2021 AMC 12A Problem 23

Problem 23 of 25HarderCounting & Probability

Frieda the frog begins a sequence of hops on a

3\times 3

grid of squares, moving one square on each hop and choosing at random the direction of each hop: up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she “wraps around” and jumps to the opposite edge. For example, if Frieda begins in the center square and makes two hops “up,” the first hop places her in the top row middle square, and the second hop causes her to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

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Solution

Classify squares as center

C,

edge-middle

E,

or corner (absorbing). From

C,

every hop lands on an

E

square. From an

E

square, two of the four neighbors are corners, one is the center, and one is another

E

square, so

P(\text{corner}) = \tfrac12,

P(\text{center}) = \tfrac14,

P(E) = \tfrac14.

Let

a_n

be the probability of reaching a corner within

n

hops starting from an edge square, and let

c_n

be the corresponding probability from the center. Set

a_0=c_0=0.

Then

c_n=a_{n-1}

and

a_n=\tfrac12+\tfrac14c_{n-1}+\tfrac14a_{n-1}.

Thus

a_1=\tfrac12,

a_2=\tfrac12+\tfrac14\cdot\tfrac12=\tfrac58,

and

a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}.

Starting from the center with four hops available, the probability equals

a_3 = \tfrac{25}{32}

(the first hop reaches an edge, leaving three hops). Thus, the correct answer is D.

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