AMC 10 · 2023 · #7

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicsystematic-enumerationfraction-arithmetic caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

Janet rolls a fair 6-sided die four times. After each roll she keeps a running total. What is the probability that at some moment among those four steps the running total equals exactly 3?

Pick an answer.

(A)

$frac{2}{9}$

(B)

$frac{49}{216}$

(C)

$frac{25}{108}$

(D)

$frac{17}{72}$

(E)

$frac{13}{54}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Because rolls only add (never subtract), the running total hits 3 on at most one specific roll number. That makes the event split into a few clean cases: "first hit on roll 1", "first hit on roll 2", "first hit on roll 3", "first hit on roll 4". Tool #2 (Systematic List) is built for "list all sequences that…" — list every starting prefix that sums to 3 at exactly that step. Tool #7 (Identify Subproblems) gives the case split, and Tool #3 (Eliminate Possibilities) trims roll 4 before any work — since each roll is ≥ 1, the minimum total after 4 rolls is 4 ＞ 3, so the first-hit-on-roll-4 case is impossible. The remaining sub-probabilities add cleanly because the cases are mutually exclusive.

1STEP 1

Each roll is at least 1, so after four rolls the total is at least 4 — roll 4 can never be the first hit, leaving only rolls 1, 2, 3.

after 4 rolls: total ≥ 4 → roll 4 cannot be the hit

2STEP 2

Case 1 — the total is 3 on roll 1 only by rolling a 3 first: one face out of six, so $\frac{1}{6}$ .

P₁ =

\frac{1}{6}

3STEP 3

Case 2 — first reaches 3 on roll 2 via (1, 2) or (2, 1), neither starting with 3: two ordered pairs give $\frac{1}{18}$ .

P₂ = 2 ·

\frac{1}{36}

\frac{2}{36}

\frac{1}{18}

4STEP 4

Case 3 — first reaches 3 on roll 3 only via (1, 1, 1), one triple out of 216 equally likely ones, so $\frac{1}{216}$ .

P₃ =

\frac{1}{6}

³ =

\frac{1}{216}

5STEP 5

The three cases are mutually exclusive, so add over the common denominator 216: $\frac{36 + 12 + 1}{216}$ = $\frac{49}{216}$ , choice (B).

P =

\frac{1}{6}

\frac{1}{18}

\frac{1}{216}

\frac{36 + 12 + 1}{216}

\frac{49}{216}

→ (B)

Answer

frac{49}{216}

Cross-check by brute force on 6⁴ = 1296 ordered roll sequences. The count of sequences whose running total hits 3 at some step is 6³ + 2 · 6² + 1 · 6 = 216 + 72 + 6 = 294 — that is, 216 sequences starting (3, *, *, *), 72 starting (1, 2, *, *) or (2, 1, *, *), and 6 starting (1, 1, 1, *). Probability =

\frac{294}{1296}

\frac{49}{216}

. Matches (B). Magnitude check:

\frac{49}{216}

≈ 0.227 is between

\frac{1}{6}

≈ 0.167 (the chance the very first roll is 3) and

\frac{1}{4}

— exactly where you'd expect, since extra rolls add small "second chance" probabilities.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability of compound events — list every roll prefix that lands exactly on 3, multiply $\frac{1}{6}$ for each roll, and add the cases that can't happen at the same time.

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