So I'm having trouble writing down a rigorous proof of something that seems very clear.
Consider the following Markov chain on a ring: with probability $1/2$, it stays where it is, with probability $1/2-1/n$ it moves counterclockwise, and with probability $1/n$ it jumps to a random node. When the latter has happened, we'll say a jump has occurred.
Let $A$ be the event that by time $10n$, exactly one jump has occurred. Let $B$ be the event that the random walk has never taken the self loop at node $1$.
I want to prove that $P(B|A)$ is lower bounded by some constant independent of $n$. I'm actually guessing that $P(B|A) \geq 1/2^{11}$.
This seems obvious to me: conditional on $A$, the number of times the chain transitions into node $1$ by time $10n$ is at most $11$, and the probability of not taking the self-loop is $1/2$ each time. But how to say this formally?
Edit: Actually the above sketch is somewhat problematic: as Craig points out in the comments, taking exactly one jump changes the probabilities of taking the self loop whenever the chain is at node $1$. But, presumably, the chance of taking the counterclockwise transition should only increase, since the number of jumps is below what you'd expect in $10n$ transitions. So I'd be perfectly satisfied with the bound $(1/2-1/n)^{11}$ instead.