AMC 10 · 2021 · #18

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicconditional-probabilitypermutations-basicsymmetry-argument easier-related-problemsymmetry-argument ↑ Prerequisites: probability-basic

📏 Short solution 💡 3 insights

Problem

Roll a fair 6-sided die over and over until an odd number (1, 3, or 5) shows up. What is the probability that, before that first odd roll, every one of the three even faces (2, 4, 6) has appeared at least once?

Pick an answer.

(A)

$~\frac{1}{120}$

(B)

$~\frac{1}{32}$

(C)

$~\frac{1}{20}$

(D)

$~\frac{3}{20}$

(E)

$~\frac{1}{6}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem) — re-frame the question so the rolls form a much simpler structure. Instead of tracking every roll (with repeats), focus on the order in which the six faces first appear. By symmetry the six faces are exchangeable, so the order of first appearances is a uniformly random permutation of {1, 2, 3, 4, 5, 6}. The event "all three evens appear before any odd" is the event "in this random permutation, 2, 4, 6 are the first three entries" — a clean counting problem. Tool #5 (Pattern) helps justify the symmetry from small cases, and Tool #6 (Guess & Check) verifies the answer against a direct sequential-product computation.

1STEP 1

Track only the order the six faces first appear; by the die's symmetry all 6! orderings are equally likely.

First-appearance order ∼ Uniform on S₆

2STEP 2

Favorable orderings put 2, 4, 6 in the first three slots: 3! · 3! = 36 out of 6! = 720 total.

#favorable = 3! · 3! = 6 · 6 = 36; #total = 6! = 720

3STEP 3

Divide favorable by total: $\frac{36}{720}$ = $\frac{1}{20}$ , which is choice (C).

P =

\frac{3! · 3!}{6!}

\frac{36}{720}

\frac{1}{20}

→ (C)

4STEP 4

Cross-check sequentially: each new distinct face is even with probability $\frac{3}{6}$ , then $\frac{2}{5}$ , then $\frac{1}{4}$ ; multiply to get $\frac{1}{20}$ .

\frac{3}{6}

\frac{2}{5}

\frac{1}{4}

\frac{6}{120}

\frac{1}{20}

Answer

\frac{1}{20}

Both approaches — the symmetry / permutation count

\frac{3! · 3!}{6!}

\frac{1}{20}

and the sequential product

\frac{1}{2}

\frac{2}{5}

\frac{1}{4}

\frac{1}{20}

— agree, so the answer is robust. The number

\frac{1}{20}

= 5% also feels right: hitting all three even faces before any of three odds appear is harder than the

\frac{1}{6}

chance of just rolling a specific value, but easier than the

\frac{1}{120}

chance of getting a specific ordering of all six faces.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability you already know! Pretend the die forgets repeats and just records the order in which it first shows each face — by symmetry every ordering of the six faces is equally likely. "All three evens before any odd" means the evens fill the first three slots, which happens $\frac{3! · 3!}{6!}$ = $\frac{1}{20}$ of the time.

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