How to expect a surprising exam

Brian Kim; Anubav Vasudevan

doi:10.1007/s11229-016-1096-y

Synthese

August 2017, Volume 194, Issue 8, pp 3101–3133 | Cite as

How to expect a surprising exam

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Brian Kim
Anubav Vasudevan

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First Online: 10 May 2016

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Abstract

In this paper, we provide a Bayesian analysis of the well-known surprise exam paradox. Central to our analysis is a probabilistic account of what it means for the student to accept the teacher’s announcement that he will receive a surprise exam. According to this account, the student can be said to have accepted the teacher’s announcement provided he adopts a subjective probability distribution relative to which he expects to receive the exam on a day on which he expects not to receive it. We show that as long as expectation is not equated with subjective certainty there will be contexts in which it is possible for the student to accept the teacher’s announcement, in this sense. In addition, we show how a Bayesian modeling of the scenario can yield plausible explanations of the following three intuitive claims: (1) the teacher’s announcement becomes easier to accept the more days there are in class; (2) a strict interpretation of the teacher’s announcement does not provide the student with any categorical information as to the date of the exam; and (3) the teacher’s announcement contains less information about the date of the exam the more days there are in class. To conclude, we show how the surprise exam paradox can be seen as one among the larger class of paradoxes of doxastic fallibilism, foremost among which is the paradox of the preface.

Keywords

Surprise exam paradox Probability Fallibilism Preface paradox Bayesianism

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Appendix 1

For a given probability P, let $\sigma ^0_P=\sigma _P$ and let:

Open image in new window

Given, the event $S_{m-1}(P)$ ($m\ge 1$), put

Open image in new window

and

Open image in new window

In this appendix, we will prove the following result:

Theorem 5

For any $m\ge 0$, there exists a probability function P satisfying:

Open image in new window

for all $i=0,\ldots ,m$, if and only if $N>m+1$.

We first establish the following lemma:

Theorem 6

For any probability P and any $m\ge 0$, $S_{m+1}(P)\models S_m(P)$.

Proof

The proof will proceed by induction on m. Suppose, for contradiction, that $E_n\models S_1(P)$ but $E_n\not \models S_0(P)$. From the fact that $E_n\models S_1(P)$, it follows that Open image in new window

. But since $E_n\not \models S_0(P)$:

Open image in new window

Thus, Open image in new window

, and so, since Open image in new window

, $E_n\models S_0(P_n)$. But since $\sigma _P(E_n)=\sigma _{P_n}(E_n)$, this contradicts the assumption that $E_n\not \models S_0(P)$. Hence, $S_1(P)\models S_0(P)$.

Now, assume that, for any probability P, $S_m(P)\models S_{m-1}(P)$ ($m\ge 1$) and let $E_n\models S_{m+1}(P)$. Then, $P(\lnot E_1\wedge \cdots \wedge \lnot E_{n-1})>0$ and Open image in new window . But since, by assumption, $S_m(P)\models S_{m-1}(P)$ it follows that Open image in new window , and so $E_n\models S_m(P)$. Hence, $S_{m+1}(P)\models S_m(P)$, which completes the induction. $\square $

We now proceed to the proof of Theorem 5. To prove the only-if direction of the claim, it suffices to show that for any probability P:

Open image in new window

The proof proceeds by induction on N. If $N=1$, then

Open image in new window

for any probability P. Now let $N>1$, and suppose that, for all $i=1,\ldots ,N-1$, there is no probability P with $P(E_1\vee \cdots \vee E_{N-i+1})=1$ satisfying

Open image in new window

(5)

