Red or Black - Futility Closet

Six sheets are set out in a room. Each identifies a different date in the same month. Weekdays are printed in black and Sundays in red. Six people will enter the room, one by one. Before the first one enters, one sheet is turned face down. Candidate 1 is then asked if she can deduce the color of the inverted sheet by examining the other sheets. Her answer, yes or no, is written on the back of the inverted sheet, followed by her number, 1. When 1 departs, a second sheet is turned face down. Candidate 2 enters and is asked whether she can deduce the color of the second sheet by considering her predecessor’s answer and the four face-up sheets. Her answer is noted in its turn, and this process continues — when the sixth candidate enters, she sees six face-down sheets, the first five of which bear the answers of the first five candidates. If all five of these answers are no, can Candidate 6 answer yes?

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Yes. The fact that Candidate 1 wrote no means that she can’t have seen more than four red sheets — if she’d seen five then she would have written yes, because a month cannot have more than five Sundays, and this fact would have told her that her own sheet was black. Similarly, the fact that Candidate 2 wrote no means that she can’t have seen more than three red sheets, because if she’d seen four red sheets (sheets 3, 4, 5, and 6) then she could infer that sheet 2 must be black, as Candidate 1 could not have seen five red sheets, per the reasoning above. And so on: Since Candidates 3, 4, and 5 all wrote no, they can have seen at most 2, 1, and 0 red sheets. And if Candidate 6 knows that Candidate 5 saw 0 red sheets, she can announce that her own sheet was black. (From Roland Sprague, Recreation in Mathematics, 1963.)

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