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](https://mathoverflow.net/a/345458/32660) demonstrates a counterexample for a stronger claim.


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



Comparison of hitting probability of two Markov chains both with only one absorbing state version 3
Let $P=(p\_{ij})$ and $Q=(q\_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, 
$$P=\frac1{1000}\left(
\begin{array}{ccccc}
 1000 & 0 & 0 & 0 & 0 \\
 1 & 241 & 260 & 38 & 460 \\
 22 & 75 & 283 & 448 & 172 \\
 389 & 67 & 103 & 158 & 283 \\
 0 & 0 & 0 & 0 & 1000 \\
\end{array}
\right),
$$
$$Q=\frac1{1000}\left(
\begin{array}{ccccc}
 1000 & 0 & 0 & 0 & 0 \\
 4 & 241 & 259 & 37 & 459 \\
 23 & 531 & 283 & 1 & 162 \\
 390 & 197 & 194 & 158 & 61 \\
 0 & 0 & 0 & 0 & 1000 \\
\end{array}
\right).
$$


Then $f\_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f\_{Q;21}$, where $f\_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f\_{Q;ij}$ is defined similarly. 


This disproves your conjecture, because there is no state to the left of $1$. 


**Remark.** Using the [formula](https://mathoverflow.net/a/345458/32660) $f=\lim\_{t\uparrow1}(I-tR)^{-1}g$, we see that, for $n=5$, 
$$f\_{P;21}=\big[-p\_{24} \left(\left(1-p\_{33}\right) p\_{41}+p\_{31} p\_{43}\right)-p\_{23} \left(p\_{34} p\_{41}+p\_{31}
 \left(1-p\_{44}\right)\right)+p\_{21} \left(-p\_{44} p\_{33}+p\_{33}+p\_{34} p\_{43}+p\_{44}-1\right)\big] \\ 
\big/\big[p\_{23}
 p\_{32}+p\_{24} p\_{43} p\_{32}-p\_{23} p\_{44} p\_{32}+p\_{33}+p\_{24} p\_{42}-p\_{24} p\_{33} p\_{42}+p\_{23} p\_{34}
 p\_{42}+p\_{34} p\_{43}-p\_{33} p\_{44}+p\_{44}-p\_{22} \left(p\_{34} p\_{43}-p\_{33}
 \left(p\_{44}-1\right)+p\_{44}-1\right)-1\big].
$$ 
This expression depends on the $p\_{ij}$'s in a very complicated way, which makes simple comparisons such as the one in the OP unlikely to hold. 


