AMC 10 · 2023 · #5
Grade 7 probabilityPick an answer.
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.
The fourth roll is too late
By the fourth it is already too late.
If you only add positive numbers, the smallest total after k rolls is k — Grade 3 reasoning about repeated addition.
3.OA.D.8Eliminate PossibilitiesReaching it in one roll
Compute the one-roll chance.
One out of six equally likely faces — the Grade 7 "equally likely outcomes" probability model.
7.SP.C.7Make A Systematic ListReaching it in two rolls
There are two ways in two rolls.
Two independent dice rolls give 36 equally likely ordered pairs — Grade 7 compound-event counting.
7.SP.C.8Make A Systematic ListReaching it in three rolls
Only one way works in three rolls.
Three independent rolls → 6 × 6 × 6 = 216 outcomes; only (1,1,1) matches — Grade 7 fundamental counting.
Three independent rolls give a grid of equally likely triples, and only one of them matches.
▸ Why?
Each roll is made without regard to the others, so the option counts multiply.
▸ Why?
Every triple is just as likely as any other, so the chance is a count over the total count.
Add them up
Adding gives forty-nine over two hundred sixteen.
Add fractions with unlike denominators by rewriting over the common denominator 216 — Grade 5 fraction addition.
5.NF.A.1Identify SubproblemsThis AMC 12 problem only needs Grade 7 probability of compound events — list every roll prefix that lands exactly on 3, multiply 1/6 for each roll, and add the cases that can't happen at the same time.