AMC 10 · 2006 · #25

Grade 7 probabilitygeometry-3d

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

probability-basicsystematic-enumerationspatial-visualization casework ↑ Prerequisites: probability-basic

📏 Long solution 💡 4 insights

Problem

A bug sits on a corner of a cube. It makes 7 moves; each move it slides along one of the 3 edges leaving its current corner, picking each edge with equal chance and independently of past moves. Find the probability that these 7 moves carry the bug through all 8 corners, hitting each corner exactly once.

Pick an answer.

(A)

$\frac{1}{2187}$

(B)

$\frac{1}{729}$

(C)

$\frac{2}{243}$

(D)

$\frac{1}{81}$

(E)

$\frac{5}{243}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Because every 7-move sequence is equally likely, the probability is just (number of successful walks) divided by (number of all walks) — that is the subproblem split (Tool #7). Counting all walks is easy: 3 choices per move. The hard half is counting the successful walks, which are exactly the routes that touch all 8 corners without repeating one. To count those cleanly you must name the corners, so draw and label the cube (Tool #1) and hold its shape in mind (Tool #17). Then walk through the possibilities in an orderly way (Tool #2, the 'how many ways' tool): fix the first move by symmetry, and follow the forced branches until every good route is listed. A systematic list is the right instrument because the successful routes are few and highly constrained — most partial walks paint themselves into a corner.

1STEP 1

Turn probability into a count of routes

Every 7-move route is equally likely, so the probability is just good routes over all routes — two counting jobs.

P(success) = (#{good routes})/(#{all routes})

2STEP 2

Count all possible routes

Each of the 7 moves has 3 independent choices, so multiplying gives 3⁷ = 2187 possible routes — the denominator.

#{all routes} = 3⁷ = 2187

3STEP 3

Label the cube and pin down the first move

Label bottom A, B, C, D and top E, F, G, H above them; start at A. The 3 first moves are symmetric, so count A→ B routes and triple.

A → {B, D, E}, all 3 symmetric → count A→ B routes, then × 3

4STEP 4

List every good route after A to B

After A→ B→ C the never-repeat rule forces the rest: exactly 3 good routes. A→ B→ F mirrors it, so A→ B starts 6 good routes.

A→ B: (B→ C) 3 + (B→ F) 3 = 6 good routes

5STEP 5

Total the good routes and form the probability

The 3 symmetric first moves give 3 × 6 = 18 good routes, so the probability is 18/2187 = 2/243 — choice (C).

P = (3 × 6)/3⁷ = 18/2187 = 2/243 → (C)

Answer

2/243

The count 18 is believable: a successful route is a Hamiltonian path of the cube from a fixed start, and the cube is symmetric enough that these come in a small, tidy number. A quick sanity check on the size — 18/2187 is a bit under 1%, which fits: touching all 8 corners in exactly 7 non-repeating moves is a demanding, rare event, so a probability near a hundredth is sensible. The value also lands exactly on an answer choice, 2/243, and no arithmetic was fudged to get there: 2187 = 3⁷ and 18/9 = 2, 2187/9 = 243.

💡Key takeaway

Every 7-move path is equally likely, so the answer is just (paths that hit all 8 corners once) over (all 3⁷ = 2187 paths); careful listing shows only 18 good paths, giving 18/2187 = 2/243.

Turn probability into a count of routes
Count all possible routes
Label the cube and pin down the first move
List every good route after A to B
Total the good routes and form the probability

Turn probability into a count of routes

Count all possible routes

Label the cube and pin down the first move

List every good route after A to B

Total the good routes and form the probability

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