For clarification, the people can all see the color of other people’s hats, right? If so, then:
The people agree in advance (using the axiom of choice!) on a set of representatives for all equivalence classes in 2N/∼, where two bit sequences are equivalent under ∼ iff they differ in only finitely many positions.
When they are guessing, they all know the equivalence class they are in and they guess the color for their hat that is implied by the agreed upon representative of that equivalence class. By construction, only finitely many of them can guess incorrectly.
For clarification, the people can all see the color of other people’s hats, right? If so, then:
The people agree in advance (using the axiom of choice!) on a set of representatives for all equivalence classes in 2N/∼, where two bit sequences are equivalent under ∼ iff they differ in only finitely many positions.
When they are guessing, they all know the equivalence class they are in and they guess the color for their hat that is implied by the agreed upon representative of that equivalence class. By construction, only finitely many of them can guess incorrectly.
Yes they can see each others hats.