, 2019

Solution:

Let $P(m, n)$ be the probability that she ascends to heaven from point $(m, n)$. Then $P(6m, 6n) = 1$ and $P(6m+3, 6n+3) = 0$ for all integers $m, n \in \mathbb{Z}$. At all other points,
$$4P(m, n) = P(m-1, n) + P(m+1, n) + P(m, n-1) + P(m, n+1)$$
This gives an infinite system of equations. However, we can apply symmetry arguments to cut down the number of variables to something more manageable. We have $P(m, n) = P(m+6a, n+6b)$ for $a, b \in \mathbb{Z}$, and $P(m, n) = P(n, m)$, and $P(m, n) = P(-m, n)$, and $P(m, n) = 1 - P(m+3, n+3)$ (since any path from the latter point to heaven corresponds with a path from the former point to hell, and vice versa).
Thus for example we have
$$P(1,2) = P(-1,-2) = 1 - P(2,1) = 1 - P(1,2),$$
so $P(1,2) = 1/2$.
Applying Equation (1) to points $(1,1)$, $(0,1)$, and $(0,2)$, and using the above symmetries, we get the equations
$$\begin{gathered} 4P(1,1) = 2P(0,1) + 1 \\ 4P(0,1) = P(0,2) + 2P(1,1) + 1 \\ 4P(0,2) = P(0,1) + 3/2 \end{gathered}$$
Solving yields $P(1,1) = 13/22$.