AMC 10 · 2020 · #25

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

probability-basicexpected-valuecaseworksystematic-enumerationconditional-probability caseworkidentify-subproblemscomplementary-counting ↑ Prerequisites: probability-basicexpected-value

📏 Long solution 💡 4 insights

Problem

Jason rolls three fair six-sided dice, sees the result, then chooses which subset of the three to reroll so as to maximize his probability of ending with sum exactly 7. What is the probability — over the random initial roll — that the optimal subset to reroll contains exactly two dice?

Pick an answer.

(A)

$\frac{7}{36}$

(B)

$\frac{5}{24}$

(C)

$\frac{2}{9}$

(D)

$\frac{17}{72}$

(E)

$\frac{1}{4}$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): four strategies (reroll 0, 1, 2, or 3) — compute each one's win probability as a separate subproblem, then compare. Tool #9 (Easier Problem): the count of two-dice sums equaling s is a tiny lookup (s-1 for s ≤ 7). Tool #2 (Systematic List): enumerate sorted triples (a, b, c) with a ≤ b ≤ c satisfying the derived conditions, multiplying by permutation counts. Tool #3 (Eliminate Possibilities): inequality reroll-2 vs. reroll-1 rules out a whole family of states; another inequality vs. reroll-3 restricts the kept die to a ∈ {1,2,3}.

1STEP 1

Reroll 0 wins only if the roll already sums to 7. Reroll 3: the triples summing to 7 give $\frac{15}{216}$ = $\frac{5}{72}$ .

P(reroll 3) =

\frac{15}{216}

\frac{5}{72}

2STEP 2

Reroll 1: keep the two smallest dice; the rerolled die hits its one target only if a+b ≤ 6, giving $\frac{1}{6}$ (else 0).

P(reroll 1) =

\frac{1}{6}

& if a+b ≤ 6 ; 0 & if a+b ≥ 7

3STEP 3

Reroll 2: keep the smallest die a; the pair must sum to 7-a, which has 6-a ways, so win probability $\frac{6-a}{36}$ .

P(reroll 2) =

\frac{6 - a}{36}

4STEP 4

Reroll 2 peaks at $\frac{5}{36}$ , below $\frac{1}{6}$ , so reroll 1 always wins when available — reroll 2 needs it killed: a+b ≥ 7.

\frac{6-a}{36}

＜

\frac{6}{36}

\frac{1}{6}

→ need a+b ≥ 7

5STEP 5

Reroll 2 beats reroll 3's $\frac{5}{72}$ only when 12-2a exceeds 5, i.e. a below 3.5 — so the kept die is a ∈ {1, 2, 3}.

a ∈ {1, 2, 3}

6STEP 6

Reroll 0 needs an initial sum of 7, but a+b ≥ 7 pushes the total to at least 8 — so reroll 0 is automatically out.

a + b ≥ 7 → a+b+c ≥ 8 → initial sum ≠ 7

7STEP 7

Casework on a under a+b ≥ 7: a=1 → (1,6,6) is 3; a=2 → 12; a=3 → 27 ordered triples.

|a=1|=3, |a=2|=12, |a=3|=27

8STEP 8

Add the counts: 3+12+27 = 42 favorable out of 216, so the probability is $\frac{42}{216}$ = $\frac{7}{36}$ , choice (A).

P =

\frac{42}{216}

\frac{7}{36}

→ (A)

Answer

\frac{7}{36}

Quick sanity.

\frac{42}{216}

simplifies via gcd(42, 216) = 6 to

\frac{7}{36}

. The cases are mutually exclusive (a sorted triple has one a value), so no double-counting. Spot-check (1,6,6): a=1, reroll-2 keeps the 1, need pair to sum to 6, which has 5 ordered ways ((1,5),(2,4),(3,3),(4,2),(5,1)), prob

\frac{5}{36}

≈ 0.139. Reroll-1 alternative would keep (6,6) summing to 12 ＞ 6, infeasible — confirming reroll-1 ruled out. Reroll-3 gives

\frac{5}{72}

≈ 0.069, lower than

\frac{5}{36}

. Reroll-0 has sum 13 ≠ 7, gives 0. So reroll-2 wins. Spot-check (4,5,6): a=4 ∉ {1,2,3} — so reroll-2 NOT optimal here. Reroll-1 feasible? a+b = 9 ＞ 6, infeasible. Reroll-3 gives

\frac{5}{72}

. Reroll-2 keeping 4 gives

\frac{6-4}{36}

\frac{2}{36}

\frac{4}{72}

＜

\frac{5}{72}

— so reroll-3 wins. Confirms filter a ≤ 3.

💡Key takeaway

This AMC 10 problem only needs Grade 7 dice-probability and inequality comparisons you already know — rerolling 2 dice is optimal only when (1) every pair of two dice exceeds sum 6 (a+b ≥ 7) and (2) the smallest die is in {1,2,3}; listing the 42 such ordered triples out of 216 gives $\frac{42}{216}$ = $\frac{7}{36}$ .

More like this