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A riddle was posted in this mathoverflow question: https://mathoverflow.net/questions/85439/how-does-intuitionism-handle-this-riddle

A riddle: You and another person are kidnapped and knocked unconscious by a demented villain. When you wake up, you are told that some of you may have an ink dot on your forehead. You can see the other person's forehead but not your own. You must each privately guess as to the status of your forehead. At least one of you must be right, or you will both be killed. No talking or signaling is allowed; this also will result in death.

Now, if you survive this, you and the other person will be transported elsewhere, never to see each other again. You will never know what was on your own forehead.

The solution to this riddle relies on the fact that there are exactly four possibilities: You both have a dot, neither has a dot, you do and she doesn't, she does and you don't. I'll leave it to the reader to figure out the strategy.

Even with the hint, and assuming that the configuration of dots was chosen uniformly at random (which is not stated in the problem), I don't see how to do better than just randomly guessing whether or not I have a dot, with a 25% chance of death.

If I'm allowed to communicate in advance with the other prisoner and devise a strategy, we can guarantee freedom by e.g. one of us guessing what he sees, and the other guessing the opposite of what he sees. But communication is explicitly forbidden in the question.

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  • $\begingroup$ Yes, but I assume that's referring to the intuitionism angle of the question and not the riddle itself. (At least, there's nothing in the comments suggesting that the riddle is ill-posed, though of course that's a possibility.) $\endgroup$ – user7530 Jan 11 '12 at 21:08
  • $\begingroup$ This riddle has no solution. Voting to close. $\endgroup$ – TonyK Jan 12 '12 at 0:20
  • $\begingroup$ Problem has been resolved: the OP (on mathoverflow) admitted that he meant that you were allowed to strategize beforehand. (Before the dots were drawn? Before you were kidnapped? Who knows.) Which makes the problem trivial. $\endgroup$ – mjqxxxx Jan 12 '12 at 5:00
  • $\begingroup$ If you are allowed to formulate a strategy beforehand, one player should agree to guess the same as what he sees on the other's forehead, and the other should guess the opposite of what she sees on the other's forehead. Exactly one of the two people will be right no matter the combination. $\endgroup$ – Marconius Jul 5 '15 at 22:23
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If the villain isn't sufficiently demented to tell you things you already knew, you could reason like this: Since the villain had to tell you that some of you may have an ink dot on your forehead, you don't both have one; thus if you both guess that you don't have one, one of you must be right.

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  • $\begingroup$ Ah, that must be it. Thanks! $\endgroup$ – user7530 Jan 11 '12 at 21:14
  • $\begingroup$ That's nonsense. This riddle is broken. $\endgroup$ – TonyK Jan 11 '12 at 21:33
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    $\begingroup$ @Tony: Why is it rubbish? I agree the riddle may well be broken; I was just trying to offer a solution under the assumption that there is one. $\endgroup$ – joriki Jan 11 '12 at 21:52
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    $\begingroup$ @Tony: You're not distinguishing between the explicit content of the statement and what we can infer from it being uttered in a certain social situation. I didn't say that the statement can be construed to have that explicit content, only that we can infer something from the fact that the villain utters it. We do that all the time with everyday speech. $\endgroup$ – joriki Jan 12 '12 at 0:21
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    $\begingroup$ @mjqxxxx: I didn't take that paragraph to be part of the riddle. Certainly "I'll leave it to the reader to figure out the strategy." isn't part of the riddle, and the reference to "this riddle" also seems to indicate that the entire last paragraph is a comment on the posing of the riddle. Also the four mutually exclusive and exhaustive possibilities were listed in the context of asking how intuitionism handles the riddle; the focus was on the fact that they're mutually exclusive and exhaustive in classical logic, which doesn't preclude one of them being ruled out by the concrete situation. $\endgroup$ – joriki Jan 12 '12 at 20:15

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