Consider a discrete time Markov chain with state space $S =\left\{A, B, C, D, E \right\}$ and transition matrix $$P= \left ( \begin{array}{c|ccccc} & A & B & C & D & E\\ \hline A & 0 & 1/3 & 1/3 &1/3 &0 \\ B & 1/3 & 0 & 1/3 &0 &1/3\\ C & 1/2 & 1/2 & 0 &0 &0\\ D & 1/2 & 0 & 0 &0 &1/2\\ E & 0& 1/2 & 0 &1/2 &0\\ \end{array} \right )$$
What is the expected number of times passing through state $B$ before reaching state $A$, assuming you start from state $C$ ? The hint only provides a system of linear recurrence equations including the following ：
$$\mathbb{E}(\tau_B<\tau_A\mid X_0=B)=1/3\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B)+1/3\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=E),$$
where $\tau_{x}:=\inf\left\{n\in\mathbb{N}:X_n=x\right\},x\in S.$
I thought that $\mathbb{E}(\tau_B<\tau_A\mid X_0=B)=P_{BA}\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=A)+P_{BB}\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=B)+P_{BC}\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=C)+P_{BD}\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=D)+P_{BE}\cdot\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=E).$
I don't why $\mathbb{E}(\tau_B<\tau_A\mid X_0=B,X_1=A)=0.$
If $X_1=A$ then $\tau_A=1$. We cannot have $\tau_B<\tau_A(=1)$ since $\tau_B$ is a positive integer.
If you interpret $\mathbb N$ as $\{0,1,2...\}$ instead of $\{1,2...\}$ then the conditionl probability of $\tau_B<\tau_A$ given $X_0=B, X_1=A$ is $1$.