Charged particles 
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?
 A: As it was pointed out in the comments, the solution below is for the case when the particles are arranged in a line, rather than circularly. So please modify the proof below. 
To simplyfy the definition of the process, in each step we have the following possibilities. 
Either with probability $1/4$ the length decreases by 1, or with probability $3/4$, the length increases by 1. So this is an unbalanced Drunkard walk (one of the most basic Markov chains there is) on infinitely many states indexed by the non-negative integers. Your initial state is $4$, and the only absorbing state is 0. Your question is the probability of absorption. 
This can be computed using standard theory in Markov chains. 
Let $p$ be the probability that from a given state $n\geq 1$ you at some point reach $n-1$. Note that this $p$ is independent from $n$. So from the law of total probability (based on the case distinction that the first move is to the left or to the right) we have 
$$p= \frac{1}{4}\cdot 1 + \frac{3}{4}\cdot p^2$$ 
Proof: if the first move is to the left, then we are done (i.e., the probability is 1 in this case).  If the first move is to the right, then we have to reach $n-1$ from $n+1$. This is equivalent to reaching $n$ from $n+1$ and then reaching $n-1$ from $n$, hence the probability is $p^2$. 
So $3p^2-4p +1 = 0$, or equivalently, $(3p-1)(p-1)=0$. Thus $p=1$ or $p=1/3$. It is easy to see that $p$ is not $1$, so $p=1/3$. 
(There is another argument: consider the  finite state Drunkard walk with the same transition probabilities starting from state $1$. Then there is a well-known closed formula for the probability of absorption, which tends to $1/3$ as the number of states tends to infinity.) 
So we can move one step to the left with probability $1/3$. Thus, to ever reach the state that is four steps to the left from the initial state, has probability $(1/3)^4=1/81$.  
A: 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.
