AMC 10 · 2006 · #21
Grade 7 probabilityFor a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5, and 6, on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice?
Pick an answer.
AMC 10 2006 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: A single die is weighted so that the chances of rolling $1,2,3,4,5,6$ are in the ratio $1:2:3:4:5:6$ — bigger faces are more likely, in exact proportion to the number. Two such identical dice are rolled. Find the probability that the two faces add up to $7$.
Givens: On one die, $P(1):P(2):P(3):P(4):P(5):P(6)=1:2:3:4:5:6$; The two dice are identical and rolled independently; Answer choices: (A) $\frac{4}{63}$, (B) $\frac{1}{8}$, (C) $\frac{8}{63}$, (D) $\frac{1}{6}$, (E) $\frac{2}{7}$
Unknowns: The probability that the sum of the two rolled faces equals $7$
Understand
Restated: A single die is weighted so that the chances of rolling $1,2,3,4,5,6$ are in the ratio $1:2:3:4:5:6$ — bigger faces are more likely, in exact proportion to the number. Two such identical dice are rolled. Find the probability that the two faces add up to $7$.
Givens: On one die, $P(1):P(2):P(3):P(4):P(5):P(6)=1:2:3:4:5:6$; The two dice are identical and rolled independently; Answer choices: (A) $\frac{4}{63}$, (B) $\frac{1}{8}$, (C) $\frac{8}{63}$, (D) $\frac{1}{6}$, (E) $\frac{2}{7}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #7 Identify Subproblems
The ratio $1:2:3:4:5:6$ does not name the probabilities directly, so Tool #4 (Introduce a Variable) supplies the missing scale: call the smallest probability $p$, write the six probabilities as $p,2p,\dots,6p$, and use the fact that they add to $1$ to pin down $p$. Once each face has a known probability, Tool #2 (Make a Systematic List) sweeps out every face pair that totals $7$ so none is missed and none double-counted. Tool #7 (Identify Subproblems) splits the job cleanly: first find the single-die probabilities, then find each pair's probability by multiplying, then add the pieces — a sum-of-products that collapses to one fraction.
Execute — Answer: C
6.RP.A.3 Step 1 Name the scale with a variable
- The ratio $1:2:3:4:5:6$ says the six probabilities are equally spaced multiples of one common piece.
- Let that piece be $p=P(1)$.
- Then $P(2)=2p$, $P(3)=3p$, and so on up to $P(6)=6p$.
- This turns the ratio into six honest probabilities that all share the single unknown $p$.
💡 A ratio only fixes the relative sizes, so one variable sets the common unit that scales them all.
6.EE.B.7 Step 2 Use total probability to solve for p
- The six outcomes are the only possibilities for one die, so their probabilities must add to $1$.
- Adding the multiples gives $p(1+2+3+4+5+6)=21p$, and setting this equal to $1$ solves the scale: $21p=1$, so $p=\tfrac{1}{21}$.
- Therefore each face has probability $P(k)=\tfrac{k}{21}$.
💡 All the chances for one die must sum to a whole $1$, which turns the ratio into a single equation for the unit piece.
7.SP.C.8 Step 3 List every face pair that totals 7
- Write out the ordered pairs (first die, second die) that add to $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — six of them.
- Because the two dice roll independently, the probability of a specific pair $(a,b)$ is the product $P(a)\,P(b)=\tfrac{a}{21}\cdot\tfrac{b}{21}=\tfrac{a\,b}{441}$.
- So each listed pair contributes a term of the form $\tfrac{a\,b}{441}$.
💡 Independent rolls multiply, so a pair's chance is just the two face chances multiplied together.
7.SP.C.8 Step 4 Add the pair probabilities and simplify
- Sum the six terms, factoring out the common $\tfrac{1}{441}$: the numerators are $1\cdot6,\,2\cdot5,\,3\cdot4,\,4\cdot3,\,5\cdot2,\,6\cdot1$, which are $6,10,12,12,10,6$ and total $56$.
- So the probability is $\tfrac{56}{441}$.
- Since $441=7\cdot63$ and $56=7\cdot8$, the common factor $7$ cancels: $\tfrac{56}{441}=\tfrac{8}{63}$.
- This is choice (C).
💡 The total chance of hitting a sum is just the chances of all the ways to make it, added up.
6.RP.A.3 The ratio $1:2:3:4:5:6$ says the six probabilities are equally spaced multiples 6.EE.B.7 The six outcomes are the only possibilities for one die, so their probabilities 7.SP.C.8 Write out the ordered pairs (first die, second die) that add to $7$: $(1,6),(2,5 7.SP.C.8 Sum the six terms, factoring out the common $\tfrac{1}{441}$: the numerators are Review
Reasonableness: A quick sanity check on size: with a fair die, a total of $7$ has probability $\tfrac{6}{36}=\tfrac16\approx0.167$. Here $\tfrac{8}{63}\approx0.127$ is a bit smaller, which fits — the weighting pushes rolls toward high faces like $6$, and a high roll on one die needs a low roll (probability only $\tfrac{1}{21}$ or $\tfrac{2}{21}$) on the other to still total $7$, so sevens become slightly harder. The value also lands between the two closely spaced trap answers $\tfrac{4}{63}$ and $\tfrac{8}{63}$, and $\tfrac{4}{63}$ is exactly half of ours — the number you would get by counting only three pairs instead of all six ordered pairs, a reminder to count both orders.
Alternative: Instead of listing pairs, use a generating function. One die's face-value weights are $1,2,3,4,5,6$ (before dividing by $21$), captured by $x+2x^2+3x^3+4x^4+5x^5+6x^6$. The total for two dice comes from squaring that polynomial; the coefficient of $x^7$ collects every way two faces sum to $7$ and equals $1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1=56$. Dividing by $21^2=441$ again gives $\tfrac{56}{441}=\tfrac{8}{63}$.
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Reading the ratio $1:2:3:4:5:6$ as scaled multiples $p,2p,\dots,6p$ of a common unit probability.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $21p=1$ so that each face gets probability $P(k)=\tfrac{k}{21}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the six ordered face pairs summing to $7$, multiplying independent single-die probabilities, and adding them to $\tfrac{56}{441}=\tfrac{8}{63}$.)
⭐ Turn a ratio into probabilities by giving it a variable and making them add to $1$; then a compound total is the sum of every ordered way to make it, each way being the two independent chances multiplied.
⭐ Turn a ratio into probabilities by giving it a variable and making them add to $1$; then a compound total is the sum of every ordered way to make it, each way being the two independent chances multiplied.
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