to a -0 or a 1 particle next to a -1, they both vanish. Assuming it is equally likely to select each of the 4 kinds, what is the probability that at some point the hub vanishes completely?

I assume it is not as easy as 
$\frac {1}{4}.\frac {1}{3}.\frac {1}{2}$, right?
I can't figure out of anything else... 
Any help?

REPLY [2 votes]: I understand the procedure as follows:
We have a sequence of particles that starts out as $0$, $1$, $-0$, $-1$. We keep adding particles to the end of this sequence. The sequence is circular in the sense that the last particle is considered to be adjacent to the first particle, so they can potentially vanish together if they match. Even if particles from either end of the sequence vanish, new particles are always added to the end of the sequence.
Then the probability that all particles will eventually vanish is $1$. In each step, we have probability $\frac12$ of at least one pair of particles vanishing (since the ends of the sequence are of two different types and we uniformly randomly choose one of four types). More than one pair of particles might vanish in a single step if the new ends also match, but we can ignore that since a probability of $\frac12$ for a single match per step is already enough. The number of particles in the sequence is a biased random walk with probability $\frac12$ of taking a step $-2$ and probability $\frac12$ of taking a step $+1$. A biased random walk almost surely diverges towards the side of the bias, so the sequence almost surely vanishes completely. This would even be true if there were an additional pair, $2$ and $-2$, since we'd then take a step $-2$ with probability $\frac13$ and a step $+1$ with probability $\frac23$, so the expected value of the step would be $0$ and the random walk would still almost surely reach any given point.