AMC 8 · 2011 · #18
Easy mode Grade 7Problem
Imagine rolling a fair 6-sided die twice in a row.
Look at the number that comes up on the first roll, and the number on the second roll.
What is the probability that the first number 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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