AMC 10 · 2020 · #23
Grade 7 probabilityPick an answer.
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}.
Rerolling all three
Rerolling all three is low.
List the four unordered triples that sum to 7 and count permutations.
7.SP.C.8Identify SubproblemsRerolling exactly one
When possible it is a solid one sixth.
Keep the smallest two; the rerolled die has one winning face out of six.
7.SP.C.7Identify SubproblemsRerolling exactly two
A smaller kept die is better.
Keep the smallest die so the pair has the widest target sum.
7.SP.C.8Solve An Easier Related ProblemWhen two wins
The one-die option must be blocked.
If reroll-1 is on the table, it always wins — so we need to kill it first.
If one option always beats the others, it has to be ruled out before the remaining cases mean anything.
▸ Why?
An option that dominates every rival is chosen whenever it is available, so no other case can occur.
▸ Why?
The cases never overlap, so once the dominant one is removed the rest can be counted separately.
Which die to keep
Only three values can be kept.
Bigger kept die hurts — too small a target sum gives too few rerolled pairs.
7.RP.A.3Eliminate PossibilitiesCheck the initial sum is not seven
In this case the initial sum cannot be seven.
Once a+b ≥ 7, the total is at least 8, so reroll-0 wins with probability 0.
7.SP.C.8Eliminate PossibilitiesCount the cases
Count the cases for each value.
Systematic by a, then b, then c — each gives a small number of multisets.
7.SP.C.8Make A Systematic ListTake the probability
Dividing gives seven thirty-sixths.
Divide favorable by total — simplify the fraction.
7.SP.C.7Identify SubproblemsThis AMC 12 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 42/216 = 7/36.