AMC 10 · 2021 · #18
Grade 7 probabilityProblem
A fair 6-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?
Pick an answer.
AMC 10 2021 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: 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?
Givens: Fair $6$-sided die, faces $\{1, 2, 3, 4, 5, 6\}$; Roll repeatedly; stop the moment the first odd appears; Want: every even face ($2, 4, 6$) appears at least once among the rolls before that first odd; Choices: (A) $\tfrac{1}{120}$, (B) $\tfrac{1}{32}$, (C) $\tfrac{1}{20}$, (D) $\tfrac{3}{20}$, (E) $\tfrac{1}{6}$
Unknowns: The probability of the event "all three even faces appear before the first odd"
Understand
Restated: 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?
Givens: Fair $6$-sided die, faces $\{1, 2, 3, 4, 5, 6\}$; Roll repeatedly; stop the moment the first odd appears; Want: every even face ($2, 4, 6$) appears at least once among the rolls before that first odd; Choices: (A) $\tfrac{1}{120}$, (B) $\tfrac{1}{32}$, (C) $\tfrac{1}{20}$, (D) $\tfrac{3}{20}$, (E) $\tfrac{1}{6}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #6 Guess and Check
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.
Execute — Answer: C
7.SP.C.7 Step 1 - Track only the order in which the six faces $1, 2, 3, 4, 5, 6$ first appear.
- By symmetry of the die, every one of the $6!$ possible orderings is equally likely (each face is the next-new face with probability proportional to whether it has been seen, but all unseen faces are symmetric).
💡 Forget the repeats — only the order of new faces matters, and the die treats all faces the same.
7.SP.C.8 Step 2 - The event "all three evens appear before the first odd" translates to "in this first-appearance order, $2, 4, 6$ are the first three (in some order) and $1, 3, 5$ are the last three." Count favorable orderings: $3!$ ways to order the evens in slots $1$-$3$, times $3!$ ways to order the odds in slots $4$-$6$.
- Total orderings: $6!$.
💡 Count the orderings where evens come first, divide by all orderings.
7.SP.C.8 Step 3 - Compute the probability: $\dfrac{36}{720} = \dfrac{1}{20}$.
- This matches choice (C).
💡 $36$ favorable orderings out of $720$ — the probability is $1/20$.
7.SP.C.8 Step 4 - Cross-check with a sequential argument (Tool #6).
- First roll: probability it is one of the three evens is $\tfrac{3}{6} = \tfrac{1}{2}$ (otherwise we lose immediately).
- Given the first roll was even, the second "new" roll must be one of the two remaining evens out of the five non-first-roll faces (since repeating the first even just delays, doesn't change the experiment) — by the same symmetry the probability the next *distinct* face seen is even is $\tfrac{2}{5}$.
- Given those, the third distinct face is even with probability $\tfrac{1}{4}$ ($1$ remaining even among $4$ unseen faces).
- Multiply: $\tfrac{1}{2} \cdot \tfrac{2}{5} \cdot \tfrac{1}{4} = \tfrac{2}{40} = \tfrac{1}{20}$.
💡 Three hurdles ($\tfrac{3}{6}$, $\tfrac{2}{5}$, $\tfrac{1}{4}$) — multiply and the answer is the same $\tfrac{1}{20}$.
7.SP.C.7 Track only the order in which the six faces $1, 2, 3, 4, 5, 6$ first appear. By 7.SP.C.8 The event "all three evens appear before the first odd" translates to "in this f 7.SP.C.8 Compute the probability: $\dfrac{36}{720} = \dfrac{1}{20}$. This matches choice 7.SP.C.8 Cross-check with a sequential argument (Tool #6). First roll: probability it is Review
Reasonableness: Both approaches — the symmetry / permutation count $\tfrac{3! \cdot 3!}{6!} = \tfrac{1}{20}$ and the sequential product $\tfrac{1}{2} \cdot \tfrac{2}{5} \cdot \tfrac{1}{4} = \tfrac{1}{20}$ — agree, so the answer is robust. The number $\tfrac{1}{20} = 5\%$ also feels right: hitting all three even faces before any of three odds appear is harder than the $\tfrac{1}{6}$ chance of just rolling a specific value, but easier than the $\tfrac{1}{120}$ chance of getting a specific ordering of all six faces.
Alternative: Tool #16 (Change Focus / Complement) — instead of P(all evens appear first), expand the event by the position of the first odd: $P = \sum_{k=3}^{\infty} P(\text{first odd is on roll } k \text{ and rolls } 1,\ldots, k-1 \text{ together cover all $3$ evens})$. This direct sum equals the same $\tfrac{1}{20}$ but requires an inclusion-exclusion or recursion — much messier than the symmetry shortcut.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling the die rolls so the first-appearance order is a uniform random permutation — the equiprobable-outcome model.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting favorable orderings ($3! \cdot 3!$) over all orderings ($6!$), and the equivalent sequential product $\tfrac{3}{6} \cdot \tfrac{2}{5} \cdot \tfrac{1}{4}$ for the compound event.)
⭐ 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 $\tfrac{3! \cdot 3!}{6!} = \tfrac{1}{20}$ of the time.
⭐ 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 $\tfrac{3! \cdot 3!}{6!} = \tfrac{1}{20}$ of the time.
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