Maximimal Independent Set on Ring and Path
if a node terminates the algorithm after an iteration with probability $1/c$, then after $k$ iterations this probability is $c^{-k}$. and the number of iterations required is $\ln _c n$. This is done by arguing that the expected number of nodes after $k$ iterations is $n c ^{-k}$, and we want to make this value less or equal to 1. We achieve this by setting $k$ to $\ln _c n$. 

Therefore, replacing $k = \ln _c n$ leads to $c^{- \ln _c n} = 1/n$, and whence the probability of correctness is $1 - 1/n$.

How to find $c$ ? As I said in my comment, I truly dont know the exact value. Here is a solution. Someone else may correct me.

We should calculate the probability that a node $v$ terminates the algorithm in any given iteration. This happens if $v$ is greater than its two neighbors OR if the left neighbor of $v$ is greater than all its neighbors OR if the right neighbor of $v$ is greater than all its neighbors. The probability of any of these events is 1/2*1/2 = 1/4. The probability of any of these events is 1/4 + 1/4 + 1/4 = 3/4. Therefore $c = 3/4$.