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? 
$$\text{Pr}\big(X^{(1)}\text{ reaches } b|X^{(1)}\_0=a\big)>\text{Pr}\big(X^{(2)}\text{ reaches }b|X^{(2)}\_0=a\big), \,\forall 1<a<b,$$
and 
$$\text{Pr}(X^{(1)}\text{ reaches }b|X^{(1)}\_0=a)<\text{Pr}(X^{(2)}\text{ reaches }b|X^{(2)}\_0=a), \,\forall n>a>b.$$


[This mathoverflow.net answer](https://mathoverflow.net/a/345444/32660) demonstrates a counterexample for a weaker condition.


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 2 under stronger condition
Let $P=(p\_{ij})$ and $Q=(q\_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains, where $n\ge2$. Your conditions imply the following:  

\begin{gather}p\_{nn}=q\_{nn},\\ 
p\_{nj}<q\_{nj}\text{ if }1\le j\le n-1.
\end{gather}
Hence, $1=\sum\_{j=1}^n p\_{nj}<\sum\_{j=1}^n q\_{nj}=1$, which is a contradiction (which makes any conclusion whatsoever true). 




---


**Added:** In a comment, the OP suggested that the conjecture be modified by now assuming $n$ to be an absorbing state as well, so that the conditions become 
\begin{gather}p\_{11}=q\_{11}=p\_{nn}=q\_{nn}=1,\\ 
p\_{ii}=q\_{ii}\text{ if }1\le i\le n, \\
p\_{ij}q\_{ij}>0\text{ if }1<i<n,\\ p\_{ij}>q\_{ij}\text{ if }1<i<j\le n,\\ 
p\_{ij}<q\_{ij}\text{ if }1\le j<i<n.
\end{gather}


The conjecture then becomes that
\begin{gather}f\_{P;ij}>f\_{Q;ij}\text{ if }1<i<j\le n,\\ 
f\_{P;ij}<f\_{Q;ij}\text{ if }1\le j<i<n,
\end{gather}
where $f\_{P;ij}$ is the probability that the first chain ever reaches $j$ from $i$, and $f\_{Q;ij}$ is defined similarly. 


This conjecture, too, is false in general. E.g., suppose that $n=5$, 
$$P=\frac1{20}
 \left(
\begin{array}{ccccc}
 20 & 0 & 0 & 0 & 0 \\
 4 & 4 & 4 & 4 & 4 \\
 2 & 2 & 2 & 12 & 2 \\
 4 & 4 & 4 & 4 & 4 \\
 0 & 0 & 0 & 0 & 20 \\
\end{array}
\right),\quad
Q=\frac1{20}\left(
\begin{array}{ccccc}
 20 & 0 & 0 & 0 & 0 \\
 7 & 4 & 3 & 3 & 3 \\
 12 & 4 & 2 & 1 & 1 \\
 5 & 5 & 5 & 4 & 1 \\
 0 & 0 & 0 & 0 & 20 \\
\end{array}
\right). 
$$
Then
$$f\_{P;32}=\frac13\not<\frac{69}{283}=f\_{Q;32}.$$ 


**Added more:** This is in response to a request by the OP to provide details. Fix any $j\in[n]:=\{1,\dots,n\}$. Let $g$ denote the $j$th column of $P$ and let $R$ denote the matrix obtained from matrix $P$ by replacing the $j$th column in $P$ by the zero column. Then for the column matrix $f:=(f\_{P;ij}\colon i\in[n])^T$ we have 
$$f=\sum\_{n=0}^\infty (R^n g);
$$
cf. e.g. the 5th display on page 85 and formula (2.6.3') on page 90 in [Resnick](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=2ahUKEwjezNzWsdnlAhVERKwKHeK7CgsQFjADegQIBBAC&url=http%3A%2F%2Fwww.fcfm.buap.mx%2Fjzacarias%2Fcursos%2Fprocesos%2Flibros%2Fbook3pe.pdf&usg=AOvVaw2S9tF8Bs3QpzA0FLECZZkJ). If all entries of the column matrix $g$ are nonzero, then the norm $\|R\|\_{\infty,\infty}$ of the matrix $R$ considered as a linear operator from $\ell\_n^\infty$ to $\ell\_n^\infty$ will be $<1$, whence we will have $f=(\sum\_{n=0}^\infty R^n)g=(I-R)^{-1}g$. However, in our case, because we have two absorbing states, every column of $P$ must have a zero entry. This difficulty is easy to circumvent, as follows. By the monotone convergence theorem, for $t\in(0,1)$
$$(I-tR)^{-1}g=\sum\_{n=0}^\infty (tR)^n g\;\Big\uparrow\;\sum\_{n=0}^\infty (R^n g)=f
$$
as $t\uparrow1$. So, 
$$f=\lim\_{t\uparrow1}(I-tR)^{-1}g. 
$$
Since $(I-tR)^{-1}g$ is rational in $t$, the latter limit is easy to compute. 


In particular, for $n=5$ we get
$$f\_{P;32}=\frac {p\_{34}p\_{42}+p\_{32}(1-p\_{44})} 
{(1-p\_{33}) (1-p\_{44})-p\_{34} p\_{43}}, 
$$
with the corresponding expression for $f\_{Q;32}$. We see that $f\_{P;32}$ is increasing in $p\_{34}$. Also, one of the OP's conditions is $p\_{34}>q\_{34}$. So, to get $f\_{P;32}>f\_{Q;32}$, we should try to make $p\_{34}$ large and $q\_{34}$ small. At the same time, we may choose $p\_{42}$, $p\_{32}$, and $p\_{43}$ close enough to $q\_{42}$, $q\_{32}$, and $q\_{43}$ (respectively), also keeping in mind the conditions $p\_{33}=q\_{33}$ and $p\_{44}=q\_{44}$. This should (and does) result in $f\_{P;32}>f\_{Q;32}$, which disproves the conjecture. 


**Added yet more:** The matter becomes much more transparent if we ignore, at least for a moment, the strictness of the inequality restrictions on the $p\_{ij}$'s and $q\_{ij}$'s. Then we may assume that $p\_{34}=q\_{31}=p\_{42}=q\_{42}=1$. Then clearly $f\_{P;32}=1\not\le0=f\_{Q;32}$. If you still insist on the strictness of the inequality restrictions, then you can have it by the continuity of $f\_{P;32}$ on the appropriate domain. 


