On the Surprise Examination

Ottilia Kasbergen

June 19, 2012

Bachelor thesis

Supervisor: dr. Alexandru Baltag

The Institute for Logic, Language and Computation

Faculty of Science

University of Amsterdam

AbstractIn this thesis several approaches and solutions to the surprise examination arediscussed, in which a teacher tells the students that the exam given next weekwill come as a surprise. By a backward induction argument the students areable to rule out every day of the week as a possibility, reaching a contradictionand thereby enabling the teacher to surprise them by giving the exam at anyday of the week. Three known logical puzzles, Smaller or Bigger, the MuddyChildren and Number+1, are made self-referential and the several solutionsare in an adjusted form applied to these puzzles to gain knowledge aboutthe concept of surprise. In this thesis the different solutions to the surpriseparadox are compared and shown to be more similar than one at first sightwould think.

InformationTitle: On the Surprise ExaminationAuthor: Ottilia Kasbergen, 59399992@science.uva.nl, 5939992Supervisor: dr. Alexandru BaltagSecond reader: dr. Sonja SmetsEnd date: June 19, 2012

The Institute for Logic, Language and ComputationUniversity of AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.illc.uva.nl

Contents

1 Introduction 31.1 The Surprise Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Self-Referentiality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Discussing the Surprise Examination . . . . . . . . . . . . . . . . . . . . . 7

2 Logical Preliminaries 9

3 Solutions to the surprise paradox 153.1 Quine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Gerbrandy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Baltag and Smets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3.1 Surprise in Terms of Knowledge . . . . . . . . . . . . . . . . . . . . 213.3.2 Surprise in Terms of Belief . . . . . . . . . . . . . . . . . . . . . . . 22

3.4 Kritchman and Raz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.5 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 The Surprise Paradox Elsewhere 364.1 Looking Elsewhere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.2 Smaller or Bigger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.2.1 The Story . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.2.2 Solutions to ‘Smaller or Bigger’ . . . . . . . . . . . . . . . . . . . . 37

4.3 Muddy Children . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.3.1 The Story . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.3.2 Solutions to ‘Muddy Children’ . . . . . . . . . . . . . . . . . . . . . 42

4.4 Number+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.4.1 The Story . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.4.2 Solutions to ‘Number+1’ . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Conclusion and Relations to Other Work 505.1 Relating to Other Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.1.1 Truth Revision versus Belief Revision . . . . . . . . . . . . . . . . . 51

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5.1.2 Non-Monotonic Logic . . . . . . . . . . . . . . . . . . . . . . . . . . 525.2 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Popular Summary 54

Appendix 565.3 A New Theory of Truth Explained . . . . . . . . . . . . . . . . . . . . . . 56

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Chapter 1

Introduction

During World War II on the Swedish radio the following announcement was made:“A civil defense exercise will be held this week. In order to make sure that the civil defenseunits are properly prepared, no one will know in advance on what day this exercise will takeplace.”[18]A Swedish mathematician, Lennart Ekbom, recognized the contradictory nature of thisannouncement. He discussed it with his students and afterwards this puzzling story spreadaround the world.

In 1948 the phenomenon first appeared in print, discussed by O’Connor in the Britishmagazine Mind. After that it appeared in many other publications all over the world, whiletaking several forms. From the civil defense exercise to the unexpected egg, in which anegg was hidden in one of ten boxes and it was announced that you cannot expect whereyou will find the egg. From the unexpected hanging, in which a judge tells a convict thathe will be hanged next week but will remain ignorant about this fact until the morning onwhich the degree will be fulfilled, to the surprise examination.

1.1 The Surprise Examination

In this thesis the surprise examination is taken as a starting point, from where we will beable to discuss the several solutions raised in the last decennia in a more standardized way.The story goes as follows:

“A teacher announces to his class: ‘Next week, there will be a surprise exam.’ It iscommonly understood that an exam comes as a surprise if you do not know the eveningbefore that it is given the next day.”

