Question

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1,$ and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0, 0),$ $(0, 4),$ $(4, 4),$ and $(4, 0).$ What is the probability that the sequence of jumps ends on a vertical side of the square?

$\displaystyle \frac{1}{2}$

$\displaystyle \frac{5}{8}$

$\displaystyle \frac{2}{3}$

$\displaystyle \frac{3}{4}$

$\displaystyle \frac{7}{8}$

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Written solution:

Let $p(x,y)$ be the probability of eventually hitting a vertical side first from point $(x,y)$ . By symmetry, set $a=p(1,1)=p(1,3)$ , $b=p(2,1)=p(2,3)$ , $c=p(1,2)$ , and $d=p(2,2)$ .

The averaging equations are $a=\dfrac{1+b+c}{4}$ , $b=\dfrac{2a+d}{4}$ , $c=\dfrac{1+2a+d}{4}$ , and $d=\dfrac{b+c}{2}$ . Solving gives $c=\dfrac58$ , which is the desired probability from $(1,2)$ . Thus, B is the correct answer.

2020 AMC 10A Problem 13

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Problem 13 in Other Years