Let us consider random walk on integers {0,1,...,N} where $P(N,N)=1$,$P(0,1)=1$, $P(N,N-1)=0$ and all other connections have probability $\frac{1}{2}$. Using first step analysis, compute $p_{00}$ for the cases $N=1,2,3$. What about when $N\rightarrow \infty$? [$p_{xy}$ is the probability that the markov chain visits state $y$ given that it starts at $x$]
Attempt: I have done the first part, it's easy, e.g when $N=2$, $p_{00}=\frac{1}{2}$; when $N=3$, it is $\frac{2}{3}$.
Now for the second part, I thought I would be able to see a pattern if I do the cases $N=4,5$; BUT I cannot see any such pattern. For example, when $N=4$, $p_{00}=\frac{1}{2}+\frac{1}{2^2}p_{10}+\frac{1}{2^2}\frac{1}{3}p_{10}$; when $N=5$, $p_{00}=\frac{1}{2}+\frac{1}{2^2}p_{10}+\frac{1}{2^3}p_{20}+\frac{1}{2^3}p_{40}$ and we observe that $p_{00}=p_{10}$ in this problem, so we can find the value by solving these equations in respective cases, but as I said, this does not help me to do the case as $N\rightarrow \infty$.
What should I do here?
Thank you for your time and help.
For every $N$, $p_{00}$ is the probability to visit state $0$ before state $N$ starting from state $1$. For every state $x$, consider $h_x$ the probability to visit state $0$ before state $N$ starting from state $x$. Then the arch-classical one-step analysis of random walks shows that $h_x=\frac12(h_{x-1}+h_{x+1})$ for every $1\leqslant x\leqslant N-1$. This means that the function $x\mapsto h_x$ is affine. Since $h_0=1$ and $h_N=0$, one gets $h_x=1-\frac{x}N$ for every $x$, in particular $p_{00}=h_1=$ $____$.