The contradictory part in this announcement lies in a backward induction argumentwhich the students can practise. They reason:“If the exam is given on Friday, then on Thurday evening we will not have gotten an exam

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yet and therefore know that it must be given on Friday. But then it would not come asa surprise, and therefore it cannot be given on Friday. But then, suppose the exam isgiven on Thursday. Then on Wednesday evening we will not have gotten an exam yet,hence the exam has to be given on either Thursday or Friday. We know the exam is noton Friday. Thus the exam has to be given on Thursday. Again this would not come as asurprise, so we can rule out Thursday as well. If we repeat this argument we can also ruleout Wednesday, Tuesday and Monday. There are no days on which we can have a surpriseexam: we will have no exam at all !?”

And so the students infer a contradiction after hearing the teacher’s announcement.But now, the teacher can give the exam on Wednesday1, thereby completely surprising thestudents in his class. So the teacher’s words became true after all.

Now where lies the fault? Was there something wrong in the students’ reasoning?Or can this contradiction be blamed on something else? Or is the surprise examinationperhaps a genuine paradox? And, paradox or not, is there a satisfying solution? What willthis solution be?

1.2 Paradox

Let us first have a look at what ‘paradox’ really means. What conditions should a puzzlingstory such as the surprise examination meet for us to call it a paradox? Intuitively, weregard as a paradox a seemingly sound piece of reasoning based on an apparently trueassumption that leads to a contradiction.

The Encyclopedia of Philosophy defines:“A logical paradox consists of two contrary or even contradictory propositions to which weare led by apparently sound arguments. The arguments are considered sound because whenused in other contexts they do not seem to create any difficulty.”[24]

Among the paradoxes we can distinguish three main categories: the semantic, set-theoretic and epistemic paradoxes. The first categorie speaks of paradoxes based on thesemantic notion of truth, the second on the formal definition of a set and the third onthe concept of knowledge. Well-known paradoxes for these categories are, respectively, theLiar’s paradox, Russell’s Paradox and the Knower paradox. I will give a brief explanationof them and the reasoning involved to improve our intuition for paradoxes.

The Liar’s paradox[7] is said to stem from the ancient Greek, where Epimedes of Crete2

stated “Cretans are always liars”.

1Or on any other day of the week.2Thus, he is a Cretan himself.

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Nowadays this paradox is more commonly known as the sentence

This sentence is false.

If “This sentence is false” is true, then the sentence is false. This would then mean thatit is actually true, but that would mean that it is false, and so on ad infinitum. Similarly,if “This sentence is false” is false, then the sentence is true. This would then mean that itis actually false, but that would mean that it is true, and so on ad infinitum. We see thatthe sentence, when uttered and assigned a truth value, changes this truth value again andagain ad infinitum.

The Liar’s paradox is an example of a semantic paradox. Let us regard now the set-theoretic paradox Russell put forward in the beginning of the previous century. The Britishlogician formulated a set R which consists of elements that contain itself[10]:

R := {x | x 6∈ x}

Now we want to know if R belongs to itself. Suppose it does: R ∈ R. All elements of Rdo not contain themself, thus R 6∈ R. But this would mean it actually does contain itself,which in turn would mean does not, and so on ad infinitum. Similarly, suppose R does notbelong to itself: R ∈ R. All elements of R do not contain themself, thus because R is notan element of R it does contain itself: R ∈ R. But this would mean it actually does notcontain itself, which in turn would mean it does, and so on ad infinitum. We see that thefollowing holds:

if R := {x | x 6∈ x}, then R ∈ R⇔ R 6∈ R,

which is clearly a contradiction.

The third paradox, the epistemic Knower paradox[10], is given by

This sentence is not known to anyone.

The reasoning for this paradox is similar to the Liar paradox introduced above, althoughit is a bit more complex. By assuming to obtain a contradiction that “This sentence is notknown to anyone.” is not true we can deduce that it has to be true. But from stating that“This sentence is not known to anyone.” is true we can infer that it cannot be true. Thisis a contradiction.

