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 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 would place her in the top row middle square, and the second hop would cause Frieda 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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Frieda the frog starts at the center of a $3 \times 3$ grid. Each hop moves her one square up, down, left, or right with probability $\tfrac{1}{4}$ 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.

Givens: Grid is $3 \times 3$ with 1 center, 4 edge midpoints, 4 corners; Each hop direction is up/down/left/right with probability $\tfrac{1}{4}$; Out-of-grid hops wrap to the opposite edge; Frieda starts at the center and stops upon landing on a corner; At most $4$ hops; Answer choices: (A) $\tfrac{9}{16}$, (B) $\tfrac{5}{8}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{25}{32}$, (E) $\tfrac{13}{16}$

Unknowns: $P(\text{Frieda lands on a corner within 4 hops})$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #5 Look for a Pattern, #3 Eliminate Possibilities

Tool #1 (Diagram) — draw the $3 \times 3$ 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 $32$.

Execute — Answer: D

#9 Solve an Easier Related Problem 7.SP.C.7 Step 1

Identify the three cell types using grid symmetry.
Center C (1 cell), Edge E (4 midpoints), Corner V (4).
From C, all four moves go to an E (the four E-cells are exactly the C's neighbors): $P(C \to E) = 1$.
From an E (say the right-middle cell): up goes to a corner V, down goes to a corner V, left goes to C, right wraps around to the left-middle E.
So $P(E \to V) = \tfrac{2}{4} = \tfrac{1}{2}, P(E \to C) = \tfrac{1}{4}, P(E \to E) = \tfrac{1}{4}$.

$$P(C \to E)=1; \;P(E \to V)=\tfrac{1}{2}, P(E \to C)=\tfrac{1}{4}, P(E \to E)=\tfrac{1}{4}$$

💡 Reduce from 9 cells to 3 types — each type behaves the same by symmetry.

#7 Identify Subproblems 7.SP.C.8 Step 2

Track two quantities: $C_n$ = probability that after $n$ hops Frieda has NOT stopped and is at the center; $E_n$ = same but at an edge.
Initial: Frieda starts at the center pre-hop, so $C_0 = 1, E_0 = 0$.
The probability of stopping for the first time on hop $n$ is $p_n = E_{n-1} \cdot P(E \to V) = \tfrac{1}{2} E_{n-1}$ (only E-cells can hop into a corner; from C she cannot stop).
The total probability of stopping within $4$ hops is $p_1 + p_2 + p_3 + p_4$.

$$p_n = \tfrac{1}{2} E_{n-1}, \quad \text{total} = p_1 + p_2 + p_3 + p_4$$

💡 Sum of disjoint "first stop at hop $n$" probabilities = total stopping probability.

#5 Look for a Pattern 7.SP.C.5 Step 3

Hop 1.
Start: $C_0 = 1, E_0 = 0$.
So $p_1 = \tfrac{1}{2} E_0 = 0$.
After hop 1 (not stopped): $C_1 = P(E \to C) E_0 = 0; E_1 = P(C \to E) C_0 + P(E \to E) E_0 = 1 \cdot 1 + 0 = 1$.
So after hop 1 she is certainly at an E cell, not stopped.

$$p_1 = 0, \;C_1 = 0, \;E_1 = 1$$

💡 Center hop goes to E with certainty — no corner reachable on hop 1.

#5 Look for a Pattern 7.SP.C.7 Step 4

Hop 2.
$p_2 = \tfrac{1}{2} E_1 = \tfrac{1}{2}$.
Survivors (not stopped) update as $C_2 = \tfrac{1}{4} E_1 = \tfrac{1}{4}; E_2 = 1 \cdot C_1 + \tfrac{1}{4} E_1 = 0 + \tfrac{1}{4} = \tfrac{1}{4}$.

$$p_2 = \tfrac{1}{2}, \;C_2 = \tfrac{1}{4}, \;E_2 = \tfrac{1}{4}$$

💡 From E with probability $1$, half the time we land on V (stop), the rest splits evenly into C and E.

#5 Look for a Pattern 7.SP.C.7 Step 5

Hop 3.
$p_3 = \tfrac{1}{2} E_2 = \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{8}$.
Update: $C_3 = \tfrac{1}{4} E_2 = \tfrac{1}{4} \cdot \tfrac{1}{4} = \tfrac{1}{16}; E_3 = 1 \cdot C_2 + \tfrac{1}{4} E_2 = \tfrac{1}{4} + \tfrac{1}{16} = \tfrac{5}{16}$.

$$p_3 = \tfrac{1}{8}, \;C_3 = \tfrac{1}{16}, \;E_3 = \tfrac{5}{16}$$

💡 Same recursion — only E-cells can stop next hop, so we track $E_n$ carefully.

