What about something like “The pupil won’t find a proof by start-of-day, that the day is exam day, if the day is in fact exam day.”
This way, the teacher isn’t denying “for any day”, only for the one exam day.
Can such a statement be true?
Well, the teacher could follow a randomized strategy. If the teacher puts 1/5th probability on each weekday, then there is a 1/5th chance that the exam will be on Friday, so the teacher will “lose” (will have told a lie), since the students will know it must be exam day. But this leaves a 4/5ths chance of success.
Perhaps the teacher should exclude Friday from the distribution, instead placing a 1/4th chance on each weekday before Friday. If we treat the situation game-theoretically, so we make the typical assumption that agents can know each other’s mixed strategies, then this would be a foolish mistake for the teacher—there’s now a 1/4th probability of lying rather than a 1/5th. (Instead, the teacher should place arbitrarily small but positive probability on Friday, to minimize chances of lying.)
But so long as we are staying in the deductive, realm, there is no reason to make that game-theoretic assumption. If the students and teacher are both reasoning in PA, then the students do not trust the teacher’s reasoning to be correct; so there is not common knowledge of rationality.
So it seems to me that in the purely deductive version of the problem, the teacher can keep their word; and in the game-theoretic version, the teacher can keep their word with arbitrarily high probability (so long as we are satisfied with arbitrarily small “surprise”).
What about something like “The pupil won’t find a proof by start-of-day, that the day is exam day, if the day is in fact exam day.”
This way, the teacher isn’t denying “for any day”, only for the one exam day.
Can such a statement be true?
Well, the teacher could follow a randomized strategy. If the teacher puts 1/5th probability on each weekday, then there is a 1/5th chance that the exam will be on Friday, so the teacher will “lose” (will have told a lie), since the students will know it must be exam day. But this leaves a 4/5ths chance of success.
Perhaps the teacher should exclude Friday from the distribution, instead placing a 1/4th chance on each weekday before Friday. If we treat the situation game-theoretically, so we make the typical assumption that agents can know each other’s mixed strategies, then this would be a foolish mistake for the teacher—there’s now a 1/4th probability of lying rather than a 1/5th. (Instead, the teacher should place arbitrarily small but positive probability on Friday, to minimize chances of lying.)
But so long as we are staying in the deductive, realm, there is no reason to make that game-theoretic assumption. If the students and teacher are both reasoning in PA, then the students do not trust the teacher’s reasoning to be correct; so there is not common knowledge of rationality.
So it seems to me that in the purely deductive version of the problem, the teacher can keep their word; and in the game-theoretic version, the teacher can keep their word with arbitrarily high probability (so long as we are satisfied with arbitrarily small “surprise”).
Huh? It seems to me that in the deductive version the student will still, every day, find proofs that the exam is on all days.