Random lattice descent from (4,4) hitting the origin

A particle begins at $(4,4)$. On each move it chooses, with probability $\tfrac13$ each, to step one unit left, one unit down, or one unit diagonally (left and down together). It stops as soon as it first lands on either coordinate axis. What is the probability that it stops at the origin?

$245/2187$. The origin can only be reached by a diagonal step from $(1,1)$. Reaching $(1,1)$ means making three net steps in each direction with $0$–$3$ diagonals: $\tfrac{\binom63}{3^6}+\tfrac{30}{3^5}+\tfrac{12}{3^4}+\tfrac1{3^3}=\tfrac{245}{729}$. Multiply by $\tfrac13$ for the final diagonal.