An urn contains 5 black or white balls. At time 0, all the balls are white. At each step, we pick a ball uniformly in the urn, remove it and replace it with a ball of the opposite colour. Let $B_n$ be the number of black balls at time $n$.
Compute the long-run fraction of time where $B_n$=3
What is the expected value of the first time where the urn contains 5 white balls again?
Is it true that $\lim_{n \rightarrow \infty} P(B_n=3)$ exists?
For 1, I want to use time reversibility. But I'm not sure how to proceed from here.
For 2, I think I've got it using one-step analysis.
For 3, I'm not sure how do proceed, since I'm not even sure about 1.
Please help, thank you!
This is a random walk on the $5$-dimensional hypercube, with edges between adjacent vertices. A random walk on a connected graph spends equal proportion of time on each edge, and thus the proportion of time spent on a vertex is proportional to its degree. Here all $2^5=32$ vertices have the same degree, so the proportion of time on each vertex is $\frac1{32}$. Since $\binom53=10$ vertices correspond to $B_n=3$, the proportion of time where $B_n=3$ is $\frac{10}{32}=\frac5{16}$.
For $2$, you don’t need any further analysis. Since the proportion of time where all $5$ balls are white is $\frac1{32}$, the expected return time from $5$ white balls back to $5$ white balls is $32$.
For $3$, the limit doesn’t exist because $B_n=3$ can obtain only for odd $n$; in fact we have $\lim_{n\to\infty}P(B_{2n+1}=3)=\frac5{8}$ and $P(B_{2n}=3)=0$ (no limit required on the second one).