Problem 21 · 2026 AMC 8
Stretch
Counting & Probability
markov-chaincasework
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Answer: B — 1/4.
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Hint 1 of 2
Ten points is too many to track one by one — but notice every point is one of just two types. An outer tip touches only 2 inner points (degree 2); an inner point touches 2 outer tips and 2 inner points (degree 4). So collapse the whole web into ‘outer vs. inner’.
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Hint 2 of 2
Now it's a 2-state chain. From outer you always step inner; from inner you step outer with probability 2/4 = ½. Track just ‘chance of being outer’ move by move.
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Approach: collapse 10 points to two states (outer/inner) and track the probability
- The 10 points come in only two kinds, so lump them: an outer tip has both edges going inward (so outer → inner for sure), while an inner point has 4 edges, 2 to outer tips and 2 to inner points (so inner → outer with probability 2/4 = ½). By symmetry it doesn't matter which tip we start at.
- Start outer. Move 1: forced inward, so the spider is surely inner.
- Move 2: from inner it goes outer with probability ½, inner with probability ½.
- Move 3: to finish on an outer tip, the only route is to have stayed inner on move 2 (½), then step outward on move 3 (½). (Being outer after move 2 forces it back inner on move 3 — a dead end.) So the chance is ½ × ½ = 1/4.
- Why this transfers: when a random walk lives on a symmetric graph, group the vertices into a few ‘types’ that behave alike — the problem shrinks from many points to a tiny state machine you can track by hand.
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