Black and white hats
A group plays a game: each person gets a white or black hat at random (at least one black hat). Everyone sees others’ hats, not their own.
The host asks them to infer their own color from what they see. The procedure: lights off—anyone who believes they wear a black hat shouts. If nobody shouts, lights on for observation, then lights off again; repeat.
Known:
- First dark period: silence;
- After lights on, second dark period: still silence;
- Third dark period: someone shouts.
How many black hats are there? (Assume perfect rationality and everyone knows the rules and “at least one black hat.”)
Three black hats.
Inductive reasoning
Let k be the number of black hats. Only someone who believes they wear black will shout when the lights go off.
If k = 1: the sole black-hat wearer sees only white hats and knows “at least one black” → it must be themselves → they shout on the first dark period.
That contradicts the puzzle (silence on round 1). After the first silence, everyone infers k ≥ 2.
If k = 2: each black-hat person sees one black hat. They reason:
- If k = 1, that person would have shouted in round 1;
- No shout in round 1 → k ≥ 2;
- I see one black → k = 2 and I am the other black → shout in round 2.
Silence in round 2 → k ≥ 3.
If k = 3: each black-hat person sees two black hats. After two silent rounds, k ≥ 3; if k = 3, “I see two blacks, so the third must be me” → shout in round 3.
This matches the puzzle.
If k ≥ 4: each black-hat wearer sees at least two blacks; after only two silent rounds they cannot yet tell whether k = 3 or they are a fourth black—so they would not all shout on round 3. So k cannot be ≥ 4.
Therefore k = 3.
General rule
With k black hats under the same rules, the first shout happens on dark period k (the first k−1 rounds are silent). A shout on round 3 means three black hats.
(Same structure as the “blue eyes” village puzzle: silence length encodes common knowledge about how many black hats there are.)