Let $N_n:=\{1,2,\cdots,n\}$. Given two finite states Markov chains $\big(X^{(j)}_t\in N_n\}\big)_{t=0}^\infty$ for $j\in\{1,2\}$, both of which have two absorbing states at $1$ and $n$. Define $p_{i,j}(t):=\text{Pr}\big(X^{(1)}_{t+1}$ and $q_{i,j}(t):=\text{Pr}\big(X^{(2)}_{t+1}=j|X^{(2)}_t=i\big), \,\forall i,j\in N_n$. Dropping the variable $t$ for the brevity of notaion, we stipulate that $$p_{1,1}=p_{n,n}=q_{1,1}=q_{n,n}=1;$$ $$p_{i,j}>q_{i,j}, \forall 1<i<j, i,j\in N_n;$$ $$p_{i,j}>q_{i,j}, \forall n>i>j, i>1, i,j\in N_n;$$ $$p_{i,i}=q_{i,i}, \forall n>i>1.$$

Are the following inequalities true? $$\text{Pr}\big(X^{(1)}\text{ reaches } b \text{ or above}|X^{(1)}_0=a\big)>\text{Pr}\big(X^{(2)}\text{ reaches }b\text{ or above}|X^{(2)}_0=a\big), \,\forall 1<a<b,$$ and $$\text{Pr}(X^{(1)}\text{ reaches }b\text{ or below}|X^{(1)}_0=a)<\text{Pr}(X^{(2)}\text{ reaches }b\text{ or below}|X^{(2)}_0=a), \,\forall n>a>b.$$

This mathoverflow.net answer demonstrates a counterexample for a stronger claim.

Would a coupling argument help to prove the inequalities if they are true?