Expected hitting time of a certain state in a Markov chain 
Let's consider the following Markov chain. If we start at state $2$, compute the probability to hit $4$ and the expected time until it happens.


The probability to hit $4$ in $n$ steps starting at $2$ is $Pr(X_n=4|X_0=2)=(\frac{1}{2})^n$ if $n$ is even, and $0$ if $n$ is odd. Then, the total probability of hitting 4 if we start at 2 will be$\sum_{n\geq1}Pr(X_n=4|X_0=2)=\sum_{i\geq1}(\frac{1}{2})^{2i}=\sum_{i\geq1}\frac{1}{4^i}=\frac{1}{3}$
How can I calculate $\mathbb{E}(T_{24})?$ I have tried considering the mean hitting times $m_{ij}$, so $m_{24}=1+\frac{1}{2}m_{14}+\frac{1}{2}m_{34}$ and $m_{34}=1+\frac{1}{2}m_{44}+\frac{1}{2}m_{24}=1+\frac{1}{2}m_{24}$.
The solution in my book is $m_{24}=m_{34}=3$, which does not seem intuitive to me.

The general way for computing mean hitting times of any absorbing state is well known, rather easily proven and in this Wikipedia article.
To get the mean hitting time of a specific state (or subset of states), we simply condition the outcome on hitting that state. For this question, the modified chain looks like:

This gives us the transition matrix $\Pi$, where $\Pi_{ij}$ gives the probability of transitioning to state j from state i:
$$
\begin{pmatrix}
0 & 1 & 0\\ 
\frac{1}{2} & 0 & \frac{1}{2}\\ 
0 & 0 & 1
\end{pmatrix}
$$
And Q:
$$
\begin{pmatrix}
0 & 1 \\ 
\frac{1}{2} & 0
\end{pmatrix}
$$
So that
$$
\begin{align}
m &= (I-Q)^{-1}1 \\
&= 
\begin{pmatrix}
2 & 2 \\ 
1 & 2
\end{pmatrix}1 \\
&= 
\begin{pmatrix}
4 \\ 
3
\end{pmatrix}
\end{align}
$$
Indeed, $m_{24} = 4$ and $m_{34} = 3$.
