AMC 10 · 2014 · #16
Grade 7 probabilityFour fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?
Pick an answer.
AMC 10 2014 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: Four fair six-sided dice are rolled at once. Find the probability that some value appears on at least three of the four dice.
Givens: Four dice are rolled, each fair with faces $1$ through $6$; The dice are independent, so all outcomes are equally likely; "At least three the same" means three dice match, or all four match; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{7}{72}$, (C) $\frac{1}{9}$, (D) $\frac{5}{36}$, (E) $\frac{1}{6}$
Unknowns: The probability that at least three of the four dice show the same value
Understand
Restated: Four fair six-sided dice are rolled at once. Find the probability that some value appears on at least three of the four dice.
Givens: Four dice are rolled, each fair with faces $1$ through $6$; The dice are independent, so all outcomes are equally likely; "At least three the same" means three dice match, or all four match; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{7}{72}$, (C) $\frac{1}{9}$, (D) $\frac{5}{36}$, (E) $\frac{1}{6}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
For equally likely rolls, probability is (favorable outcomes) divided by (all outcomes). Tool #7 (Identify Subproblems) splits the fuzzy phrase "at least three the same" into two clean, non-overlapping cases: all four dice equal, and exactly three equal with one different. Tool #2 (Make a Systematic List) then counts each case by choosing the repeated value, the odd die's position, and its value in order. Adding the two counts and dividing by $6^4$ gives the answer.
Execute — Answer: B
7.SP.C.7 Step 1 Count all rolls and split the goal
- Each die shows one of $6$ values independently, so there are $6^4 = 1296$ equally likely outcomes.
- The event "at least three dice show the same value" happens in exactly two ways that cannot overlap: all four dice equal, or exactly three equal with the fourth different.
- Count each separately, add, then divide by $1296$.
💡 When every roll is equally likely, probability is just a fraction: good rolls over all rolls.
7.SP.C.8 Step 2 Count all four the same
- If all four dice match, the only choice is which value they all show.
- There are $6$ values, and each value fixes the whole roll (like $3333$).
- So there are $6$ such outcomes.
💡 One matching value decides the entire roll, so there is nothing left to choose.
7.SP.C.8 Step 3 Count exactly three the same
- Build such a roll by three ordered choices.
- First pick the value shown three times: $6$ ways.
- Then pick which one of the four dice is the odd one out: $4$ ways.
- Finally pick the odd die's value, which must differ from the tripled value: $5$ ways.
- Multiplying gives $6 \times 4 \times 5 = 120$ outcomes.
💡 Choosing the repeated value, then which die breaks the pattern, then its different value counts each roll exactly once.
7.SP.C.8 Step 4 Add the cases and divide
- The two cases are disjoint, so add their counts: $6 + 120 = 126$ favorable outcomes out of $1296$.
- Simplify the fraction by dividing top and bottom by $18$: $\frac{126}{1296} = \frac{7}{72}$.
- So the probability is $\frac{7}{72}$, which is answer (B).
💡 Non-overlapping cases can be counted separately and simply added before dividing by the total.
7.SP.C.7 Each die shows one of $6$ values independently, so there are $6^4 = 1296$ equall 7.SP.C.8 If all four dice match, the only choice is which value they all show. There are 7.SP.C.8 Build such a roll by three ordered choices. First pick the value shown three tim 7.SP.C.8 The two cases are disjoint, so add their counts: $6 + 120 = 126$ favorable outco Review
Reasonableness: The answer $\frac{7}{72}\approx 0.097$ is small but not tiny, which fits: getting three matches among four dice is uncommon yet far from rare. A sanity check on the counts: the "exactly three" case ($120$) should dwarf the "all four" case ($6$), and it does, because forcing a fourth match is much harder than allowing one odd die. The favorable total $126$ is under $1296$, so the probability is safely below $1$. All of this agrees with choice (B).
Alternative: Fix the first die's value as a reference and use conditional probability. The chance a specific three of the four dice all match the pattern can be built from single-die probabilities of $\frac{1}{6}$, then multiplied by the $\binom{4}{3}=4$ positions for the tripled group and adjusted for the all-four overlap. Working it through gives $6\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot 4\cdot\frac{5}{6}$ for exactly three plus $6\cdot\left(\frac{1}{6}\right)^4$ for all four, which again totals $\frac{7}{72}$.
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the $6^4$ rolls as equally likely so probability equals favorable outcomes over total outcomes.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and counting (Counting the all-four-same ($6$) and exactly-three-same ($120$) cases and combining them into $\frac{126}{1296}=\frac{7}{72}$.)
⭐ Split "at least three match" into "all four" ($6$ ways) plus "exactly three" ($6\times4\times5=120$ ways), then divide $126$ by $1296$ to get $\frac{7}{72}$.
⭐ Split "at least three match" into "all four" ($6$ ways) plus "exactly three" ($6\times4\times5=120$ ways), then divide $126$ by $1296$ to get $\frac{7}{72}$.
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