AMC 10 · 2024 · #24

Grade 7 probabilitygeometry-3dcounting

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

probability-basicspatial-visualizationsystematic-enumerationcombinations-basic identify-subproblemscaseworkeasier-related-problem ↑ Prerequisites: probability-basicspatial-visualization

📏 Long solution 💡 4 insights

Problem

A bee starts at (0,0,0) in 3D space. A fair 6-sided die is rolled 4 times; each face A^±, B^±, C^± shifts the bee by ± 1 along the matching axis. What is the probability that the four edges the bee traces are all distinct and all lie on the same unit cube?

Pick an answer.

(A)

$frac{1}{54}$

(B)

$frac{7}{54}$

(C)

$frac{1}{6}$

(D)

$frac{5}{18}$

(E)

$frac{2}{5}$

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The success condition stacks two rules on top of a 4-step sequence, so the natural attack is Tool #7 (Identify Subproblems): count the legal choices at each of the four moves separately, then multiply. Tool #9 (Solve an Easier Related Problem) reduces the work even further — the cube has full symmetry, so we may fix M₁ = A^+ and M₂ = B^+ without loss of generality and just multiply by the number of equivalent starts. Tool #1 (Draw a Diagram) keeps the two candidate cubes (one above, one below the xy-plane after the first two moves) straight in our head when we check which neighbors of P₂ are on a common cube.

1STEP 1

Four independent rolls give 6⁴ = 1296 equally likely sequences — this is the denominator.

|Ω| = 6⁴ = 1296

2STEP 2

From (0,0,0) every direction is symmetric, so all 6 moves are valid starts — fix M₁ = A^+ and scale by 6 later.

#{M₁} = 6, P₁ = (1,0,0)

3STEP 3

From (1,0,0) the same axis is banned (retrace or collinear), so only the 4 perpendicular moves B^±, C^± are legal — fix M₂ = B^+.

#{M₂} = 4, P₂ = (1,1,0)

4STEP 4

With P₀, P₁, P₂ pinned, exactly 2 unit cubes hold all three vertices — one stacked above z = 0, one below.

Cube_+ = {0,1}³, Cube_- = {0,1} × {0,1} × {0,-1}

5STEP 5

From (1,1,0) three moves stay on a cube: A^- is planar (1), C^± are vertical (2), each committing to one cube.

#{M₃} = 3 = 1_planar (A^-) + 2_vertical (C^+, C^-)

6STEP 6

Planar branch: from P₃ = (0,1,0) three moves stay on a cube — B^- closes the square, C^± exit up or down.

#{M₄ ∣ M₃ = A^-} = 3

7STEP 7

Vertical branch: once M₃ = C^± commits to one cube, exactly 2 unused edges of (1,1,1) remain — A^- and B^-.

#{M₄ ∣ M₃ = C^+} = 2, #{M₄ ∣ M₃ = C^-} = 2

8STEP 8

Per fixed (M₁, M₂): 1·3 + 2·2 = 7 completions; times the 6·4 folded-away starts gives 168 favorable.

#{favorable} = 6 · 4 · (1 · 3 + 2 · 2) = 24 · 7 = 168

9STEP 9

Reduce $\frac{168}{1296}$ by their common factor 24 to get $\frac{7}{54}$ → (B).

P =

\frac{168}{1296}

\frac{168 ÷ 24}{1296 ÷ 24}

\frac{7}{54}

→ (B)

Answer

\frac{7}{54}

Decimal sanity:

\frac{7}{54}

≈ 0.130, comfortably between

\frac{1}{54}

≈ 0.019 and

\frac{1}{6}

≈ 0.167. The success condition is fairly tight — about 1 in 8 four-roll sequences — which fits the kind of restriction the problem imposes (no axis repeat at Move 2, no reverse anywhere, plus a cube-membership condition at Moves 3 and 4). Cross-check the counts directly: ignoring the cube rule, distinct-edge sequences number 6 · 5 · 5 · 5 = 750 (Move 1 free, then never repeat the previous edge's reverse), and our favorable count 168 is well under that — exactly as expected because the cube-membership rule is strictly stronger than "distinct edges".

💡Key takeaway

When a path problem stacks several rules, count the legal choices one move at a time and use the shape's symmetry to fix the first moves "without loss of generality". Here the count is 6 · 4 · 7 = 168 favorable out of 1296 total, which reduces to $\frac{7}{54}$ .

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