Let $N_n:=\{1,2,\cdots,n\}$. Given two finite states Markov chains $\big(X^{(j)}_i\in N_n\}\big)_{i=0}^\infty$ for $j\in\{1,2\}$, both of which have two absorbing states at $1$ and $n$. 
$\text{Pr}\big(X^{(1)}_{i+1}=1|X^{(1)}_i=1\big)=\text{Pr}\big(X^{(1)}_{i+1}=n|X^{(1)}_i=n\big)=\text{Pr}\big(X^{(2)}_{i+1}=1|X^{(2)}_i=1\big)=\text{Pr}\big(X^{(2)}_{i+1}=n|X^{(2)}_i=n\big)=1, \,\forall a\in N_n$.  $$\text{Pr}\big(X^{(1)}_{i+1}=b|X^{(1)}_i=a\big)>\text{Pr}\big(X^{(2)}_{i+1}=b|X^{(2)}_i=a\big)>0, \,\forall 1<a<b, a,b\in N_n.$$
$$0<\text{Pr}\big(X^{(1)}_{i+1}=b|X^{(1)}_i=a\big)< \text{Pr}\big(X^{(2)}_{i+1}=b|X^{(2)}_i=a\big), \,\forall n>a>b, a>1, a,b\in N_n,$$
$$\text{Pr}\big(X^{(1)}_{i+1}=a|X^{(1)}_i=a\big)= \text{Pr}\big(X^{(2)}_{i+1}=a|X^{(2)}_i=a\big), \,\forall n>a>1, a\in N_n.$$ 
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][1] demonstrates a counterexample for a stronger claim.

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


  [1]: https://mathoverflow.net/a/345458/32660