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 one absorbing state $1$. 
Pr$(X^{(1)}\_{i+1}=1|X\_i=1)=$Pr$(X^{(2)}\_{i+1}=1|X\_i=1)=1, \,\forall a\in N\_n$. $$\text{Pr}\big(X^{(1)}\_{i+1}=b|X\_i=a\big)>\text{Pr}\big(X^{(2)}\_{i+1}=b|X\_i=a\big)>0, \,\forall 1<a<b, a,b\in N\_n,$$
$$0<\text{Pr}\big(X^{(1)}\_{i+1}=b|X\_i=a\big)\le \text{Pr}\big(X^{(2)}\_{i+1}=b|X\_i=a\big), \,\forall a\ge b, a>1, a,b\in N\_n.$$ 
Is the following true? 
$$\text{Pr}\big(X^{(1)}\text{ reaches } b|X^{(1)}\_0=a\big)>\text{Pr}\big(X^{(2)}\text{ reaches }b|X^{(2)}\_0=a\big), \,\forall a<b,$$
and 
$$\text{Pr}(X^{(1)}\text{ reaches }b|X^{(1)}\_0=a)\le\text{Pr}(X^{(2)}\text{ reaches }b|X^{(2)}\_0=a), \,\forall a
\ge b.$$


[This math.stackexchange.com answer](https://math.stackexchange.com/a/3424918/64809) states that it is not true for an irreducible transition probability matrix compared to a reducible transition probability matrix. But is it true for two chains each of which has only one absorbing state?


Would a coupling argument help to resolve this problem?




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[Here is the version 2](https://mathoverflow.net/q/345450/32660) of this question under a more stringent condition.



Comparison of hitting probability of two Markov chains both with only one absorbing state
Let $P=(p\_{ij})$ and $Q=(q\_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains, such that 
\begin{gather}p\_{11}=q\_{11}=1,\\ p\_{ij}q\_{ij}>0\text{ if }i>1,\\ p\_{ij}>q\_{ij}\text{ if }1<i<j,\\ 
p\_{ij}\le q\_{ij}\text{ if }i>1\text{ and }1\le j\le i.
\end{gather}
The conjecture was that then 
\begin{gather}f\_{P;ij}>f\_{Q;ij}\text{ if }1<i<j,\\ 
f\_{P;ij}\le f\_{Q;ij}\text{ if }i>1\text{ and }1\le j\le i,
\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 is false in general. E.g., suppose that $n=3$, 
$$P=\frac1{16}
 \left(
\begin{array}{ccc}
 16 & 0 & 0 \\
 1 & 3 & 12 \\
 4 & 4 & 8 \\
\end{array}
\right),\quad
Q=\frac1{16}\left(
\begin{array}{ccc}
 16 & 0 & 0 \\
 4 & 4 & 8 \\
 4 & 4 & 8 \\
\end{array}
\right). 
$$
Then
$$f\_{P;22}=\frac9{16}\not\le\frac8{16}=f\_{Q;22}.$$ 


