question regarding a Markov chain probability of not having visited a state 
Hi all, I have a question regarding the difference between $P(X_3=0)$ and $P(X_3=0 |T>3)$ here. Are they the same or not? 
The reason I am asking this is because the transition matrix indicates that state 2 is an absorbing state, so I thought it is enough just to calculate $P_{00}^3$ i.e. the transition matrix to the power of 3, and then take the (0,0) component to be the probability that the chain starts at 0 and ends at 0 right after 3 transitions without ever entering state 2?
But how come we need to divide by $P(X_3=0) + P(X_3=1)$? I thought calculating $P_{00}^3$ alone already gives the answer?

The probability $\mathbb P(X_3=0)$ is not the same as the conditional probability $\mathbb P(X_3=0\mid T>3)$. Indeed,
$$
\mathbb P(X_3 = 0\mid X_0=0) = \frac{457}{1000} \approx 0.457
$$
while
$$
\mathbb P(X_3 = 0\mid X_0=0, T>3) = \frac{15215}{23328}\approx 0.652221.
$$
Note that we can simplify the computation of $\mathbb P(X_3=0\mid X_0=0,T>3)$ a bit by considering the transition matrix without the row/column corresponding to state $3$, rescaled so that the rows sum to one:
$$
Q = \left(
\begin{array}{cc}
 \frac{7}{9} & \frac{2}{9} \\
 \frac{3}{8} & \frac{5}{8} \\
\end{array}
\right).
$$
Then
$$
Q^3 = \left(
\begin{array}{cc}
 \frac{15215}{23328} & \frac{8113}{23328} \\
 \frac{8113}{13824} & \frac{5711}{13824} \\
\end{array}
\right),
$$
and the desired probability is simply the $(1,1)$-entry of this matrix.
