Abstract
The surprise exam paradox is an apparently sound argument to the apparently absurd conclusion that a surprise exam cannot be given within a finite exam period. A closer look at the logic of the paradox shows the argument breaking down immediately. So why do the beginning stages of the argument appear sound in the first place? This paper presents an account of the paradox on which its allure is rooted in a common probabilistic mistake: the base rate fallacy. The account predicts that the paradoxical argument should get less and less convincing as it goes along—a prediction I take to be welcome. On a bleaker note, the account suggests that the base rate fallacy may be more widespread than previously thought.
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Notes
Quine (1966) classifies paradoxes into those that are genuine antinomies and those that are not. Among those that are not, some are falsidical and others are veridical. The conclusion of a veridical paradox is apparently absurd but actually true.
The student’s belief is neither safe from error in nearby worlds nor sensitive to the truth value of the proposition believed in those worlds. At any rate, the point that the announcement is consistent does not depend on any particular account of knowledge. Reflection on the ease with which real teachers fulfill real announcements (not relevantly dissimilar from the one in the paradox) suffices.
Proof: suppose \(K_{4}\)Exam. By Conjunction, we derive \(K_4(\)Exam \(\wedge \lnot E_1 \wedge \lnot E_2 \wedge \lnot E_3 \wedge \lnot E_4)\). Since (Exam \( \wedge \lnot E_1 \wedge \lnot E_2 \wedge \lnot E_3 \wedge \lnot E_4) \supset E_5\) is a tautology, Logic entails \(K_4((\)Exam \(\wedge \lnot E_1 \wedge \lnot E_2 \wedge \lnot E_3 \wedge \lnot E_4) \supset E_5)\). The desired conclusion follows by Entailment.
I say ‘presumably’ because nothing in what follows turns on the claim that, in a Friday-exam world, the student knows the announcement on Sunday. That claim is false if the following ‘Confidence’ principle is true: if one knows p, one knows that one will continue to know p. For what it is worth, my opinion is that the student may know the announcement on Sunday, even in a world in which that knowledge is later defeated. Future defeat is consistent with present knowledge. However, while Confidence is false, there is a true principle in the vicinity: see Hall (1999). Anyway, the issue is orthogonal to the solution defended in this section, which relies only on the point that if the exam is given Friday, the student cannot know the announcement on Thursday, provided that she knows on Thursday that no exam was given Monday through Thursday.
The reference is to Quine (1953).
This is Ayer (1973)’s response to Quine. His story is that the student witnesses the teacher shuffle an ace of spades into a pile of four other cards, where the protocol is that the teacher will keep the cards in view of the student throughout the week, draw a card without replacement from the top of the deck each day, and give the exam on the day the ace of spades is drawn. The ‘rule of the school’ idea is Kripke (2011)’s.
Note that the first step is sound on and only on Ayer’s enrichment. For suppose that Ayer’s enrichment is false. Then there are two possible cases: in \(\alpha \), the student’s total evidence supports Exam and Surprise to an equal degree, as in the unenriched paradox, where the student’s total evidence is exhausted by the teacher’s testimony. In \(\beta \), the student’s total evidence supports Exam to a lesser degree than it supports Surprise. Suppose that \(\alpha \) obtains. Then (ii) and (iii) are either both true or both false, since in \(\alpha \) the student’s evidence does not distinguish between Exam and Surprise. If (ii) and (iii) are both false, the first step is unsound. If (ii) and (iii) are both true, then by Quine’s bind (i) is false and the first step is again unsound. Suppose, on the other hand, that \(\beta \) obtains. Then (ii) is true only if (iii) is as well, since in \(\beta \) the student’s evidence for Surprise is even stronger than her evidence for Exam. By Quine’s bind, (ii) and (iii) are true only if (i) is false. Hence, in \(\beta \), (ii) is true only if (i) is false. Since the first step relies on both, it is unsound in \(\beta \). In either case, then, the first step is unsound. By conditional introduction, it follows that if Ayer’s enrichment is false, the first step is unsound. By contraposition, it follows that if the first step is sound, Ayer’s enrichment is true: i.e. the first step is sound only on Ayer’s enrichment.
The account sketched in this section draws from Kripke (2011), Jackson (1987), and Wright and Sudbury (1977). A virtue of the account is that it can explain several different versions of the paradox. Some commentators define “surprise” in terms of justified belief (J), rather than knowledge. Even though J is non-factive, the account still applies, provided that J abides by an enkratic constraint according to which, for any i, \(\lnot J_{i}(p \wedge \lnot J_{i}p)\). Other commentators have shown that the diachronic aspect of the paradox is inessential: see Sorensen (1988)’s version of the paradox involving multiple students facing forward in a line. The account still applies in the synchronic case. As in the original paradox, the student relies on an illicit assumption concerning an epistemic state distinct from her own at the time of her reasoning. The only difference is that, in the synchronic case, the assumption concerns the epistemic state of another student at the same time, rather than the student’s own epistemic state at a later time. See Williamson (2000), p. 138, on this point.
Let it be understood that, in what follows, I am taking on board the canonical solution to the paradox, as presented in Sect. 2. Now, this solution follows from a purely logical, non-probabilistic interpretation of the paradox. In subsequent sections, I give a particular probabilistic interpretation of the paradox, and I use it to explain the allure of the student’s reasoning. But the solution to the paradox given in Sect. 2 does not depend on my particular probabilistic interpretation of the paradox or any probabilistic interpretation whatsoever. Further, let it be understood that my specific probabilistic interpretation of the paradox is not the only one available. For a different probabilistic account, see Hall (1999).
