AMC 10 · 2021 · #23

Grade 11 probability

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

geometric-series-infinitesequences-arithmeticpermutations-basicindependent-events convert-to-algebraidentify-subproblemspattern-recognition ↑ Prerequisites: geometric-series-infiniteindependent-events

📏 Long solution 💡 4 insights

Problem

Three balls are tossed one at a time into bins numbered 1, 2, 3, and so on, independently. Each toss lands in bin i with probability two to the minus i. Find the probability that the three land in three distinct bins that are evenly spaced — once sorted, the two gaps are equal — then report the sum of the numerator and denominator in lowest terms.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Tool #13 (Convert to Algebra) first: "evenly spaced" is ordinary English for "the sorted bins form an arithmetic progression". Tool #4 (Introduce a Variable) is the load-bearing move: name the smallest bin a and the common gap d. Every favourable bin-set is then {a, a+d, a+2d} for exactly one pair (a, d) with a ≥ 1 and d ≥ 1, so the infinitely many favourable outcomes get indexed cleanly with no double counting and no side conditions. Tool #7 (Identify Subproblems) splits the work into two independent pieces — the probability of one fixed arrangement, and the number of arrangements. Tool #2 (Systematic List) supplies that count: 3! ways to hand three distinct bins to three separate balls. Tool #15 (Organize Information in More Ways) finishes it: the double sum's term is a product of an a-part and a d-part, so the double sum factors into two ordinary geometric series.

1STEP 1

Translate the condition

Sorted, the two gaps are equal.

{3, 17, 10} ⟶ 3, 10, 17 with 10 - 3 = 17 - 10 = 7

2STEP 2

Name the start and the gap

Describe the three bins with two letters.

bins = {a, a+d, a+2d}, a ≥ 1, d ≥ 1

3STEP 3

Probability of one landing

Multiply the three probabilities.

2^-a · 2^-(a+d) · 2^-(a+2d) = 2^-(3a+3d) = 1/8^a 8^d

4STEP 4

Count the ball arrangements

The balls are distinguishable, so multiply by the orderings.

P(a, d) = 3! · 2^-(3a+3d) = 6/8^a 8^d

5STEP 5

Split the double sum

It becomes a product of two geometric series.

Σ_a ≥ 1Σ_d ≥ 16/8^a 8^d = 6(Σ_a ≥ 11/8^a) (Σ_d ≥ 11/8^d) = 6·1/7·1/7

6STEP 6

Reduce, then add

Adding gives 55.

p/q = 6/49, gcd(6, 49) = 1, p + q = 6 + 49 = 55 → (A)

Answer

Size check first. 6/49 ≈ 0.122 — small, but not vanishingly small, which fits the model: the balls land in low-numbered bins most of the time, and low bins are exactly where the short arithmetic triples live. As a floor, the single triple {1, 2, 3} alone contributes 6 · 2⁻¹2⁻²2⁻³ = 6/64 ≈ 0.094, already most of the total, with all remaining triples adding a modest amount on top. That is the right shape for the answer. Second, an independent parametrization reproduces the same number: condition on the middle bin m. The smallest bin can be any of b = 1, 2, …, m-1, and then the largest is forced to 2m - b, so every such triple has probability 6 · 2^-b 2^-m 2^-(2m-b) = 6 · 2⁻³m — again independent of b. Summing gives 6Σ_m ≥ 2(m-1) 2⁻³m = 6 · 1/49 = 6/49, matching. Finally, the answer choices 55 through 59 are consecutive, so no digit-level slip would land on a neighbouring choice by accident; the reduction gcd(6, 49) = 1 is what fixes p + q = 55 rather than an unreduced larger sum.

💡Key takeaway

"Evenly spaced" is an arithmetic sequence in disguise: name the smallest bin a and the gap d, and every winning triple is a, a+d, a+2d with probability 6 · 1/8^a 8^d — a double sum that factors into two copies of the same geometric series, each equal to 1/7. That gives 6/49, so p + q = 55, choice (A).

Translate "evenly spaced"
Name the start and the gap
Probability of one fixed landing
Count the ball arrangements
Split the double sum
Reduce, then add