AMC 10 · 2024 · #24
Grade 7 probabilitygeometry-3dcountingProblem
A bee is moving in three-dimensional space. A fair six-sided die with faces labeled A+,A−,B+,B−,C+, and C− is rolled. Suppose the bee occupies the point (a,b,c). If the die shows A+, then the bee moves to the point (a+1,b,c) and if the die shows A−, then the bee moves to the point (a−1,b,c). Analogous moves are made with the other four outcomes. Suppose the bee starts at the point (0,0,0) and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?
Pick an answer.
AMC 10 2024 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.
Toolkit + CCSS Solution
Understand
Restated: A bee starts at $(0,0,0)$ in 3D space. A fair $6$-sided die is rolled $4$ times; each face $A^\pm, B^\pm, C^\pm$ shifts the bee by $\pm 1$ 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?
Givens: Start at $P_0 = (0,0,0)$; Six equally likely moves: $\pm \hat{x}, \pm \hat{y}, \pm \hat{z}$; The die is rolled $4$ times, producing $4$ edges $P_0P_1, P_1P_2, P_2P_3, P_3P_4$; Success condition: the $4$ edges are pairwise distinct AND all five vertices $P_0, \dots, P_4$ lie on a common unit cube; Answer choices: (A) $\tfrac{1}{54}$, (B) $\tfrac{7}{54}$, (C) $\tfrac{1}{6}$, (D) $\tfrac{5}{18}$, (E) $\tfrac{2}{5}$
Unknowns: The probability of a successful $4$-roll sequence
Understand
Restated: A bee starts at $(0,0,0)$ in 3D space. A fair $6$-sided die is rolled $4$ times; each face $A^\pm, B^\pm, C^\pm$ shifts the bee by $\pm 1$ 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?
Givens: Start at $P_0 = (0,0,0)$; Six equally likely moves: $\pm \hat{x}, \pm \hat{y}, \pm \hat{z}$; The die is rolled $4$ times, producing $4$ edges $P_0P_1, P_1P_2, P_2P_3, P_3P_4$; Success condition: the $4$ edges are pairwise distinct AND all five vertices $P_0, \dots, P_4$ lie on a common unit cube; Answer choices: (A) $\tfrac{1}{54}$, (B) $\tfrac{7}{54}$, (C) $\tfrac{1}{6}$, (D) $\tfrac{5}{18}$, (E) $\tfrac{2}{5}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #1 Draw a Diagram
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 $M_1 = A^+$ and $M_2 = B^+$ 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 $P_2$ are on a common cube.
Execute — Answer: B
7.SP.C.8 Step 1 - Set up the denominator.
- Four independent die rolls give $6 \times 6 \times 6 \times 6$ equally likely move sequences.
- Every favorable count will be over this same denominator.
💡 Independent trials multiply — the Grade 7 "sample space of compound events" rule.
7.SP.C.7 Step 2 - Sub-count for Move 1.
- From $(0,0,0)$ every direction is symmetric, so all $6$ moves are valid starts.
- By the cube's symmetry, the number of good completions is the same for each.
- Fix $M_1 = A^+$ as a representative and multiply by $6$ at the end.
💡 Equally likely starts let us solve one easier representative case and scale — the Tool #9 "WLOG" reduction.
7.SP.C.8 Step 3 - Sub-count for Move 2.
- From $P_1 = (1,0,0)$ we may not repeat the same axis: $A^+$ would push to $(2,0,0)$ (no unit cube fits all three points), and $A^-$ would retrace the edge (not distinct).
- The two perpendicular axes give $4$ legal moves: $B^\pm, C^\pm$.
- Fix $M_2 = B^+$ as the representative and multiply by $4$ later.
💡 "Not the same axis" is the combined translation of both rules at Move 2: same-axis means either repeat ($\to$ collinear) or reverse ($\to$ duplicate edge).
5.G.A.1 Step 4 - Identify the two candidate cubes at $P_2$.
- With $P_0, P_1, P_2 = (0,0,0), (1,0,0), (1,1,0)$ already pinned, the only unit cubes containing all three vertices are the one with $z \in \{0, 1\}$ and the one with $z \in \{0, -1\}$.
- Drawing both cubes (they share the $xy$-face we just traced) makes the next move's options visible.
💡 A quick 3D sketch of the two stacked cubes makes the case-split for Move 3 obvious — Grade 5 coordinate-grid thinking lifted to 3D.
7.SP.C.8 Step 5 - Sub-count for Move 3 (split by cube residency).
- From $P_2 = (1,1,0)$ each of the $6$ moves leads to one of six points; check membership in $\text{Cube}_+$ or $\text{Cube}_-$ and the distinct-edge rule.
- Reverse $B^-$ is forbidden; $A^+$ to $(2,1,0)$ and $B^+$ to $(1,2,0)$ leave both cubes.
- The legal moves are $A^-$ to $(0,1,0)$ (vertex of both cubes — a planar choice), and $C^\pm$ to $(1,1,\pm 1)$ (each commits the path to exactly one cube — a vertical choice).
💡 Splitting Move 3 into "stay in the $xy$-face" and "jump to a stacked face" prepares the right cases for Move 4.
7.SP.C.8 Step 6 - Sub-count for Move 4 — planar branch.
- If $M_3 = A^-$, the path lives on the $z = 0$ face and either cube is still possible.