#5 Look for a Pattern 7.SP.C.7 Step 6

Hop 4.
$p_4 = \tfrac{1}{2} E_3 = \tfrac{1}{2} \cdot \tfrac{5}{16} = \tfrac{5}{32}$.

$$p_4 = \tfrac{5}{32}$$

💡 Only need the $E_3$ value; $C_4, E_4$ don't matter past hop $4$.

#3 Eliminate Possibilities 5.NF.A.1 Step 7

Sum the disjoint events.
Total $= p_1 + p_2 + p_3 + p_4 = 0 + \tfrac{1}{2} + \tfrac{1}{8} + \tfrac{5}{32}$.
Common denominator $32$: $0 + \tfrac{16}{32} + \tfrac{4}{32} + \tfrac{5}{32} = \tfrac{25}{32}$.

$$0 + \tfrac{16}{32} + \tfrac{4}{32} + \tfrac{5}{32} = \tfrac{25}{32}$$

💡 Add disjoint stopping events with a common denominator — the answer drops out directly.

[1] #9 7.SP.C.7 Identify the three cell types using grid symmetry. Center C (1 cell), Edge E (4

[2] #7 7.SP.C.8 Track two quantities: $C_n$ = probability that after $n$ hops Frieda has NOT sto

[3] #5 7.SP.C.5 Hop 1. Start: $C_0 = 1, E_0 = 0$. So $p_1 = \tfrac{1}{2} E_0 = 0$. After hop 1 (

[4] #5 7.SP.C.7 Hop 2. $p_2 = \tfrac{1}{2} E_1 = \tfrac{1}{2}$. Survivors (not stopped) update a

[5] #5 7.SP.C.7 Hop 3. $p_3 = \tfrac{1}{2} E_2 = \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{8}$

[6] #5 7.SP.C.7 Hop 4. $p_4 = \tfrac{1}{2} E_3 = \tfrac{1}{2} \cdot \tfrac{5}{16} = \tfrac{5}{32

[7] #3 5.NF.A.1 Sum the disjoint events. Total $= p_1 + p_2 + p_3 + p_4 = 0 + \tfrac{1}{2} + \tf

Review

Reasonableness: Bounds check. $P(\text{stop within 4 hops}) \ge p_2 = \tfrac{1}{2}$ since hop 2 alone is at least $\tfrac{1}{2}$. And $\le 1$ obviously. Our $\tfrac{25}{32} \approx 0.78$ sits between, matching the intuition that the corners are reasonably reachable but not guaranteed. Another check: $p_n$ should be decreasing — we got $0, \tfrac{1}{2}, \tfrac{1}{8}, \tfrac{5}{32}$. Wait, $\tfrac{5}{32} > \tfrac{4}{32} = \tfrac{1}{8}$ — actually $p_4 > p_3$! That's because $E_3 = \tfrac{5}{16} > E_2 = \tfrac{1}{4} = \tfrac{4}{16}$: the survivor mass at E grew between hops 2 and 3 (since some C-mass flowed back to E). All consistent. Choice (D) $\tfrac{25}{32}$ matches.

Alternative: Tool #16 (Complement): compute $P(\text{NOT stopped in 4 hops}) = C_4 + E_4 = \tfrac{1}{4} E_3 + (C_3 + \tfrac{1}{4} E_3) = \tfrac{1}{4} \cdot \tfrac{5}{16} + (\tfrac{1}{16} + \tfrac{5}{64}) = \tfrac{5}{64} + \tfrac{4}{64} + \tfrac{5}{64} = \tfrac{14}{64} = \tfrac{7}{32}$; then $1 - \tfrac{7}{32} = \tfrac{25}{32}$. Same answer.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators (Combining $\tfrac{1}{2} + \tfrac{1}{8} + \tfrac{5}{32}$ via common denominator $32$.)
7.SP.C.5 Understand that the probability of a chance event is a number between 0 and 1 (Each move probability is $\tfrac{1}{4}$, transition probabilities $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{4}$ all in $[0, 1]$.)
7.SP.C.7 Develop a probability model and use it to find probabilities of events (Building the 3-state (C, E, V) model with explicit transition probabilities by grid symmetry.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Computing the compound event "stop within 4 hops" as $p_1 + p_2 + p_3 + p_4$ (disjoint first-stop hops).)

⭐ This AMC 10 problem only needs Grade 7 probability you already know — collapse the $3 \times 3$ grid to three cell types (Center, Edge, Corner) by symmetry, work out the transition probabilities $P(E \to V) = \tfrac{1}{2}, P(E \to C) = P(E \to E) = \tfrac{1}{4}$, and add the four hop-by-hop stopping probabilities $0 + \tfrac{1}{2} + \tfrac{1}{8} + \tfrac{5}{32} = \tfrac{25}{32}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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