2020 AMC 10A Problem 13

Let $p(x, y)$ be the probability of ending on a vertical side ($x = 0$ or $x = 4$) from the interior point $(x, y)$. By the square's symmetries, $p(2,2) = \frac12$ (vertical and horizontal look the same from the center), $p(3, y) = p(1, y)$, and $p(x, 3) = p(x, 1)$.

Set $a = p(1,2)$, $c = p(1,1)$, $d = p(2,1)$. Conditioning on the first jump:

- From $(1,2)$: left hits $x = 0$ (win), right goes to $(2,2)$, up or down goes to a point with value $c$:
$$a = \tfrac14 \cdot 1 + \tfrac14 \cdot \tfrac12 + \tfrac12 c = \tfrac38 + \tfrac12 c.$$
- From $(1,1)$: left wins, down hits $y = 0$ (lose), up goes to $(1,2)$, right to $(2,1)$:
$$c = \tfrac14 + \tfrac14 a + \tfrac14 d.$$
- From $(2,1)$: down loses, up goes to $(2,2)$, left or right goes to a point with value $c$:
$$d = \tfrac14 \cdot \tfrac12 + \tfrac12 c = \tfrac18 + \tfrac12 c.$$

Substitute $d$ into the second equation: $c = \tfrac14 + \tfrac14 a + \tfrac{1}{32} + \tfrac18 c$, so $\tfrac78 c = \tfrac{9}{32} + \tfrac14 a$, i.e. $c = \tfrac{9}{28} + \tfrac{2}{7} a$.

Then $a = \tfrac38 + \tfrac12\left(\tfrac{9}{28} + \tfrac27 a\right) = \tfrac38 + \tfrac{9}{56} + \tfrac17 a$, so $\tfrac67 a = \tfrac{21 + 9}{56} = \tfrac{30}{56}$ and $a = \tfrac{30}{56} \cdot \tfrac76 = \tfrac58$.

The answer is $\boxed{\textbf{(B)}\ \frac{5}{8}}$.

A random walk that stops at a boundary is a state machine: the probability from each interior point is the average of the probabilities of its four neighbors, with boundary points fixed at $1$ or $0$. Symmetry is what makes it tractable: nine unknowns become three, and the center's value is free.

• Assuming the frog is equally likely to hit either kind of side because the start is one step from the left edge and two from the top and bottom, or setting up states without using symmetry and drowning in equations.
• Forgetting that the frog can wander back toward the start; a one- or two-step tree analysis undercounts and gives a value below the true $\frac58$.