I hope that someone could help me with understanding the exercise.
In a cycle shaped house there are n chambers. In this house there is a mouse and each chamber has cheese except the room where the mouse starts. Every chamber has also two adjoining chambers, and each of those two chambers has the same probability to be visited. I need to show that each cheese has the same chance of being eating by the mouse with the probability of 1/n.
I thought about myself to proof this I will first try it with 7 chambers.
As shown in this picture:
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To calculate the probability I thought about the following:
$$ c0 = 0 (no cheese) $$ $$ cn = 1 (lastCheese) $$ $$ c1 = {1\over2}(c0 + c2) $$ $$ c2 = {1\over2}(c1 + c3) $$ $$ c3 = {1\over2}(c2 + c4) $$ $$ c4 = {1\over2}(c3 + c5) $$ $$ c5 = {1\over2}(c4 + c6) $$ $$ c6 = {1\over2}(c0 + c5) $$
But I am not sure if this correct and I have no idea how I can proof for n chambers.
I'm not sure about what you mean by the cheese "being eaten by the mouse". I'm guessing the mouse wanders around the house for a "long" time, and stops in one room to eat the cheese.
In this case, this would be a question from stochastic processes where you are attempting to find the stationary distribution, i.e. find $\pi$ such that $\pi=\pi P$, where $\pi$ is your stationary distribution vector (of length $n$) and $P$ is the transition matrix which is the zero matrix with $1/2$ on the off-diagonals (except for the first and last rows which are $(0,\frac{1}{2},0,0,...,0,\frac{1}{2})$ and $(\frac{1}{2},0,0,...,0,\frac{1}{2},0)$ respectively), in which case you have a system of equations that is solved by $\pi=(\frac{1}{n},...,\frac{1}{n})$.
In particular:
$$ \begin{align} \pi_1 &=\frac{1}{2}(\pi_2+\pi_n)\\ \pi_k &=\frac{1}{2}(\pi_{k-1}+\pi_{k+1}) \quad k=2,...,n-1\\ \pi_n &=\frac{1}{2}(\pi_1+\pi_{n-1})\\ \end{align}$$
such that $\pi_1+...+\pi_n=1$ is solved by $\pi_m=\frac{1}{n}$ for $m=1,...,k$.