Corner Absorption From a 3x3 Center
On a $5 \times 5$ grid, label the columns $0$-$4$ and the rows $0$ (bottom) to $4$ (top). Pick a starting square uniformly at random from the central $3 \times 3$ block (columns $1$-$3$, rows $1$-$3$).
At each step the token moves diagonally upward: up-left or up-right, each with probability $\tfrac{1}{2}$. If one of those moves would take the token off the board, only the legal move is made (with probability $1$). The walk ends when the token reaches row $4$.
What is the probability that the walk finishes in one of the two top corners -- column $0$ or column $4$ of row $4$? Show how you aggregate over all nine equally likely starting squares.
Hints
- The token always moves upward, so you can compute the corner-reaching probability for each cell by working backward from row 4.
- Each step changes the column by $1$ and the row by $1$, so the parity of column $+$ row never changes. Both corners have column $+$ row even: which starting cells can never reach a corner? At the edge columns the move is forced inward, so handle those cells separately.
- Set $P(0,4) = P(4,4) = 1$ and $P(c,4) = 0$ for interior columns, then apply $P(c,r) = \tfrac{1}{2}P(c-1,r+1) + \tfrac{1}{2}P(c+1,r+1)$ (with boundary adjustments) down to row 1. Average the nine central-block values.
Worked Solution
How to Think About It: This is a bounded random walk on a 5-wide grid, and the token always moves upward, so it reaches row 4 in a fixed number of steps (depending on its starting row). The key is that boundary reflections break the usual diffusion symmetry -- the edges funnel probability toward or away from the corners. Rather than setting up a big system of equations, just compute the corner-absorption probability $P(c, r)$ for every cell, working backward from row 4.
Quick Estimate: By symmetry, columns 1 and 3 are mirror images, so their corner probabilities match. The center column (column 2) "sees" both corners equally and the boundary effects cancel. A rough guess: about half the starting squares feed into a corner and about half do not, so the average should be somewhere around $\tfrac{1}{4}$ to $\tfrac{1}{3}$. The exact answer will be $\tfrac{5}{18} \approx 0.278$.
Approach: Define $P(c, r)$ as the probability of finishing in column 0 or column 4 at row 4 when starting from $(c, r)$. Work backward row by row from the terminal conditions.
Formal Solution:
*Terminal row (row 4):*
$$P(0,4) = 1,\quad P(1,4) = 0,\quad P(2,4) = 0,\quad P(3,4) = 0,\quad P(4,4) = 1$$
*Row 3 (one step from the top):*
- $P(0,3)$: left edge, forced right $\to (1,4)$. $P = 0$.
- $P(1,3) = \tfrac{1}{2}P(0,4) + \tfrac{1}{2}P(2,4) = \tfrac{1}{2}$.
- $P(2,3) = \tfrac{1}{2}P(1,4) + \tfrac{1}{2}P(3,4) = 0$.
- $P(3,3) = \tfrac{1}{2}P(2,4) + \tfrac{1}{2}P(4,4) = \tfrac{1}{2}$.
- $P(4,3)$: right edge, forced left $\to (3,4)$. $P = 0$.
*Row 2:*
- $P(0,2)$: forced right $\to (1,3)$. $P = \tfrac{1}{2}$.
- $P(1,2) = \tfrac{1}{2}P(0,3) + \tfrac{1}{2}P(2,3) = 0$.
- $P(2,2) = \tfrac{1}{2}P(1,3) + \tfrac{1}{2}P(3,3) = \tfrac{1}{2}$.
- $P(3,2) = \tfrac{1}{2}P(2,3) + \tfrac{1}{2}P(4,3) = 0$.
- $P(4,2)$: forced left $\to (3,3)$. $P = \tfrac{1}{2}$.
*Row 1:*
- $P(0,1)$: forced right $\to (1,2)$. $P = 0$.
- $P(1,1) = \tfrac{1}{2}P(0,2) + \tfrac{1}{2}P(2,2) = \tfrac{1}{2}$.
- $P(2,1) = \tfrac{1}{2}P(1,2) + \tfrac{1}{2}P(3,2) = 0$.
- $P(3,1) = \tfrac{1}{2}P(2,2) + \tfrac{1}{2}P(4,2) = \tfrac{1}{2}$.
- $P(4,1)$: forced left $\to (3,2)$. $P = 0$.
*Aggregation over the central $3 \times 3$ block:*
| Cell | $P$ | |---|---| | $(1,1)$ | $\tfrac{1}{2}$ | | $(2,1)$ | $0$ | | $(3,1)$ | $\tfrac{1}{2}$ | | $(1,2)$ | $0$ | | $(2,2)$ | $\tfrac{1}{2}$ | | $(3,2)$ | $0$ | | $(1,3)$ | $\tfrac{1}{2}$ | | $(2,3)$ | $0$ | | $(3,3)$ | $\tfrac{1}{2}$ |
Five of the nine cells have probability $\tfrac{1}{2}$ and four have probability $0$. The overall probability is:
$$\frac{1}{9}\sum_{\text{block}} P(c,r) = \frac{1}{9}\left(5 \cdot \frac{1}{2} + 4 \cdot 0\right) = \frac{5}{18}$$
Answer: The probability that the walk is absorbed in either top corner is $\dfrac{5}{18} \approx 0.2778$.
Intuition
The striking pattern here is that every cell's corner probability is either $\tfrac{1}{2}$ or $0$, with no intermediate values. The $0$s come from parity: each step changes the column and the row by $1$, so column $+$ row keeps its parity, and both corners have column $+$ row even. A cell with odd column $+$ row can never reach a corner. The $\tfrac{1}{2}$s come from the last step: from an even cell the token always arrives at row $3$ in column $1$ or $3$, and from there the final move is a fair coin between a corner and the center column. Five of the nine central cells are even, so the average is $\tfrac{5}{2} \cdot \tfrac{1}{9} = \tfrac{5}{18}$.
This is a nice illustration of how boundary conditions in bounded random walks can create sharp, non-intuitive partitions of the state space. In practice, similar parity arguments show up when analyzing lattice-based option models (binomial trees) where the reachable terminal nodes depend on the number of steps and the starting position. Recognizing parity early saves you from grinding through unnecessary algebra.