Suppose we have a random walk on $\mathbb{Z}^2$ as seen on the picture below. We can walk to north, west, south and east each with probability of $\frac{1}{4}$
Now we start at point 3 and we want to know the probability of leaving the set $\{1,2,3\}$ in north or east direction. My idea was to assign to each vertex the probability and solve it a system of linear equation systems.
$\omega(x)=\text{probability of leaving the set \{1,2,3\} in north or east direction}$
Note that $\omega(x)=1:x\in \{10,6,4\}$ because these are the vertices where we leave the set $\{1,2,3\}$ in north and east direction and $\omega(y)=0:y\in \{9,8,7,5\}$ because these are the vertices where we leave at south or west.
$$\omega(3)=\underbrace{\frac{1}{4}\omega(6)}_{=1}+\underbrace{\frac{1}{4}\omega(4)}_{=1}+\frac{1}{4}\omega(1)+\frac{1}{4}\omega(5)$$ $$\omega(1)=\underbrace{\frac{1}{4}\omega(8)}_{=0}+\frac{1}{4}\omega(2)+\underbrace{\frac{1}{4}\omega(7)}_{=0}+\frac{1}{4}\omega(3)$$ $$\omega(2)=\underbrace{\frac{1}{4}\omega(10)}_{=1}+\underbrace{\frac{1}{4}\omega(9)}_{=0}+\underbrace{\frac{1}{4}\omega(6)}_{=1}+\frac{1}{4}\omega(1)$$
If we solve this system of linear equations we get
$\omega(3)=\frac{4}{7}$, $\omega(1)=\frac{2}{7}$, $\omega(2)=\frac{4}{7}$
Is this a correct approach?