Competition · AMC preparation · step 4 of 4

AMC 10 · 2020A · #13

Grade 7 probability
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 14\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

Draw the grid and start point

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

Use the diagonal symmetry

Reflecting across y = x swaps vertical and horizontal exits, so every diagonal point is even: P_V(1,1) = P_V(2,2) = 12\frac{1}{2}.

P_V(x,x) = 1/2 → P_V(1,1) = P_V(2,2) = 1/2
3STEP 3

Use the anti-diagonal symmetry

Reflecting across x = 2 keeps vertical sides vertical, so the anti-diagonal is pinned too: P_V(1, 3) = 12\frac{1}{2}.

P_V(1, 3) = 1 - P_V(3, 1) = 1 - 1/2 = 1/2
4STEP 4

Split the first jump

Split the first jump into four 14\frac{1}{4} cases: P(1,2) = 14\frac{1}{4}[P(0,2) + P(2,2) + P(1,3) + P(1,1)].

P(1,2) = 1/4[P(0,2) + P(2,2) + P(1,3) + P(1,1)]
5STEP 5

Read off the four probabilities

Read off the four values: P(0,2) = 1 (already on x = 0), and P(2,2) = P(1,3) = P(1,1) = 12\frac{1}{2} (all on diagonals).

P(0,2) = 1, P(2,2) = P(1,3) = P(1,1) = 1/2
6STEP 6

Combine the four cases

Combine: P(1,2) = 14\frac{1}{4}(1 + 12\frac{1}{2} + 12\frac{1}{2} + 12\frac{1}{2}) = 14\frac{1}{4} · 52\frac{5}{2} = 58\frac{5}{8}.

P(1,2) = 1/4 · 5/2 = 5/8
7STEP 7

Match to the choices

Match 58\frac{5}{8} to the list: choice (B). The other options come from ignoring that the start sits closer to a vertical side.

5/8 → (B)
Answer
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 12\frac{1}{2} (vertical is favored) but not by a huge amount. The exact value 58\frac{5}{8} = 0.625 is just above 12\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 = 12\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 = 12\frac{1}{2} each). Average them: 14\frac{1}{4}(1 + 32\frac{3}{2}) = 58\frac{5}{8}, answer (B).

  • Draw the grid and start point
  • Use the diagonal symmetry
  • Use the anti-diagonal symmetry
  • Split the first jump
  • Read off the four probabilities
  • Combine the four cases
  • Match to the choices

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