AMC 10 · 2021 · #23

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsymmetry-argumentrecursive-sequencecomplementary-counting identify-subproblemscasework ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A frog starts at the centre of a three by three grid. Each hop moves it one square up, down, left, or right with probability one quarter each; stepping off the grid wraps around to the opposite edge. It stops the instant it lands on a corner. Find the probability that within at most four hops it lands on a corner.

Pick an answer.

(A)

$\frac{9}{16}$

(B)

$\frac{5}{8}$

(C)

$\frac{3}{4}$

(D)

$\frac{25}{32}$

(E)

$\frac{13}{16}$

How to solve

Strategy Solve an Easier Related Problem

Tool #1 (Diagram) — draw the 3 × 3 grid, color-code C/E/V, and sketch the wrap-around to understand each move type. Tool #9 (Easier Problem) — reduce from "track 9 positions" to "track 3 cell types" via the grid's symmetry; this collapses the state space dramatically. Tool #7 (Subproblems) — compute hop-by-hop: the probability of (not stopped yet, currently at C) and (not stopped yet, currently at E), since V means stopped. Tool #5 (Pattern) — the recursion is short (4 hops), so iterating gives the answer directly. Tool #3 (Eliminate) — answer choices all have denominator dividing 32, so we expect a denominator of 32.

1STEP 1

Sort the squares

Sort the squares into three kinds.

P(C → E)=1; P(E → V)=1/2, P(E → C)=1/4, P(E → E)=1/4

2STEP 2

Write the transition rules

The centre cannot reach a corner directly.

p_n = 1/2 E_n-1, total = p₁ + p₂ + p₃ + p₄

3STEP 3

After the first hop

After one hop it sits on an edge square.

p₁ = 0, C₁ = 0, E₁ = 1

4STEP 4

After the second hop

Half the chance lands on the second hop.

p₂ = 1/2, C₂ = 1/4, E₂ = 1/4

5STEP 5

After the third hop

Update the surviving cases.

p₃ = 1/8, C₃ = 1/16, E₃ = 5/16

6STEP 6

After the fourth hop

Compute the last contribution.

p₄ = 5/32

7STEP 7

Add them up

Adding gives twenty-five thirty-seconds.

0 + 16/32 + 4/32 + 5/32 = 25/32

Answer

25/32

Bounds check. P(stop within 4 hops) ≥ p₂ = 1/2 since hop 2 alone is at least 1/2. And ≤ 1 obviously. Our 25/32 ≈ 0.78 sits between, matching the intuition that the corners are reasonably reachable but not guaranteed. Another check: p_n should be decreasing — we got 0, 1/2, 1/8, 5/32. Wait, 5/32 ＞ 4/32 = 1/8 — actually p₄ ＞ p₃! That's because E₃ = 5/16 ＞ E₂ = 1/4 = 4/16: the survivor mass at E grew between hops 2 and 3 (since some C-mass flowed back to E). All consistent. Choice (D) 25/32 matches.

💡Key takeaway

This AMC 12 problem only needs Grade 7 probability you already know — collapse the 3 × 3 grid to three cell types (Center, Edge, Corner) by symmetry, work out the transition probabilities P(E → V) = 1/2, P(E → C) = P(E → E) = 1/4, and add the four hop-by-hop stopping probabilities 0 + 1/2 + 1/8 + 5/32 = 25/32.