AMC 10 · 2009 · #22
Grade 7 probabilityTwo cubical dice each have removable numbers 1 through 6. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is 7?
Pick an answer.
AMC 10 2009 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: Take the numbers $1$ through $6$ off of two dice — twelve numbers total, two copies of each value $1$–$6$. Shuffle all twelve in a bag and glue them back on at random, one number per face, six faces per die. Roll both dice and add the two numbers on top. Find the probability that this sum equals $7$.
Givens: Two dice; each originally carries $1,2,3,4,5,6$, so the bag holds twelve numbers: two copies of each value $1$–$6$.; The twelve numbers are reattached to the twelve faces uniformly at random, one number per face, six faces per die.; Both dice are rolled; each shows a uniformly random one of its six faces on top.; We add the two top numbers and want that sum to be $7$.; Answer choices: (A) $\frac{1}{9}$, (B) $\frac{1}{8}$, (C) $\frac{1}{6}$, (D) $\frac{2}{11}$, (E) $\frac{1}{5}$.
Unknowns: The probability that the two top numbers add up to $7$.
Understand
Restated: Take the numbers $1$ through $6$ off of two dice — twelve numbers total, two copies of each value $1$–$6$. Shuffle all twelve in a bag and glue them back on at random, one number per face, six faces per die. Roll both dice and add the two numbers on top. Find the probability that this sum equals $7$.
Givens: Two dice; each originally carries $1,2,3,4,5,6$, so the bag holds twelve numbers: two copies of each value $1$–$6$.; The twelve numbers are reattached to the twelve faces uniformly at random, one number per face, six faces per die.; Both dice are rolled; each shows a uniformly random one of its six faces on top.; We add the two top numbers and want that sum to be $7$.; Answer choices: (A) $\frac{1}{9}$, (B) $\frac{1}{8}$, (C) $\frac{1}{6}$, (D) $\frac{2}{11}$, (E) $\frac{1}{5}$.
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #4 Introduce a Variable, #2 Make a Systematic List
The random gluing and the rolling look like a lot of moving parts, but they only decide which single number ends up on top of each die. Tool #9 (Solve an Easier Related Problem) strips the story down to its core: it is the same as drawing two numbers out of the bag, one for each die's top. Tool #4 (Introduce a Variable) names die 1's top number $n$, which instantly fixes the partner $7-n$ that die 2 needs. Tool #2 (Make a Systematic List) counts how many of the remaining numbers hit that target by walking through the sum-to-$7$ pairs $\{1,6\},\{2,5\},\{3,4\}$. The whole thing then collapses to one division.
Execute — Answer: D
7.SP.C.7 Step 1 Strip away the rolling
- All the gluing and rolling do is decide which single number sits on top of each die.
- Because every number is glued on at random and each die is equally likely to land on any of its six faces, every one of the twelve numbers is equally likely to be die 1's top, and the same holds for die 2.
- So the whole contraption is just this easier experiment: pull one number from the bag for die 1's top, then a second (different) number for die 2's top.
- Only their sum matters.
💡 The faces you never see can't change the answer, so ignore them and just draw the two tops.
6.EE.A.2 Step 2 Name die 1's top number
- Let $n$ be whatever number ends up on top of die 1.
- Its exact value doesn't matter yet — what matters is that for the sum to be $7$, die 2's top must be $7-n$.
- Since $n$ is between $1$ and $6$, the needed partner $7-n$ is also between $1$ and $6$, and it can never equal $n$ (that would need $2n=7$, impossible for a whole number).
💡 Once you know the first top, exactly one target value makes the sum $7$.
7.SP.C.8 Step 3 Count the matching numbers left
- After die 1's top uses up one number, $11$ numbers remain available for die 2's top.
- The bag started with two copies of every value, and the value we now need, $7-n$, is different from $n$, so neither copy of $7-n$ was touched — both are still among the $11$.
- (Check the pairs: $1{+}6,\ 2{+}5,\ 3{+}4$; in each, the two values differ, so both copies of the partner always survive.)
💡 Removing an $n$ never removes a $7-n$, so both partners are always still there.
7.SP.C.7 Step 4 Divide to get the probability
- By the symmetry from Step 1, die 2's top is equally likely to be any one of the $11$ remaining numbers.
- Exactly $2$ of them equal the needed $7-n$, so the probability of a sum of $7$ is $\tfrac{2}{11}$.
- This holds no matter what $n$ turned out to be, so it is the overall probability.
- That is choice (D).
💡 Two winning numbers out of eleven equally likely ones gives $2/11$ right away.
7.SP.C.7 All the gluing and rolling do is decide which single number sits on top of each 6.EE.A.2 Let $n$ be whatever number ends up on top of die 1. Its exact value doesn't matt 7.SP.C.8 After die 1's top uses up one number, $11$ numbers remain available for die 2's 7.SP.C.7 By the symmetry from Step 1, die 2's top is equally likely to be any one of the Review
Reasonableness: Cross-check by counting unordered pairs. Treat the twelve numbers as distinct tokens; the two tops form an unordered pair, and all $\binom{12}{2}=66$ pairs are equally likely. A sum of $7$ comes from $\{1,6\},\{2,5\},\{3,4\}$, and since each value has $2$ copies, each combination gives $2\times2=4$ token pairs, for $3\times4=12$ winning pairs. That is $\tfrac{12}{66}=\tfrac{2}{11}$ — the same result. Size check too: $\tfrac{2}{11}\approx0.18$, just a hair above the $\tfrac16\approx0.17$ you get from two ordinary dice, which makes sense — a normal die has $1$ partner out of $6$ that completes a $7$, and here it is $2$ partners out of $11$, essentially the same rate.
Alternative: Fastest mental version: whatever shows on die 1, exactly $2$ of the other $11$ numbers complete a sum of $7$, so the answer is $\tfrac{2}{11}$ with no case-work at all. A slower but reassuring route is Tool #3 (Eliminate Possibilities): the value must land near the $\tfrac16$ of ordinary dice, which rules out $\tfrac19,\tfrac18,\tfrac15$ as too far off and exposes $\tfrac16$ as the 'pretend the faces are distinct' trap, leaving $\tfrac{2}{11}$.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Recognizing that symmetry makes each of the twelve numbers equally likely on top, then dividing the $2$ favorable numbers by the $11$ equally likely outcomes for die 2's top.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming die 1's top number $n$ and writing the partner die 2 needs as the expression $7-n$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the sum-to-$7$ value pairs $\{1,6\},\{2,5\},\{3,4\}$ and counting how many copies of the needed partner remain among the leftover numbers.)
⭐ Whatever the first die shows on top, exactly two of the eleven leftover numbers make the sum $7$ — so the probability is just $2/11$, and the whole gluing-and-rolling story never mattered.
⭐ Whatever the first die shows on top, exactly two of the eleven leftover numbers make the sum $7$ — so the probability is just $2/11$, and the whole gluing-and-rolling story never mattered.
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