The Liar’s paradox, Russell’s paradox and Knower paradox all seem to have a para-doxical nature: to be both true and false at the same time. We can regard them as trueparadoxes : although the argumentation steps seem to be right we get a contradictory out-come.

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1.3 Self-Referentiality

The paradoxes introduced in the previous section share one common feature: they are allself-referential: they refer to themself or their own referent. Their cyclic self-referentialnature has for a long time been seen as the true and only source of their paradoxicality.Until Yablo in 1985 came with his own paradox: ‘a paradox without self-reference’, whichmakes use of a kind of ‘unfolded’ self-referentiality. Yablo states:“Imagine an infinite sequence of sentence S1, S2, S3, . . ., each to the effect that every sub-sequent sentence is untrue:(S1) for all k > 1, Sk is untrue,(S2) for all k > 2, Sk is untrue,(S3) for all k > 3, Sk is untrue, . . .[25]

A sequence defined like this is non-self-referential and still has a paradoxical nature likedescribed above. First it can be proved that for all k Sk cannot be true. We assume forcontradiction that Si is true for some i. Thus it holds by definition that for all k > i Sk isuntrue. In particular this holds for k = i+ 1. Sentence Si+1 states that for all k > i+ 1 Skis untrue, thus this sentence being untrue implies that there has to exist a j > i + 1 suchthat Sj is true. But this contradicts that none of the sentences Sk with k > i are true.Thus we may conclude that none of the sentences Si can be true.Now this is expressed by the sentence S1, so S1 has to be true. This is in contradictionwith what we have shown before.

Yablo’s example showed us that (cyclic) self-referentiality is not needed for a paradoxto occur.3 However, most known paradoxes do exist because of their self-referentiality.Having explored the concept of self-referentiality somewhat further, we can now look atsolutions to the paradoxes.

1.4 Solution

Paradoxes are important for the understanding of the foundational concepts involved andthe solutions raised to the several paradoxes therefore have far-reaching consequences. Thedifferent categories of paradoxes had different consequences: semantical paradoxes led toseveral theories of truth, set-theoretic paradoxes to new foundations for set-theory and newdefinitions for the term set and epistemic paradoxes to new theories of knowability. Theseare however somewhat outside the scope of this thesis.

3Though his example is analogous with a self-referential paradox: the Liar paradox.

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Other solutions to paradoxes are even more rigourous: the British philosopher Priesttells us to just accept that some sentences are both true and false. He concludes this be-cause all known solutions fail to indicate which premise or argumentation step is invalid orto give an illuminating independent not ad hoc reason for rejecting it. The ideal solutionPriest describes resembles Haack’s[15] division in the formal and philosophical solution: theformal solution should indicate which apparently unexceptionable premises or principle ofinference must be disallowed, whereas the philosophical solution should be an explanationof why that premise or principle is exceptionable.

This should be what we are looking for while exploring the surprise examination. Now,according to Ditmarsch and Kooi [12] there are three ways to solve a paradox4:

• There is something wrong with the proposition whereupon it is impossible to assignit a truth value;

• There is something wrong in the argumentation of why this proposition should hold;

• There is something wrong in the argumentation of why this proposition should nothold.

The Liar paradox and Russell’s paradox mentioned before have solutions with the firstas starting point: there seems to be something wrong with the proposition whereupon itis impossible to assign it a truth value. The Liar paradox is by Revision Theory of Truthsaid to be an unstable sentence, and therefore alternates it truth value for ever.5 The setR in Russell’s Paradox is by Axiomatic Set Theory not to be a set6, just a proper class,whereby speaking of ’something containing R’ is meaningless.7

1.5 Discussing the Surprise Examination

The surprise examination, if it is indeed a real paradox, is an example of a epistemic para-dox, just like the Knower paradox mentioned in the previous sections. For the Knowerparadox it is clear that it is indeed a true paradox, for the surprise examination, even thisremains to be controversial. However, because the reasoning in the surprise examinationseems to be sound and the outcome is contradictory, in this thesis I will take the libertyof sometimes referring to it as a paradox.

