Knight Probability - Transition

From position $(r, c)$, the knight can move to 8 positions: $(r \pm 1, c \pm 2)$ and $(r \pm 2, c \pm 1)$. Each move is chosen with probability $1/8$.

The formula to compute dp[r][c][k]: $dp[r][c][k] = \frac{1}{8} \sum_{(r', c') \in moves} dp[r'][c'][k-1]$ But wait: what if $(r', c')$ is off the board? Then that move contributes $0$ to the sum. The knight fell off, so the probability of staying on-board from that path is $0$.