AMC 10 · 2020 · #13

Grade 7 geometry-2d

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsymmetry-argumentcoordinate-geometryrecursive-sequencesystems-of-equations symmetry-argumentcaseworkidentify-subproblems ↑ Prerequisites: probability-basiccoordinate-geometry

📏 Long solution 💡 3 insights

Problem

A frog starts at (1, 2) inside the square with vertices (0,0), (4,0), (4,4), (0,4). At each step it jumps one unit up, down, left, or right with equal probability

\frac{1}{4}

, and it stops the first time it lands on a side of the square. What is the probability that the stopping side is a vertical side (x=0 or x=4)?

Pick an answer.

(A)

$\frac12$

(B)

$\frac 58$

(C)

$\frac 23$

(D)

$\frac34$

(E)

$\frac 78$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch the 5 × 5 grid of lattice points and mark the start (1, 2) and the four boundary sides. Tool #15 (Reorganize): exploit the square's reflection symmetries — swapping x ⇔ y turns vertical sides into horizontal ones, so P_V(x,y) = 1 - P_V(y,x). Tool #7 (Subproblems): one jump from (1,2) goes to one of (0,2), (2,2), (1,1), (1,3); compute the four sub-probabilities, then combine with the law of total probability. Tool #3 (Eliminate): match the final fraction to the five choices.

1STEP 1

Sketch the lattice grid, mark the start at (1, 2), and label the vertical sides x = 0, x = 4.

start = (1, 2), vertical sides: x = 0, x = 4

2STEP 2

Reflecting across y = x swaps vertical and horizontal exits, so every diagonal point is even: P_V(1,1) = P_V(2,2) = $\frac{1}{2}$ .

P_V(x,x) =

\frac{1}{2}

→ P_V(1,1) = P_V(2,2) =

\frac{1}{2}

3STEP 3

Reflecting across x = 2 keeps vertical sides vertical, so the anti-diagonal is pinned too: P_V(1, 3) = $\frac{1}{2}$ .

P_V(1, 3) = 1 - P_V(3, 1) = 1 -

\frac{1}{2}

\frac{1}{2}

4STEP 4

Split the first jump into four $\frac{1}{4}$ cases: P(1,2) = $\frac{1}{4}$ [P(0,2) + P(2,2) + P(1,3) + P(1,1)].

P(1,2) =

\frac{1}{4}

[P(0,2) + P(2,2) + P(1,3) + P(1,1)]

5STEP 5

Read off the four values: P(0,2) = 1 (already on x = 0), and P(2,2) = P(1,3) = P(1,1) = $\frac{1}{2}$ (all on diagonals).

P(0,2) = 1, P(2,2) = P(1,3) = P(1,1) =

\frac{1}{2}

6STEP 6

Combine: P(1,2) = $\frac{1}{4}$ (1 + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ ) = $\frac{1}{4}$ · $\frac{5}{2}$ = $\frac{5}{8}$ .

P(1,2) =

\frac{1}{4}

\frac{5}{2}

\frac{5}{8}

7STEP 7

Match $\frac{5}{8}$ to the list: choice (B). The other options come from ignoring that the start sits closer to a vertical side.

\frac{5}{8}

→ (B)

Answer

\frac{5}{8}

The starting point (1, 2) is exactly 1 step from the nearest vertical side and 2 steps from the nearest horizontal side, so the answer should beat

\frac{1}{2}

(vertical is favored) but not by a huge amount. The exact value

\frac{5}{8}

= 0.625 is just above

\frac{1}{2}

— a moderate vertical advantage, exactly what closeness to one vertical side suggests.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability you already know! Reflecting the square across the line y = x swaps vertical and horizontal sides, which forces P = $\frac{1}{2}$ at every diagonal point like (1, 1), (2, 2), (1, 3). One step from (1, 2) gives four equally likely cases: (0, 2) wins immediately (P = 1), and the other three land on diagonals (P = $\frac{1}{2}$ each). Average them: $\frac{1}{4}$ (1 + $\frac{3}{2}$ ) = $\frac{5}{8}$ , answer (B).

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