AMC 10 · 2021 · #23

Grade 7 geometry-2d

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

Frieda the frog starts at the center of a 3 × 3 grid. Each hop moves her one square up, down, left, or right with probability

\frac{1}{4}

each; if she would step off the grid, she wraps around to the opposite edge. She stops the instant she lands on a corner. Find the probability that within at most 4 hops she lands on a corner.

Pick an answer.

(A)

$~\frac{9}{16}$

(B)

$~\frac{5}{8}$

(C)

$~\frac{3}{4}$

(D)

$~\frac{25}{32}$

(E)

$~\frac{13}{16}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

By symmetry, 3 cell types — Center C, Edge E, Corner V. C always hops to E; from an E, P(E → V) = $\frac{1}{2}$ , P(E → C) = P(E → E) = $\frac{1}{4}$ .

P(C → E)=1; P(E → V)=

\frac{1}{2}

, P(E → C)=

\frac{1}{4}

, P(E → E)=

\frac{1}{4}

2STEP 2

Track C_n, E_n = chance she's un-stopped at center/edge after n hops. First-stop chance p_n = $\frac{1}{2}$ E_n-1, and the total is p₁+p₂+p₃+p₄.

p_n =

\frac{1}{2}

E_n-1, total = p₁ + p₂ + p₃ + p₄

3STEP 3

Hop 1: from C she must land on an E, so p₁ = 0 and after one hop E₁ = 1 (C₁ = 0) — a corner is impossible on hop 1.

p₁ = 0, C₁ = 0, E₁ = 1

4STEP 4

Hop 2: p₂ = $\frac{1}{2}$ E₁ = $\frac{1}{2}$ . Survivors split evenly — C₂ = $\frac{1}{4}$ and E₂ = $\frac{1}{4}$ .

p₂ =

\frac{1}{2}

, C₂ =

\frac{1}{4}

, E₂ =

\frac{1}{4}

5STEP 5

Hop 3: p₃ = $\frac{1}{2}$ E₂ = $\frac{1}{8}$ . Survivors: C₃ = $\frac{1}{16}$ and E₃ = $\frac{5}{16}$ (center mass flowed back into E).

p₃ =

\frac{1}{8}

, C₃ =

\frac{1}{16}

, E₃ =

\frac{5}{16}

6STEP 6

Hop 4: p₄ = $\frac{1}{2}$ E₃ = $\frac{5}{32}$ — only E₃ matters, since we stop counting after hop 4.

p₄ =

\frac{5}{32}

7STEP 7

Sum the disjoint stops: p₁ + p₂ + p₃ + p₄ = 0 + $\frac{16}{32}$ + $\frac{4}{32}$ + $\frac{5}{32}$ = $\frac{25}{32}$ .

0 +

\frac{16}{32}

\frac{4}{32}

\frac{5}{32}

\frac{25}{32}

Answer

\frac{25}{32}

Bounds check. P(stop within 4 hops) ≥ p₂ =

\frac{1}{2}

since hop 2 alone is at least

\frac{1}{2}

. And ≤ 1 obviously. Our

\frac{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,

\frac{1}{2}

\frac{1}{8}

\frac{5}{32}

. Wait,

\frac{5}{32}

＞

\frac{4}{32}

\frac{1}{8}

— actually p₄ ＞ p₃! That's because E₃ =

\frac{5}{16}

＞ E₂ =

\frac{1}{4}

\frac{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)

\frac{25}{32}

matches.

💡Key takeaway

This AMC 10 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) = $\frac{1}{2}$ , P(E → C) = P(E → E) = $\frac{1}{4}$ , and add the four hop-by-hop stopping probabilities 0 + $\frac{1}{2}$ + $\frac{1}{8}$ + $\frac{5}{32}$ = $\frac{25}{32}$ .

More like this