A.A. Bennett offered this puzzle in the American Mathematical Monthly of May 1937:
A car with n (n > 2) passengers of different speeds of mental reaction passes through a tunnel and each passenger acquires unconsciously a smudge of soot upon his forehead. Suppose that each passenger
(1) laughs and continues to laugh as soon as and only so long as he sees a smudge upon the forehead of a fellow passenger;
(2) can see the foreheads of all his fellows;
(3) reasons correctly;
(4) will clean his own forehead when and only when his reasoning forces him to conclude that he has a smudge;
(5) knows that (1), (2), (3), and (4) hold for each of his fellows.
Show that each passenger will eventually wipe his own forehead.
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I don’t have Bennett’s solution, but here’s mine:
Take the simplest case, with three people, A, B, and C. Each can see smudges on his friends’ foreheads, so all are laughing. A reasons: Suppose my own forehead is clean. Then B can see only one smudge, the one on C’s forehead. But B sees that C is laughing. This should prove to B that his own forehead must have a smudge, and he should wipe it off. He isn’t doing so. This must mean that this whole supposition — that my own forehead is clean — is false.
With that realization, A wipes his own forehead. And at that point his supposition comes true, and B cleans his own forehead, following the reasoning given above. Now C sees that A and B are laughing, and that they both have clean foreheads. This tells C that his own forehead must be smudged, and he cleans it.
So that’s the case with three people. Now start over and add D. D can think: If my own forehead were clean, all of the above reasoning would hold and A would clean his forehead. He isn’t doing so. That must mean that my own forehead is dirty. D cleans his forehead and everything else plays out as above.
Then similarly you can add E, and F, and as many additional people as you like, following the same idea.
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