The standard version of the surprise exam paradox is that the teacher announces that next week there will be a surprise exam: an exam whose occurrence will surprise the students. The students are smart and reason: it can't be on Friday, since if Monday through Thursday pass without an exam, they'd know that it would be on Friday and wouldn't be surprised. But likewise it can't be on Thursday, since they know it can't be on Friday, and so if Monday through Wednesday pass without an exam, they'll know by Thursday that it's on Thursday, and won't be surprised. Repeating this reasoning, the exam has to be on Monday, but then that won't be a surprise either. So a surprise exam is impossible, which is paradoxical. Besides, then, despite all the reasoning, the exam happens on, say, Tuesday and the students really are surprised.
The above version is over a span of 5 days. Generalize by supposing a span of n days for the surprise exam. I really don't know the surprise exam literature, so there may be nothing new here.
First thought: The n=1 case is already a bit paradoxical. The teacher announces: "We will have a surprise exam on Monday." Students are puzzled. Since they know that it will be on Monday, how can it be a surprise to them when it happens on Monday? Should they conclude that the teacher has just told them something obviously false? But if so, then they don't know that there will be an exam on Monday. And then when Monday rolls around, they are quite open to being surprised by the occurrence of the exam. So maybe what the teacher told them isn't obviously false. So charity suggests that they believe the teacher—there really will be a surprise exam on Monday. But if so, then once again they won't be surprised, and so the teacher is telling them something false, and so they should dismiss it. And so on: They keep on thinking this through, and Monday rolls around, and they fail the exam because instead of studying for it, they were thinking about whether there would be an exam! So the n=1 case is paradoxical. It is interesting to ask: Does the n=1 case contain all of the paradoxicality of the n=5 case?
Second thought. Suppose we have some probability cutoffs the define assertibility and surprise: something is assertible provided it has probability at least α and it's surprising provided it had probability less than β. Maybe the values are α=0.9 and β=0.1. I'll do the examples with those. Suppose now that the teacher genuinely will set up an exam at a random date, with some probability distribution on the days 1,2,...,n. First, suppose the distribution is known to the students to be uniform. If the exam is on day n, there is no surprise. But that isn't enough to undercut the assertibility of the teacher's statement. For the probability that the exam would end up on day n is only 1/n, and as long 1−α≥1/n, the teacher might not be taking an undue risk. But we do get a constraint here: n≥1/(1−α). With our sample numbers, this means n≥1/(1−0.9)=10. So we don't have assertibility in the original version where n=5.
Let's keep ploughing through and see what other constraints there are. On the 1/β (rounded down) last days, the probability each day that there would be an exam on that day, if we get to that day examless, is greater than or equal to β, and so there is no surprise if the exam is then. So for assertibility, we better have approximately (I am ignoring rounding) (1/β)/n≤1−α, or n≥1/(β(1−α)). And if we have that approximately, then the teacher has assertibility in saying "There will be a surprise exam over the next n class days." With our sample numbers, the constraint is n≥100. So on our probabilistic understanding of surprise and assertibility (and rounding issues don't come up since in our case 1/β=10 exactly), you can honestly announce a surprise exam if there are at least 100 days that the exam might be on, and then just choose a day uniformly randomly, and even tell the students that's how you're choosing.
It would be fun to see if other distributions than the uniform one might not allow one to bring down the n≥100 constraint.
With respect to the n=1 case, the exam seems to be surprising iff the students don't believe there will be a surprise exam. That is, the teacher has asserted a P which has the logical properties of "You don't believe this proposition." That is a pretty paradoxical assertion if you think about it; when would it be rational to believe?
ReplyDeleteI think that in my probabilistic version, with the values of alpha and beta as given, the optimal distribution will let one have assertibility for n from 23 up. Below that, one can't have assertibility about surprise.
ReplyDeleteIf our assertibility level is .9, then we can assert surprise in the original n=5 case at a surprise level of .44 (very mild surprise) if we choose the right distribution.
ReplyDeleteI think the optimal distribution is one where P(k) = beta (1-beta)^(k-1) for k<n and P(n) = (1-beta)^(n-1). On this distribution an exam on every day but the last is surprising (at level beta).
Optimality here is maximization of assertibility of surprise.
ReplyDeleteFor my optimality stuff, I am assuming the criterion for surprise is probability less than our equal to beta, not less than beta as in the post.
ReplyDeleteAbout the only way to keep this exam a surprise is for the teacher to not say anything and give the students the exam! Now that would be something! The students will surely cry foul. When I was at the University of Michigan, we sometimes got to rate our professors. We would give as good as we got!
ReplyDelete"The students are smart and reason: it can't be on Friday," What I'll say to that is "Don't fool yourselves, kiddies." I once had the guy from India who was a teaching assistant for our calculus class. He gave an exam on a Friday evening of all times! How dare he! Didn't he know that Friday evenings were sacred for partying? Didn't he know this was America and not India?
Amusingly, the students can refuse to be surprised by the exam by assigning credence C(n) to the exam being on day n in a sufficiently steeply exponentially decreasing way. For then they expect the exam to be on day 1, and if it's not on day 1, on day 2, and so on.
ReplyDelete