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
$$f_{P;22}=\frac9{16}\not\le\frac8{16}=f_{Q;22}.$$