棋子随机起点向上走：到达左右上角概率

对 9 个起点分别计算到达某个上角的概率并取平均，可得总体概率为

$$\frac{1}{4}.$$

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Original Explanation

Let's look at the probability of reaching a corner from each of the center $3 \times 3$ grid. First, we have equal chance of selecting each of the $9$ inner squares. From here, we can look at the probability of reaching a corner from each square. Right away we can recognize $4$ of the $9$ spots can't reach a corner, as our movements are (up+left) or (up+right); our moves are limited by the vertical distance to a corner. With the number of moves we need the current distance from either corner minus the distance they move (left = $-1$) and (right = $+1$) to equal to $0$.

First, let's look at the top-left square in the inner $3 \times 3$ grid. We see that we have probability $\frac{1}{2}$ of immediately reaching the left corner and probability $\frac{1}{2}$ of reaching the top-center, in which case there's no chance of reaching a corner. Hence, we have overall probability $\frac{1}{2}$. By symmetry, the top-right square has the same probability.

Next up the bottom left and bottom right spots in the $3 \times 3$ grid have the same probability: the bottom-left has $3$ movements of (up+right) and by symmetry the bottom right has $3$ movements of (up+left). Lastly, looking at the center position we can see it has $2$ paths to a corner with $2$ moves of (up+right) or $2$ moves of (up+left).

This gives us: $$\frac{1}{9} \left(2\cdot\frac{1}{2} + 2\cdot\left(\frac{1}{2}\right)^2 + 2\cdot\left(\frac{1}{2}\right)^3\right) = \frac{1}{4}$$