Comparing hitting probabilities vs comparing mean hitting times of a random walk on a graph I am trying to understand random walks on graphs and whether an intuition that I have can be made rigorous mathematically, and whether it is also true.
Let $G$ be a finite, connected undirected graph with $n$ vertices and we perform a simple random walk on $G$. Assume we have three distinct vertices $a$, $b$, $c$ and that our random walk starts at $a$. Then, we can define the mean hitting times $h_{ab}$, $h_{ac}$ which are the expected number of steps until we reach $b$ and $c$ respectively. These can each be calculated by solving a linear system of equations. Now, my question is the following.

Assume again that our walk starts at $a$. If we know that the probability of reaching $b$ without passing through $c$ is strictly greater than the probability of reaching $c$ without passing through $b$, then can we say that $h_{ab} < h_{ac}$? 

The intuitive reasoning would be that if we have a higher chance that we reach one point without passing through the other, rather than the other way around, then it should make sense that we arrive at that point also faster on average.
How could this be made mathematically rigorous? Is it even true?
 A: I think it's false. This is because paths from $a$ to $b$ via $c$, though having low probability, maybe have sufficiently large mean hitting time that it overcomes the difference in probability. An example follows (sorry I can't construct a simple one). Basically, in going from $a$ to $b$, we can get "trapped" for some time in the sub-graph below node $c$ and this increases the mean hitting time $h_{ab}$.

Denote by $p_{ab} \;(p_{ac})$ the probability of walking from $a$ to $b \;(c)$ avoiding $c \;(b)$. Then,
\begin{align}
p_{ab} &= 3\cdot \dfrac{1}{4}\cdot \dfrac{1}{2} + 3\cdot \dfrac{1}{4}\cdot \dfrac{1}{2}\cdot p_{ab} = \dfrac{3}{8} + \dfrac{3}{8} p_{ab} \\
\therefore\quad p_{ab} &= \dfrac{3}{5} \\
& \\
p_{ac} &= \dfrac{1}{4} + 3\cdot \dfrac{1}{4}\cdot \dfrac{1}{2}\cdot p_{ac} = \dfrac{1}{4} + \dfrac{3}{8} p_{ac} \\
\therefore\quad p_{ac} &= \dfrac{2}{5} \lt p_{ab}.
\end{align}
Mean hitting times:
\begin{align}
h_{ac} &= \dfrac{1}{4}\cdot 1 + 3\cdot\dfrac{1}{4}\cdot\dfrac{1}{2}(2 + h_{ac}) + 3\cdot\dfrac{1}{4}\cdot\dfrac{1}{2}(2 + h_{bc}) \\
 &= \dfrac{7}{4} + \dfrac{3}{4} h_{ac} \qquad\qquad\text{since $h_{bc} = h_{ac}$} \\
\therefore\quad h_{ac} &= 7. \\
& \\
h_{ab} &= 3\cdot\dfrac{1}{4}\cdot\dfrac{1}{2}\cdot 2 + 3\cdot\dfrac{1}{4}\cdot \dfrac{1}{2}\cdot (2 + h_{ab}) + \dfrac{1}{4}(1 + h_{cb}) = \dfrac{7}{4} + \dfrac{3}{8} h_{ab} + \dfrac{1}{4} h_{cb} \\
\therefore\quad h_{ab} &= \dfrac{14}{5} + \dfrac{2}{5} h_{cb} \\
& \\
h_{cb} &= \dfrac{1}{6}(1 + h_{ab}) + \dfrac{1}{6}\cdot 1 + \dfrac{4}{6}(1 + h_{gb}) = 1 + \dfrac{1}{6}h_{ab} + \dfrac{2}{3}h_{gb} \\
& \\
h_{gb} &= \dfrac{1}{2}(1 + h_{cb}) + \dfrac{1}{2}(2 + h_{gb}) = \dfrac{3}{2} + \dfrac{1}{2}h_{cb} + \dfrac{1}{2}h_{gb} \\
\therefore\quad h_{gb} &= 3 + h_{cb} \\
& \\
h_{cb} &= 1 + \dfrac{1}{6} h_{ab} + 2 + \dfrac{2}{3} h_{cb} = 3 + \dfrac{1}{6} h_{ab} + \dfrac{2}{3} h_{cb} \\
\therefore\quad h_{cb} &= 9 + \dfrac{1}{2} h_{ab} \\
\therefore\quad h_{ab} &= \dfrac{14}{5} + \dfrac{2}{5}(9 + \dfrac{1}{2}h_{ab}) = \dfrac{32}{5} + \dfrac{1}{5} h_{ab} \\
\therefore\quad h_{ab} &= 8 \gt h_{ac}.
\end{align}