- From $P_3 = (0,1,0)$ exclude reverse $A^+$ and the two off-cube moves ($B^+$ to $(0,2,0)$ is off both cubes; $A^-$ already used as reverse).
- The survivors are $B^-$ to $(0,0,0)$, $C^+$ to $(0,1,1)$ in $\text{Cube}_+$, and $C^-$ to $(0,1,-1)$ in $\text{Cube}_-$.
💡 Closing the square ($B^-$ back to origin) is legal because it is a $4$th distinct edge of $\text{Cube}_+$ (and of $\text{Cube}_-$); the two vertical exits each commit to one of the cubes.
7.SP.C.7 Step 7 - Sub-count for Move 4 — vertical branch.
- If $M_3 = C^+$ then the path has committed to $\text{Cube}_+$ and $P_3 = (1,1,1)$.
- Exclude reverse $C^-$.
- In $\text{Cube}_+$ the edge-neighbors of $(1,1,1)$ are $(0,1,1)$ and $(1,0,1)$, reached by $A^-$ and $B^-$; the moves $A^+, B^+$ leave the cube.
- By the $z \leftrightarrow -z$ symmetry, $M_3 = C^-$ also gives exactly $2$ legal $M_4$.
💡 Once Move 3 commits the path to one specific cube, Move 4 is just "pick an unused edge from $(1,1,1)$ inside that cube".
5.OA.A.1 Step 8 - Combine the Move 3 + Move 4 counts under the fixed prefix $(M_1, M_2) = (A^+, B^+)$.
- The planar branch contributes $1 \cdot 3 = 3$ and the two vertical branches contribute $2 \cdot 2 = 4$, for $7$ completions.
- Then multiply by the $6 \cdot 4 = 24$ choices for $(M_1, M_2)$ that the WLOG reduction folded away.
💡 Multiplying the sub-counts is the standard "tree of choices" — Grade 5 order of operations on the count.
6.NS.B.4 Step 9 - Form the probability and reduce.
- Numerator $168$ and denominator $1296$ share the factor $24$.
💡 Pull the common factor out of numerator and denominator — the Grade 6 GCF move that lands the answer in lowest terms.
7.SP.C.8 Set up the denominator. Four independent die rolls give $6 \times 6 \times 6 \ti 7.SP.C.7 Sub-count for Move 1. From $(0,0,0)$ every direction is symmetric, so all $6$ mo 7.SP.C.8 Sub-count for Move 2. From $P_1 = (1,0,0)$ we may not repeat the same axis: $A^+ 5.G.A.1 Identify the two candidate cubes at $P_2$. With $P_0, P_1, P_2 = (0,0,0), (1,0,0 7.SP.C.8 Sub-count for Move 3 (split by cube residency). From $P_2 = (1,1,0)$ each of the 7.SP.C.8 Sub-count for Move 4 — planar branch. If $M_3 = A^-$, the path lives on the $z = 7.SP.C.7 Sub-count for Move 4 — vertical branch. If $M_3 = C^+$ then the path has committ 5.OA.A.1 Combine the Move 3 + Move 4 counts under the fixed prefix $(M_1, M_2) = (A^+, B^ 6.NS.B.4 Form the probability and reduce. Numerator $168$ and denominator $1296$ share th Review
Reasonableness: Decimal sanity: $7/54 \approx 0.130$, comfortably between $1/54 \approx 0.019$ and $1/6 \approx 0.167$. 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 $6 \cdot 5 \cdot 5 \cdot 5 = 750$ (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".
Alternative: Tool #2 (Make a Systematic List) on a smaller version: enumerate by computer (or by hand) all $6^4 = 1296$ sequences and filter. The 168 successes split by direction of $M_1$ into $6$ symmetric groups of $28$, by direction of $M_2$ into $24$ groups of $7$, and by the type of $M_3$ into $24$ planar completions (giving $3$ each) and $48$ vertical completions (giving $2$ each), recovering $168 = 6 \cdot 4 \cdot (3 + 4)$. The arithmetic matches the structural count and confirms (B).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Treating the four rolls as a compound event, building the move-by-move count $6 \cdot 4 \cdot 3 \cdot (\text{branch})$ as a tree of legal choices, and reading off the favorable count $168$ over the sample space $1296$.)7.SP.C.7Develop a uniform probability model by assigning equal probability to all outcomes (Justifying that all $1296$ four-roll sequences are equally likely and that, by cube symmetry, the favorable count for any choice of $(M_1, M_2)$ is the same, so we may fix one representative and scale.)5.G.A.1Use a pair of perpendicular number lines (axes) to define a coordinate system; represent points (Plotting $P_0, P_1, P_2$ in 3D coordinates and identifying the two candidate unit cubes that contain those three vertices — the geometric setup for the Move 3 case-split.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions, and evaluate them (Combining the sub-counts as $6 \cdot 4 \cdot (1 \cdot 3 + 2 \cdot 2) = 168$ with the correct order of operations.)6.NS.B.4Find the greatest common factor of two whole numbers (Reducing $\tfrac{168}{1296}$ by their common factor $24$ to land on the lowest-terms answer $\tfrac{7}{54}$.)
⭐ 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 $6 \cdot 4 \cdot 7 = 168$ favorable out of $1296$ total, which reduces to $\tfrac{7}{54}$.
⭐ 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 $6 \cdot 4 \cdot 7 = 168$ favorable out of $1296$ total, which reduces to $\tfrac{7}{54}$.
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