AMC 10 · 2014 · #17
Grade 7 probabilityThree fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
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: Three fair six-sided dice are rolled. Find the probability that two of the dice show values that add up to the value shown on the third die.
Givens: Three fair six-sided dice are rolled, each showing a value from 1 to 6 independently; We want the event that some two of the dice add up to the remaining die's value; Answer choices: (A) 1/6, (B) 13/72, (C) 7/36, (D) 5/24, (E) 2/9
Unknowns: The probability that two of the dice sum to the third die
Understand
Restated: Three fair six-sided dice are rolled. Find the probability that two of the dice show values that add up to the value shown on the third die.
Givens: Three fair six-sided dice are rolled, each showing a value from 1 to 6 independently; We want the event that some two of the dice add up to the remaining die's value; Answer choices: (A) 1/6, (B) 13/72, (C) 7/36, (D) 5/24, (E) 2/9
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Because every die is fair and independent, all $6^3 = 216$ ordered outcomes are equally likely, so the probability is just (favorable outcomes)$/216$. The phrase "two of the dice sum to the third" is slippery because it does not say which two. Tool #16 (Change Focus) fixes that: decide which single die is the total, then the other two must add up to it. That splits the job into 3 matching subproblems (Tool #7). Inside each subproblem the counting is a clean Tool #2 (Make a Systematic List): list the ordered pairs of the other two dice that add to each reachable total, which fall into the tidy staircase $1, 2, 3, 4, 5$.
Execute — Answer: D
7.SP.C.7 Step 1 Count all equally likely outcomes
- Each die is fair and rolls 1 to 6 independently, so writing the three faces in order gives $6 \times 6 \times 6 = 216$ equally likely ordered outcomes.
- Because every outcome has the same chance, the probability equals the number of favorable outcomes divided by 216.
- The whole task is now to count the outcomes where one die equals the sum of the other two.
💡 When every outcome is equally likely, probability becomes a counting job: favorable over total.
7.SP.C.8 Step 2 Pick which die is the total
- Instead of hunting for "some two that add to the third," turn it around: first choose which one of the three dice is the total, then require the other two to add up to it.
- There are 3 choices for which die is the total.
- For a fixed total die, the other two dice must show values that sum to whatever the total die shows.
- This reframing turns one fuzzy condition into 3 clear, identical subproblems.
💡 Naming which die is the total turns a vague 'some two' into a concrete target to hit.
7.SP.C.8 Step 3 List two-dice sums for one total die
- Fix the total die.
- The other two dice, read in order, must add to a value the total die can actually show, so the target sum runs from 2 to 6.
- Count the ordered pairs $(a,b)$ with $a+b = s$: for $s=2$ there is $1$ (namely $1{+}1$); for $s=3$ there are $2$; for $s=4$ there are $3$; for $s=5$ there are $4$; for $s=6$ there are $5$.
- Each such pair pins the total die to $s$, which is a legal face.
- Adding the staircase gives $1+2+3+4+5 = 15$ ordered outcomes for each choice of total die.
💡 Ordered pairs adding to a target line up as 1, 2, 3, 4, 5 — a staircase that is easy to add.
7.SP.C.8 Step 4 Combine the three cases without overlap
- Each of the 3 choices of total die gives 15 outcomes, for $3 \times 15 = 45$.
- These 45 never overlap: if one die equalled the sum of the other two and a second die also equalled the sum of the other two, subtracting those two equations would force some die to show 0, which no face allows.
- So the 45 favorable ordered outcomes are all distinct and can be added straight up.
💡 Two dice cannot each be the total of the rest at once, so the three cases stay separate.
7.NS.A.3 Step 5 Form the probability
- Divide favorable outcomes by total outcomes: $45 / 216$.
- Both share a factor of 9, so $45/216 = 5/24$.
- This matches choice (D), so the answer is (D).
💡 Divide the favorable count by the total count once, then reduce the fraction.
7.SP.C.7 Each die is fair and rolls 1 to 6 independently, so writing the three faces in o 7.SP.C.8 Instead of hunting for "some two that add to the third," turn it around: first c 7.SP.C.8 Fix the total die. The other two dice, read in order, must add to a value the to 7.SP.C.8 Each of the 3 choices of total die gives 15 outcomes, for $3 \times 15 = 45$. Th 7.NS.A.3 Divide favorable outcomes by total outcomes: $45 / 216$. Both share a factor of Review
Reasonableness: The probability $\tfrac{5}{24} \approx 0.208$ is sensible: hitting "two add to the third" is possible but far from common, so a value near one-fifth fits. A cross-check confirms the count another way: list the multisets $\{a, b, a{+}b\}$ with $a{+}b \le 6$ and weight each by its arrangements — the six all-distinct ones give $6$ each ($36$), and the three with a repeat ($\{1,1,2\}, \{2,2,4\}, \{3,3,6\}$) give $3$ each ($9$), totalling $45$ again. The near-miss wrong answers come from slips: dropping the factor of 3 for which die is the total gives $15/216 = 5/72$, and other mis-added staircases land on (B) $\tfrac{13}{72}$ or (C) $\tfrac{7}{36}$.
Alternative: Count unordered instead. List every multiset $\{a, b, c\}$ with $c = a + b$ and $c \le 6$: $\{1,1,2\}, \{1,2,3\}, \{1,3,4\}, \{1,4,5\}, \{1,5,6\}, \{2,2,4\}, \{2,3,5\}, \{2,4,6\}, \{3,3,6\}$. Weight each by how many ordered rolls it makes — $6$ for all-distinct, $3$ when two faces match — sum to $45$, then divide by $216$ to get $\tfrac{5}{24}$.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Treating all 216 ordered dice outcomes as equally likely so the probability is the favorable count over 216.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Choosing which die is the total, listing the ordered pairs that add to each reachable value, and combining the three cases without overlap.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Dividing 45 by 216 and reducing to form the final probability 5/24.)
⭐ Decide which die is the total, count the other two's ordered pairs (1+2+3+4+5 = 15), times 3 dice is 45, over 216 gives 5/24.
⭐ Decide which die is the total, count the other two's ordered pairs (1+2+3+4+5 = 15), times 3 dice is 45, over 216 gives 5/24.
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