AMC 10 · 2014 · #16

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccaseworkcombinations-basic casework ↑ Prerequisites: probability-basic

📏 Medium solution 💡 3 insights

Problem

Four fair six-sided dice are rolled at once, so each die independently shows one of the values 1 through 6. Find the probability that at least three of the four dice show the same value.

Pick an answer.

(A)

$\frac{1}{36}$

(B)

$\frac{7}{72}$

(C)

$\frac{1}{9}$

(D)

$\frac{5}{36}$

(E)

$\frac{1}{6}$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 6⁴ gives the answer.

1STEP 1

Count all rolls and split the goal

Each die is independent, so all rolls number 6⁴ = 1296; split the goal into all four equal, or exactly three equal.

total outcomes = 6⁴ = 1296

2STEP 2

Count all four the same

All four alike means picking only the shared value, and each of the 6 values fixes the whole roll like 3333: 6 outcomes.

6 values → 6 outcomes

3STEP 3

Count exactly three the same

For exactly three alike, pick the tripled value (6), the odd die out (4), then its different value (5): 6 × 4 × 5 = 120 outcomes.

6 × 4 × 5 = 120 outcomes

4STEP 4

Add the cases and divide

Disjoint cases add: 6 + 120 = 126 favorable out of 1296, which reduces to 7/72, choice (B).

(6 + 120)/1296 = 126/1296 = 7/72 → (B)

Answer

7/72

The answer 7/72≈ 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).

💡Key takeaway

Split "at least three match" into "all four" (6 ways) plus "exactly three" (6×4×5=120 ways), then divide 126 by 1296 to get 7/72.

Count all rolls and split the goal
Count all four the same
Count exactly three the same
Add the cases and divide

Count all rolls and split the goal

Count all four the same

Count exactly three the same

Add the cases and divide

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