Nine Balls, One Heavy

Question

Nine visually identical balls; exactly one is slightly heavier. Using a two-pan balance (no weights), what is the minimum number of weighings that guarantees finding the heavy ball?

Accepted Answer

Two weighings. 1. Weigh 3 vs 3 (3 stay off). Tilt → heavy is in the heavier pan's trio; balance → it's in the trio that sat out. Either way: down to 3 suspects. 2. Weigh 1 vs 1 from the suspects. Tilt → that one; balance → the third. Lower bound: one weighing has only 3 outcomes but there are 9 hypotheses, a single weighing cannot separate them, so 2 is optimal. In general k weighings distinguish at most 3^k hypotheses; 9 = 3^2 makes this case exactly tight, which is also why the 3/3/3 split is forced: any other split (say 4 vs 4) leaves a branch with 4 suspects after one weighing, and 4 > 3 breaks the second weighing. The transferable frame: a balance is a ternary question. Most people instinctively binary-search (4 vs 4, then 2 vs 2…) and get 3 weighings, recognizing the instrument's true information capacity is the entire puzzle, and the same audit ("how many distinguishable outcomes does one test really have?") is the right first move on any measurement-budget problem.

Nine Balls, One Heavy

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