AMC 10 · 2018 · #9

Grade 7 probability
Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery
probability-basicsymmetry-argument symmetry-argument ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
Seven ordinary dice (each face 1 to 6) are rolled and the top faces are added. The total 10 comes up with some probability p. Find which other total comes up with that same probability p.

Pick an answer.

(A)
$text{ 13}$
(B)
$text{ 26}$
(C)
$text{ 32}$
(D)
$text{ 39}$
(E)
$text{ 42}$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Tool #9 (Solve an Easier Related Problem): counting all the ways seven dice add to 10 is hard, so first look at a single die to see how its values pair up. Tool #16 (Change Focus / Count the Complement): instead of counting, flip every die by sending each value d to 7-d. This pairs each roll with exactly one other roll, turning the sum S into 49-S, so the two sums must be equally likely. Tool #3 (Eliminate Possibilities): the partner of 10 must match one of the five choices, which pins down the answer.

1STEP 1

Look at one die first

For one die, value d is as likely as its partner 7-d: 1⇔6, 2⇔5, 3⇔4 each add to 7.

1⇔ 6, 2⇔ 5, 3⇔ 4, d⇔ 7-d
2STEP 2

Flip all seven dice

Flip all 7 dice at once by sending each d to 7-d; this pairs every roll one-to-one with another, since flipping twice returns it.

(d₁,…,d₇) ⟼ (7-d₁,…,7-d₇)
3STEP 3

See what flipping does to the sum

If the original faces sum to S, the flipped faces sum to seven 7s minus S, which is 49-S.

Σ_i=1⁷(7-d_i)=7· 7-Σ_i=1⁷ d_i = 49 - S
4STEP 4

Equal counts mean equal probability

The flip pairs each sum-S roll with exactly one sum-(49-S) roll, so the counts match and P(S)=P(49-S).

#(sum=S)=#(sum=49-S) → P(S)=P(49-S)
5STEP 5

Apply it to the sum 10

The partner of 10 is 49-10=39, the only choice present, (D); 13, 26, 32, 42 are not the mirror of 10.

49-10=39 → (D)
Answer
text{ 39}
The pairing rule says sums S and 49-S are equally likely, so the whole picture is symmetric about 49/2=24.5. The sum 10 sits 14.5 below the center, and 39 sits 14.5 above it, so they mirror perfectly. Both 10 and 39 are inside the possible range [7,42], which a real answer must be. The other choices fail: 42 is the rare all-sixes maximum (far less likely than 10), while 13, 26, and 32 are not 49-10. Only (D) 39 fits.
💡Key takeaway

Swap every die for its opposite face (d becomes 7-d): the total S flips to 49-S, so 10 and 49-10=39 happen equally often.

  • Look at one die first
  • Flip all seven dice
  • See what flipping does to the sum
  • Equal counts mean equal probability
  • Apply it to the sum 10

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