AMC 8 · 2011 · #18
Grade 7 probabilityProblem
A fair 6 sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
Pick an answer.
AMC 8 2011 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: Roll a fair $6$-sided die twice and record the two outcomes in order. Find the probability that the first roll is greater than or equal to the second roll.
Givens: Two independent rolls of a fair $6$-sided die; Each die shows one of $1, 2, 3, 4, 5, 6$ with equal probability; Order matters: the outcome is the ordered pair (first roll, second roll); Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{5}{12}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{7}{12}$, (E) $\tfrac{5}{6}$
Unknowns: The probability that the first roll $\ge$ the second roll
Understand
Restated: Roll a fair $6$-sided die twice and record the two outcomes in order. Find the probability that the first roll is greater than or equal to the second roll.
Givens: Two independent rolls of a fair $6$-sided die; Each die shows one of $1, 2, 3, 4, 5, 6$ with equal probability; Order matters: the outcome is the ordered pair (first roll, second roll); Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{5}{12}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{7}{12}$, (E) $\tfrac{5}{6}$
Plan
Primary tool: #2 Find Symmetry
Secondary: #7 Identify Subproblems, #3 Make a Systematic List
The sample space is the $6 \times 6$ grid of ordered pairs $(a, b)$, with $36$ equally likely outcomes. Tool #2 (Find Symmetry) exploits a clean swap symmetry: the map $(a, b) \mapsto (b, a)$ pairs each "first $>$ second" outcome with a unique "first $<$ second" outcome, so those two counts are equal. Tool #7 (Identify Subproblems) splits the event "first $\ge$ second" into two disjoint cases — strict inequality $a > b$ and equality $a = b$ — that we can count separately and add. Tool #3 (Make a Systematic List) handles the small piece — directly listing the $6$ ties $(k, k)$ — and also drives the alternative full-enumeration approach in Review.
Execute — Answer: D
7.SP.C.8 Step 1 - Set up the sample space.
- Each roll is independent and uniform on $\{1, \dots, 6\}$, so the ordered pair $(a, b)$ takes $6 \times 6 = 36$ equally likely values.
💡 Listing the sample space of a compound event as ordered pairs is the Grade 7 probability standard for two-stage experiments.
7.SP.C.8 Step 2 - Split the target event into two disjoint pieces.
- The event $a \ge b$ is the union of $\{a > b\}$ and $\{a = b\}$, with no overlap, so $P(a \ge b) = P(a > b) + P(a = b)$.
💡 Breaking an event into disjoint sub-events is the Tool #7 subproblems move — count each piece, then add.
7.SP.C.7 Step 3 - Count the ties.
- The ordered pairs with $a = b$ are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ — exactly $6$ outcomes.
💡 Just listing the diagonal of the $6 \times 6$ grid uses the Grade 7 uniform-probability counting standard.
7.SP.C.8 Step 4 - Use symmetry on the non-tie outcomes.
- The remaining $36 - 6 = 30$ ordered pairs satisfy $a \ne b$.
- The swap map $(a, b) \mapsto (b, a)$ is a bijection that sends each pair with $a > b$ to a unique pair with $a < b$, so the two counts are equal.
- Hence each is half of $30$.
💡 This is the heart of Tool #2 (Find Symmetry): when a problem treats two roles the same way, swapping them must give equal counts.
7.SP.C.7 Step 5 Add the two pieces and divide by the size of the sample space to get the probability.
💡 Favorable outcomes divided by total outcomes in a uniform sample space is the Grade 7 definition of probability.
7.SP.C.8 Set up the sample space. Each roll is independent and uniform on ${1, \dots, 6\ 7.SP.C.8 Split the target event into two disjoint pieces. The event $a \ge b$ is the unio 7.SP.C.7 Count the ties. The ordered pairs with $a = b$ are $(1,1), (2,2), (3,3), (4,4), 7.SP.C.8 Use symmetry on the non-tie outcomes. The remaining $36 - 6 = 30$ ordered pairs 7.SP.C.7 Add the two pieces and divide by the size of the sample space to get the probabi Review
Reasonableness: Without ties, symmetry forces $P(a > b) = P(a < b) = \tfrac{1}{2} \cdot P(a \ne b)$. So $P(a \ge b) = P(a > b) + P(a = b) = \tfrac{1}{2}(1 - P(a = b)) + P(a = b) = \tfrac{1}{2} + \tfrac{1}{2} P(a = b)$. With $P(a = b) = \tfrac{6}{36} = \tfrac{1}{6}$, this gives $\tfrac{1}{2} + \tfrac{1}{12} = \tfrac{7}{12}$, matching (D). The answer also has to exceed $\tfrac{1}{2}$ (because ties are included on top of the symmetric strict case), which rules out (A), (B), (C).
Alternative: Tool #3 (Make a Systematic List): fix the second roll $b$ and count first rolls $a$ with $a \ge b$. For $b = 1$ there are $6$ choices; for $b = 2$, $5$; for $b = 3$, $4$; $\ldots$; for $b = 6$, $1$. Total favorable $= 6 + 5 + 4 + 3 + 2 + 1 = 21$, so the probability is $\tfrac{21}{36} = \tfrac{7}{12}$.
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Using the uniform model on the $36$ ordered pairs to compute $P(\text{event}) = \dfrac{\text{favorable outcomes}}{36}$ for both the ties and the final answer.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Treating two dice rolls as a compound event on the $6 \times 6$ sample space, and splitting the event $a \ge b$ into the disjoint sub-events $a > b$ and $a = b$.)
⭐ Two dice rolls have a built-in symmetry — swapping them shows $a > b$ and $a < b$ happen equally often, so all you have to do is add the ties on top.
⭐ Two dice rolls have a built-in symmetry — swapping them shows $a > b$ and $a < b$ happen equally often, so all you have to do is add the ties on top.
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