AMC 8 · 2011 · #18

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basicsystematic-enumerationcomplementary-counting complementary-countingsystematic-enumeration ↑ Prerequisites: probability-basic

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Problem

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.

Pick an answer.

(A)

dfrac16

(B)

$dfrac5{12}$

(C)

dfrac12

(D)

$dfrac7{12}$

(E)

dfrac56

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find Symmetry

The sample space is the 6 × 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) ↦ (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 ≥ 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.

1STEP 1

Each roll is independent and uniform on {1, …, 6}, so the ordered pair (a, b) takes 36 equally likely values.

|Ω| = 6 × 6 = 36

2STEP 2

Split a ≥ b into two disjoint pieces — strict a ＞ b and ties a = b — so P(a ≥ b) = P(a ＞ b) + P(a = b).

{a ≥ b} = {a ＞ b} a = b

3STEP 3

The ordered pairs with a = b are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — exactly 6 outcomes.

|{a = b}| = 6

4STEP 4

By the swap (a, b) ↦ (b, a), the 30 non-tie pairs split equally between a ＞ b and a ＜ b, giving |{a ＞ b}| = 15.

|{a ＞ b}| = |{a ＜ b}| =

\frac{30}{2}

= 15

5STEP 5

Add the pieces: 21 favorable outcomes out of 36 give the probability.

P(a ≥ b) =

\frac{15 + 6}{36}

\frac{21}{36}

\frac{7}{12}

→ (D)

Answer

dfrac7{12}

Without ties, symmetry forces P(a ＞ b) = P(a ＜ b) =

\frac{1}{2}

· P(a ≠ b). So P(a ≥ b) = P(a ＞ b) + P(a = b) =

\frac{1}{2}

(1 - P(a = b)) + P(a = b) =

\frac{1}{2}

\frac{1}{2}

P(a = b). With P(a = b) =

\frac{6}{36}

\frac{1}{6}

, this gives

\frac{1}{2}

\frac{1}{12}

\frac{7}{12}

, matching (D). The answer also has to exceed

\frac{1}{2}

(because ties are included on top of the symmetric strict case), which rules out (A), (B), (C).

💡Key takeaway

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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