Deduce the general form of $P^n$ in terms of $n$ and use this answer to calculate $P(X_{20}=2|X_{10}=2)$
$$P=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0&0\\ \frac{1}{2}&\frac{1}{2}&0&0 \\ 0&0&\frac{2}{3}&\frac{1}{3} \\ 0&0&0&1 \end{pmatrix}$$
Its obvious that when I split the matrices into blocks, it produces the following because each row must sum to one
$$P^n=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0&0\\ \frac{1}{2}&\frac{1}{2}&0&0 \\ 0&0&\ \left(\frac{2}{3}\right)^n&1-\left(\frac{2}{3}\right)^n \\ 0&0&0&1 \end{pmatrix}$$
It is shown by the Chapman-Kolmogorov equation that $$P_{ij}^{m+n}=P(X_{m+n}=j|X_0=i)$$
Now $P(X_{20}=2|X_{10}=2)$ = $P(X_{10}=2|X_{0}=2)$, however the result shows that $$P(X_{10}=2|X_{0}=2)=\frac{2^{10}}{3^{10}}$$
I thought that $i=2, j=2$ would mean the intersection at the second row and second column which is $\frac{1}{2}$, however this is not the case. May some kindly explain this inconsistency in my reasoning for me!