The case is from the Wikipedia entry on the base rate fallacy.
In stating his thesis, Stalnaker (1970) applies degrees of belief to conditional sentences, rather than conditional propositions. Alternatively, some understand the thesis in terms of degrees of assertability of conditional sentences. For more on this latter approach, see DeRose (2010). The differences between these versions of the thesis will not matter for present purposes, although it is worth noting that the propositional version of the thesis has been called into question by the triviality results of, for example, Lewis (1976). However, those results may be avoidable on a contextualist theory of conditionals, such as the one developed in Bacon (2015).
Note that a simple alternative to Stalnaker’s Thesis, according to which the probability of a conditional is equal to the probability of the associated material conditional, falters badly in the card case:
$$\begin{aligned} \begin{aligned} P(CLUB \rightarrow SIX)&= P(CLUB \supset SIX) = P(\lnot CLUB \vee SIX) \\&= P(\lnot CLUB) + P(SIX) - P(\lnot CLUB \wedge SIX) \\&= 39/52 + 4/52 - 3/52 = 10/13. \end{aligned} \end{aligned}$$Of course, one should not be \(77\%\) confident that the card is a six if it is a club.
A more general proof, following Edgington (2013): given any propositions X, Y and Z:
$$\begin{aligned} P(X \mid Y)&= \dfrac{P(X \wedge Y)}{P(Y)} = \dfrac{P(X \wedge Y \wedge Z) + P(X \wedge Y \wedge \lnot Z)}{P(Y)} \\&= \dfrac{P(X \mid (Y \wedge Z)) \times P(Y \wedge Z) + P(X \mid (Y \wedge \lnot Z)) \times P(Y \wedge \lnot Z)}{P(Y)} \\&= P(X \mid (Y \wedge Z)) \times P(Z \mid Y) + P(X \mid (Y \wedge \lnot Z)) \times P(\lnot Z \mid Y). \end{aligned}$$Kaufmann (2004) introduced versions of Eqs. 5 and 6 while discussing purported counterexamples to Stalnaker’s Thesis. Equation 5 is partition-invariant in the sense that the choice of \({\mathcal {Z}}\) is irrelevant to the value of \(P(X \mid Y)\). But Eq. 6 is partition-dependent. The choice of \({\mathcal {Z}}\) matters very much to the value of \(P(X \not \mid Y)\). For example, letting \({\mathcal {Z}} = \{X, \lnot X\}\), Eq. 6 entails that \(P(X \not \mid Y) = P(X)\). Douven (2008) presents another example that dramatizes the partition-dependence of Eq. 6. See Khoo (2016) for a discussion of partition selection.
My account of the first step of the student’s argument relies on some claims about conditionals. Because of this, one might worry that the account is parochial. For the first step of the student’s argument could just as easily be presented without any conditionals, as a disjunctive syllogism, in the following way: either the exam will not be given Friday or else it will not be a surprise; it will be a surprise; so it will not be given Friday. However, it is well-known that English speakers are quick to infer from \(X \vee Y\) to \(\lnot X \rightarrow Y\) (see, e.g., Stalnaker (1975)’s “direct argument”, Jackson (1987)’s “passage principle”, and Bennett (2003)’s “or-to-if inference”). This inference is so automatic that, I suspect, any temptation to accept the disjunction is, in the first instance, a temptation to accept the inferred conditional. Thanks to [removed] for helpful discussion on this point.
More carefully, the account suggests that NOSUR is false in the canonical version of the paradox, sans Ayer’s enrichment.
Why use this partition? Khoo (2016) argues that the context in which a conditional is uttered supplies a salient question which serves as the appropriate partition. The question of whether the student will know a crucial part of the announcement on Thursday strikes me as a particularly salient question in the context of the first step of the student’s argument.
In fact, \(P(K_{4}\text {Exam})\) should be greater than 2/3, since the possibility in which no exam is given deserves lower credence than the other possibilities. After all, the teacher promised an exam and, at least at the outset of the student’s argument, we have no reason to doubt that she will follow through with that promise.
This sentiment is echoed in Kripke (2011), p. 29.
Here I assume that we are confident that the student will be able to run the first step of the argument on Wednesday, so as to come to know that the exam will not be given Friday. This assumption is also implicit in the claim that \(P(K_{3}\text {Exam} \mid E_4) = 0\). Indeed, throughout this section I make similar assumptions, according to which we are confident that the student will be able to run various steps of the argument on various days of the week. For reasons discussed at the end of Sect. 2, we may not be justified in being so confident. Still, the assumptions are innocuous here, as the purpose of this section is not to identify what we are justified in believing about the paradox but rather to identify what we might be doing when we mistakenly accommodate the student’s reasoning. (The base rate fallacious probabilities are fallacious, after all.)
Thanks to an anonymous referee for raising this challenge and the related challenge below.
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Acknowledgements
My foremost thanks go to Stephen Yablo and Roger White for invaluable feedback on the ideas in this paper. I also want to thank Andrew Bacon, Alex Byrne, Nilanjan Das, Ned Hall, Justin Khoo, Bernhard Salow, Robert Stalnaker and audiences at Cambridge University, Caltech and MIT.
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Wells, I. No Surprises. Erkenn 86, 389–406 (2021). https://doi.org/10.1007/s10670-019-00110-9
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DOI: https://doi.org/10.1007/s10670-019-00110-9