AMC 10 · 2023 · #7
Grade 7 probabilityProblem
Janet rolls a standard 6-sided die 4 times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal 3?
Pick an answer.
AMC 10 2023 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: 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 $3$?
Givens: Each roll independently lands on $1, 2, 3, 4, 5,$ or $6$ with probability $\tfrac{1}{6}$; Four rolls total; Track the running total after each roll; Answer choices: (A) $\tfrac{2}{9}$, (B) $\tfrac{49}{216}$, (C) $\tfrac{25}{108}$, (D) $\tfrac{17}{72}$, (E) $\tfrac{13}{54}$
Unknowns: $P(\text{running total} = 3 \text{ at some roll among rolls } 1, 2, 3, 4)$
Understand
Restated: 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 $3$?
Givens: Each roll independently lands on $1, 2, 3, 4, 5,$ or $6$ with probability $\tfrac{1}{6}$; Four rolls total; Track the running total after each roll; Answer choices: (A) $\tfrac{2}{9}$, (B) $\tfrac{49}{216}$, (C) $\tfrac{25}{108}$, (D) $\tfrac{17}{72}$, (E) $\tfrac{13}{54}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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 $\ge 1$, 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.
Execute — Answer: B
3.OA.D.8 Step 1 - Eliminate roll $4$.
- Each roll is at least $1$, so after four rolls the running total is at least $1+1+1+1 = 4$.
- The running total cannot equal $3$ for the first time on roll $4$ — by then it has already passed $3$ (or hit it earlier).
- So only rolls $1$, $2$, $3$ can be the "hit moment".
💡 If you only add positive numbers, the smallest total after $k$ rolls is $k$ — Grade 3 reasoning about repeated addition.
7.SP.C.7 Step 2 - Case 1: total $= 3$ on roll $1$.
- List all single-roll prefixes summing to $3$: just $(3)$.
- Probability is one face out of six.
💡 One out of six equally likely faces — the Grade 7 "equally likely outcomes" probability model.
7.SP.C.8 Step 3 - Case 2: running total first equals $3$ on roll $2$.
- List all two-roll prefixes whose entries sum to $3$ and whose first entry is not $3$: $(1, 2)$ and $(2, 1)$.
- (We must exclude any prefix that already hit $3$ on roll $1$ — none of these did.) Each ordered pair has probability $\tfrac{1}{6} \cdot \tfrac{1}{6} = \tfrac{1}{36}$.
💡 Two independent dice rolls give $36$ equally likely ordered pairs — Grade 7 compound-event counting.
7.SP.C.8 Step 4 - Case 3: running total first equals $3$ on roll $3$.
- The only triple of die faces summing to $3$ is $(1,1,1)$ — and its running totals $1, 2, 3$ are first $3$ on roll $3$.
- So one ordered triple works, out of $216$ equally likely ordered triples.
💡 Three independent rolls $\to 6 \times 6 \times 6 = 216$ outcomes; only $(1,1,1)$ matches — Grade 7 fundamental counting.
5.NF.A.1 Step 5 - The three cases are mutually exclusive (the running total can be $3$ for the first time at only one moment), so the answers add.
- Use the common denominator $216$.
💡 Add fractions with unlike denominators by rewriting over the common denominator $216$ — Grade 5 fraction addition.
3.OA.D.8 Eliminate roll $4$. Each roll is at least $1$, so after four rolls the running t 7.SP.C.7 Case 1: total $= 3$ on roll $1$. List all single-roll prefixes summing to $3$: j 7.SP.C.8 Case 2: running total first equals $3$ on roll $2$. List all two-roll prefixes w 7.SP.C.8 Case 3: running total first equals $3$ on roll $3$. The only triple of die faces 5.NF.A.1 The three cases are mutually exclusive (the running total can be $3$ for the fir Review
Reasonableness: Cross-check by brute force on $6^4 = 1296$ ordered roll sequences. The count of sequences whose running total hits $3$ at some step is $6^3 + 2 \cdot 6^2 + 1 \cdot 6 = 216 + 72 + 6 = 294$ — that is, $216$ sequences starting $(3, *, *, *)$, $72$ starting $(1, 2, *, *)$ or $(2, 1, *, *)$, and $6$ starting $(1, 1, 1, *)$. Probability $= \tfrac{294}{1296} = \tfrac{49}{216}$. Matches (B). Magnitude check: $\tfrac{49}{216} \approx 0.227$ is between $\tfrac{1}{6} \approx 0.167$ (the chance the very first roll is $3$) and $\tfrac{1}{4}$ — exactly where you'd expect, since extra rolls add small "second chance" probabilities.
Alternative: Tool #16 (Change Focus / Count the Complement): instead of counting prefixes that hit $3$, count $4$-roll sequences whose running total skips $3$. The complement event is "every running total is $\ne 3$". On roll $1$ a non-$3$ face has probability $\tfrac{5}{6}$. After landing on $1$ (prob $\tfrac{1}{6}$), roll $2$ must avoid $2$; after $2$ (prob $\tfrac{1}{6}$), roll $2$ must avoid $1$; after $1$-then-$1$ (prob $\tfrac{1}{36}$), roll $3$ must avoid $1$. Tracking through gives $1 - P(\text{skip}) = \tfrac{49}{216}$. Same answer (B), with the complement made of fewer cases.
CCSS standards used (min grade 7)
3.OA.D.8Solve two-step word problems using the four operations (The eliminate-roll-$4$ argument: $1+1+1+1 = 4 > 3$, so the running total cannot first hit $3$ on roll $4$.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Single-roll case: each face has probability $\tfrac{1}{6}$ in a uniform probability model on $\{1, \ldots, 6\}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and tree diagrams (Listing the prefixes $(1,2), (2,1), (1,1,1)$ and assigning each ordered tuple probability $(\tfrac{1}{6})^k$ — compound-event counting on independent rolls.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $\tfrac{1}{6} + \tfrac{1}{18} + \tfrac{1}{216} = \tfrac{49}{216}$ via the common denominator $216$.)
⭐ This AMC 10 problem only needs Grade 7 probability of compound events — list every roll prefix that lands exactly on $3$, multiply $\tfrac{1}{6}$ for each roll, and add the cases that can't happen at the same time.
⭐ This AMC 10 problem only needs Grade 7 probability of compound events — list every roll prefix that lands exactly on $3$, multiply $\tfrac{1}{6}$ for each roll, and add the cases that can't happen at the same time.
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