AMC 10 · 2015 · #18

Grade 7 rate-ratio
Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)
expected-valueprobability-basicsequences-geometric identify-subproblems ↑ Prerequisites: probability-basicexpected-value
📏 Medium solution 💡 2 insights
Problem
Johann starts with 64 fair coins and flips them all. Every coin that shows tails gets flipped again, and any coin still showing tails after that second flip gets one more flip. After at most three flips per coin, how many coins are expected to show heads?

Pick an answer.

(A)
32
(B)
40
(C)
48
(D)
56
(E)
64

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Tracking exactly when each coin first turns up heads is messy, so use tool #9 (Solve an Easier Related Problem): figure out the chance that a single coin ends on heads, then scale up to 64 coins because each coin behaves the same way and contributes independently. For that single coin, asking directly for "heads sometime in three flips" splits into three cases, but the opposite event is one clean case — tails three times in a row. That is tool #16 (Change Focus / Count the Complement): find the probability of all-tails and subtract from 1. Tool #5 (Look for a Pattern) confirms the same answer by the steady halving of how many coins are still tails after each round.

1STEP 1

Shrink to one coin, then scale

All 64 coins act the same and independently, so expected heads = 64 × the chance one coin ends heads; solve the one-coin case first.

Expected heads = 64 × P(one coin ends heads)
2STEP 2

Flip the question to all-tails

A coin ends heads unless it lands tails all three times, so count the one opposite case: tails, tails, tails.

P(ends heads) = 1 - P(tails, tails, tails)
3STEP 3

Compute the one-coin probability

Three independent tails give (1/2)³ = 1/8, so one coin ends heads with probability 1 - 1/8 = 7/8.

P(ends heads) = 1 - (1/2)³ = 1 - 1/8 = 7/8
4STEP 4

Scale back to 64 coins

Multiply by the 64 coins: 64 × 7/8 = 56 coins are expected to show heads. The answer is (D).

64 × 7/8 = (64 × 7)/8 = 56 → (D)
Answer
56
Check by counting the tails that survive each round, which is exactly the steady halving pattern. Start with 64; about half show heads and leave, so 32 tails go to round two; about half of those leave, so 16 tails go to round three; about half of those leave, so 8 are still tails at the end. Heads = 64 - 8 = 56, matching (D). The value also sits sensibly between 32 (one flip only) and 64 (impossible to reach), as expected for three flips.
💡Key takeaway

When "heads at least once" splits into many cases, count the one opposite case instead — a coin only fails by landing tails all three times, which happens just 1 time in 8, so 7 out of every 8 coins, or 56 of the 64, end up heads.

  • Shrink to one coin, then scale
  • Flip the question to all-tails
  • Compute the one-coin probability
  • Scale back to 64 coins

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