Checker Reaching a Top Corner

Label squares $(r,c)$ with $r$ the row and $c$ the column. A move takes $(r,c)$ to $(r+1,c\pm1)$, so $r+c$ never changes parity.

Both corners $(5,1)$ and $(5,5)$ have $r+c$ even. So any start with $r+c$ odd reaches a corner with probability $0$.

Let $p(r,c)$ be the probability of finishing in a corner from $(r,c)$. In the interior,

On an edge the move is forced: $p(r,1)=p(r+1,2)$ and $p(r,5)=p(r+1,4)$. The top row gives $p(5,1)=p(5,5)=1$ and $p(5,c)=0$ for $c=2,3,4$.

Work downwards one row at a time.

• Row 4: $p(4,1)=p(5,2)=0$. Also $p(4,2)=\frac12(1+0)=\frac12$ and $p(4,3)=0$. By symmetry $p(4,4)=\frac12$ and $p(4,5)=0$.
• Row 3: $p(3,1)=p(4,2)=\frac12$. Next, $p(3,2)=\frac12(0+0)=0$ and $p(3,3)=\frac12\bigl(\frac12+\frac12\bigr)=\frac12$. By symmetry $p(3,4)=0$ and $p(3,5)=\frac12$.
• Row 2: $p(2,2)=\frac12\bigl(\frac12+\frac12\bigr)=\frac12$. Next, $p(2,3)=\frac12(0+0)=0$, and by symmetry $p(2,4)=\frac12$.

On the central block (row 4 at the top), the probabilities are therefore

The five squares with $r+c$ even each give $\frac12$, and the four odd squares give $0$. Each of the nine starting squares has probability $\frac19$, so averaging gives

Key idea. Because the parity of $r+c$ is preserved, four of the nine starts are hopeless. A short backward recursion shows that each of the other five succeeds with probability exactly $\frac12$.