A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$, $(-2, 2)$, $(-2, -2)$, $(2, -2)$. A particle starts at $(0,0)$. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$, $(x + 1, y + 1)$, $(x + 1, y)$, $(x + 1, y - 1)$, $(x, y - 1)$, $(x - 1, y - 1)$, $(x - 1, y)$, or $(x - 1, y + 1)$. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?
I assigned the central point the letter Z, the north, east, west, and south points the letter R, and the ne, se, sw, and nw points the letter Q. In the answer, it says that the probability of arriving at a corner from point Z was, Z=1/2R+1/2Q. When I tried to solve this problem I made my probability Z=4R+4Q because in my mind Z's next step has to arrive at one of the 8 lattice points and I assumed the probability of z is equal to the probability of R or R or R or R or Q or Q or Q or Q. Why is this not so?
Due to the inherent symmetries the game has only three nonterminal states: $$O=\{(0,0)\}, \quad A=\{(\pm 1,0),(0,\pm1)\},\quad B=\{(\pm1,\pm1)\}\ .$$ We consider it a win when the walk ends in one of the four points $(\pm2,\pm2)$. Denoteby $p_O$, $p_A$, $p_B$ the probability of a win when we are in state $O$, $A$, and $B$ respectively. Then we have the following equations: $$p_O={1\over2}p_A+{1\over2}p_B,\quad p_A={1\over8}p_O+{1\over4}p_A+{1\over4}p_B, \quad p_B={1\over8}p_O+{1\over4}p_A+{1\over8}\ .\tag{1}$$ The system $(1)$ has the solutions $$p_O={4\over35},\quad p_A={1\over14},\quad p_B={11\over70}\ .$$ The answer to the question therefore is $p_O={4\over35}$, so that $m+n=39$.