It follows that

Open image in new window

(6)

for $i=1,\ldots ,N-1$, for otherwise, the function defined by

$$\begin{aligned} P(E_n)=P_{i+1}(E_{n+i}) (n=1,\ldots ,N-i+1) \end{aligned}$$

would satisfy (5). But (6) implies that

$$\begin{aligned} E_{i+1}\not \models S_{N-1-i}(P), \end{aligned}$$

for $i=1,\ldots ,N-1$, and so, from Theorem 6, we have

$$\begin{aligned} E_{i+1}\not \models S_{N-1}(P), \end{aligned}$$

for $i=1,\ldots ,N-1$. Moreover, since $E_{N+1}\not \models S_0(P)$, by Theorem 6, $E_{N+1}\not \models S_{N-1}(P)$, and so we have $S_{N-1}(P)\models E_1$. Now, suppose that Open image in new window

. Then Open image in new window

, which implies that $E_1\not \models S_0(P)$. But then, by Theorem 6, $E_1\models S_{N-1}(P)$, and so $S_{N-1}(P)\equiv \bot $. Contradiction. Hence, Open image in new window

.

We now prove the if direction of Theorem 5. Suppose that $m+1<N$. We will construct a probability P satisfying:

Open image in new window

for $i=0,1,\ldots ,m$.

Let c be any value in the open interval Open image in new window

, and for each $n=1,\ldots ,N+1$, let

$$\begin{aligned} P(E_n)=\left\{ \begin{array}{ll} c^{n-1}(1-c) &{}\quad 1\le n\le N\\ c^N &{}\quad n=N+1\\ \end{array}\right. \end{aligned}$$

Then P is a probability function. Moreover, for any k such that $1\le k\le N+1$:

$$\begin{aligned} \sigma _{P_k}(E_n)=\left\{ \begin{array}{ll} c &{}\quad \text {if } k\le n\le N\\ 0 &{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

Since Open image in new window

, it follows that $S_0(P_k)= E_k\vee \cdots \vee E_N$, and so:

$$\begin{aligned} P_k(S_0(P_k)) = 1- c^{N-k+1}. \end{aligned}$$

But since c is in the interval Open image in new window

, it follows that

$$\begin{aligned} \sigma ^1_{P_{k}}(E_n)=\left\{ \begin{array}{ll} 1 &{}\quad \text {if } k\le n\le N-1\\ 0 &{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

Now suppose that

$$\begin{aligned} \sigma ^i_{P_{k}}(E_n)=\left\{ \begin{array}{ll} 1 &{}\quad \text {if } k\le n\le N-i\\ 0 &{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

(7)

for some $1<i<m $. Then $S_i(P_k)=E_k\vee \cdots \vee E_{N-i}$, and so

$$\begin{aligned} P_k(S_i(P_k))=\left\{ \begin{array}{ll} 1-c^{j-k+1} &{}\quad \text {if }k\le N-i\\ 0 &{}\quad \text {otherwise} \end{array}\right. \end{aligned}$$

But since c is in the interval Open image in new window

, it follows that

$$\begin{aligned} \sigma ^{i+1}_{P_{k}}(E_n)=\left\{ \begin{array}{ll} 1 &{}\quad \text {if } k\le n\le N-(i+1)\\ 0 &{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

Hence, by induction, (7) holds for all $i=1,\ldots ,m$. This (together with the fact that $S(P_0)= E_1\vee \cdots \vee E_N$) implies that $S_i(P)= E_1\vee \cdots \vee E_{N-i}$, for $i=0,1,\ldots ,m$. But then, since $m<N-1$:

Open image in new window

for $i=0,1\ldots ,m$.

Appendix 2

In this appendix, we will prove Theorems 3 and 4. Both propositions follow straightforwardly from the following lemma:

Theorem 7

For each Open image in new window

, let

$$\begin{aligned} P_\alpha (E_n)=\left\{ \begin{array}{ll} \alpha ^{n-1}(1-\alpha ) &{}\quad i=1,\ldots ,N \\ \alpha ^N &{}\quad i=N+1\\ \end{array}\right. \end{aligned}$$

Then, $P_\alpha (S(P_\alpha ,\alpha ))\ge P(S(P,\alpha ))$, for any probability P. Moreover, the inequality is strict unless $P= P_\alpha $.

Proof

It is easy to verify that $\sigma _{P_\alpha }(E_n)=\alpha $, for $n=1,\ldots ,N$. Hence, $S(P_\alpha ,\alpha )= E_1\vee \cdots \vee E_N$ and $P_\alpha (S(P_\alpha ,\alpha ))=1-\alpha ^N$.

Fix Open image in new window

and let P be any probability. For $n=1,\ldots ,N+1$, put

$$\begin{aligned} q_n=\sum _{j=n}^{N+1}P(E_j). \end{aligned}$$

Then, for $n=1,\ldots ,N$:

$$\begin{aligned} \sigma _P(E_n)= \left\{ \begin{array}{ll} q_{n+1}/q_n &{}\quad q_{n+1}\ne q_{n}\\ 0 &{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

Thus, $P(S(P,\alpha ))$ is equal to the probability of:

$$\begin{aligned} \bigvee \left\{ E_n: n\le N \text { and } q_{n+1}\ge \alpha q_n \right\} . \end{aligned}$$

For each $n=1,\ldots ,N$, let

$$\begin{aligned} \gamma _n=\max \left\{ \alpha , \frac{q_{n+1}}{q_{n}}\right\} \end{aligned}$$

(here we assume that $0/0=1$), and define $q_n^*$ inductively as follows:

$$\begin{aligned} q_1^*= & {} 1\\ q_{n+1}^*= & {} \gamma _n q_n^* \end{aligned}$$

A simple induction confirms that, for $n=1,\ldots ,N+1$, both (i) $q_n^*\ge q_n$; and (ii) $q_n^*\ge \alpha ^{n-1}$. Hence:

$$\begin{aligned} P(S(P,\alpha ))= & {} \sum _{\begin{array}{c} n\le N\\ q_{n+1}\ge \alpha q_n \end{array}} q_n-q_{n+1}\\= & {} \sum _{\begin{array}{c} n\le N\\ q_{n+1}\ge \alpha q_n \end{array}} \left( 1-\frac{q_{n+1}}{q_n}\right) q_n\\= & {} \sum _{\begin{array}{c} n\le N\\ q_{n+1}\ge \alpha q_n \end{array}} (1-\gamma _n)q_n\\\le & {} \sum _{\begin{array}{c} n\le N\\ q_{n+1}\ge \alpha q_n \end{array}} (1-\gamma _n)q_n^*\\= & {} \sum _{\begin{array}{c} n\le N\\ q_{n+1}\ge \alpha q_n \end{array}} q_n^*-q_{n+1}^*\\\le & {} \sum _{n=1}^N q_n^*-q_{n+1}^* = q_1^*-q_{N+1}^* \le 1-\alpha ^N = P_\alpha (S(P_\alpha ,\alpha )) \end{aligned}$$

Note that the inequality is strict unless $q_{n+1}= \alpha q_n$, for $n=1,\ldots ,N$. But this is equivalent to the claim that $P=P_\alpha $. $\square $

To prove Theorem 3, we first take note of the obvious fact that (for a given N) if $(\alpha ,\beta )$ is coherent, then $(\alpha ,\beta ')$ is coherent, for all $\beta '$ such that Open image in new window . In addition, we note that for any Open image in new window , it follows from Theorem 7 that $(\alpha ,1-\alpha ^N)$ is coherent, but that if $\beta >1-\alpha ^N$, $(\alpha ,\beta )$ is incoherent. These two facts together entail Theorem 3.

To prove Theorem 4, simply note that Theorem 3 implies that if $(\alpha ,\beta )$ satisfies the condition stated in the proposition, then $\beta =1-\alpha ^N$. The result then follows from Theorem 7.

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Brian Kim
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Anubav Vasudevan
- 2

1.Department of PhilosophyOklahoma State UniversityStillwaterUSA
2.Department of PhilosophyUniversity of ChicagoChicagoUSA

How to expect a surprising exam

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