In the teacher’s statement we can distinguish two seperate statements: “next weekthere will be an exam” and “the exam will come as a surprise”. Some scientists who raised

4Ditmarsch and Kooi proposed this to solve epistemic paradoxes, like the surprise examination.5For a further discussion on this I refer to the appendix.6Whereas according to Naive Set Theory any definable collection of sets is a set.7For a further explanation of this I refer to Devlin[11].

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a solution took the first part as a certainty, some did not. However, in the chapter 3 wewill see that this does not really change their solutions.

But then, how should “surprise” be formalized? In many solutions this is interpretedas “not knowing beforehand (the evening before) that something is the case”. Others in-terprete it in terms of belief or even in terms of provability.

There are different variations of the surprise paradox, in which mostly time plays arole. However, Sorensen proved in 1988 that this is not necessary, with his ‘designatedstudent’.[21] The solutions discussed in chapter 3 however are all based on variations inwhich time does play a role. These solutions are standardized: all deal with the surpriseexamination, taking “there will be an exam next week” as a certainty. Some say it is indeeda paradox, others say it is not: there might be something strange going on that createsunexpected results. One major question remains whether the teacher’s statement has tobe and stay true.

Before we can have a closer look at the various solutions to the surprise paradox, I willintroduce some logical concepts which will be used in the later chapters.

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Chapter 2

Logical Preliminaries

To be able to discuss the several solutions to the surprise examination, we first need to havesome logical concepts introduced. In this chapter the definitions and axioms are introducedwhich are needed in later chapters.

We are able to translate any real situation into a model of worlds, relations and avaluation, a possible worlds model. In this model the individuals spoken about are calledagents and the several worlds represent the several scenarios that may be representing theactual truth. If a world is accessible for a particular agent, agent i, by accessibility relationRi, this means that this agent considers this world possible. The relations an agent has iscalled it’s propositional attitude. We should now define the language in which the formulaethat hold in the worlds are written.

Definition 2.1. [22] (Epistemic Logic language LEL). For G a set of agents, i ∈ G andA a basic set of propositions (atoms) named by p, q, r, . . .,it is inductively defined how to construct further expressionsϕ ::= p | ¬ϕ | (ϕ ∧ ψ) | Kiϕ

1

This means that any formula in this language is constructed from atoms, negations,etc. Note that ϕ ∨ ψ and ϕ→ ψ can be constructed by using negation and conjunction2.The last concepts, knowledge by agent i that ϕ, Kiϕ, needs some further introduction.But first one more definition is needed.

Definition 2.2. [6] (Epistemic Possible Worlds Model). For A the set of propositionalatoms and G the set of agents, a possible worlds model for agents i ∈ G is a triple S =(S,Ri, ‖ · ‖) such that(i) S is a non-empty set of possible worlds;(ii) Ri is an accessibility relation between the worlds of S;(iii) the valution map ‖ · ‖ assigns to each atomic sentence p ∈ A some set ‖p‖ ⊆ S ofworlds in which p holds.

1In the original definition common knowledge is defined as well. We don’t need it in this thesis, so itis left out in all definitions in this chapter.

2ϕ ∨ ψ ≡ ¬(¬ϕ ∧ ¬ψ) and then ϕ→ ψ ≡ ¬ϕ ∨ ψ.

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Now we can define knowledge for agent i as being true in all worlds in the modelaccessible for i. Formally, the following defines the truth value of the various formulae:

Definition 2.3. [22] (Truth conditions). Every possible worlds model M and world s sat-isfy:M, s |= p⇔ s ∈ ‖p‖M, s |= ¬ϕ⇔ not M, s |= ϕM, s |= ϕ ∧ ψ ⇔ M, s |= ϕ and M, s |= ψM, s |= Kiϕ⇔ for all t with sRit: M, t |= ϕ

To make all this a bit more insightful, we shall look at an example: consider Alice, whojust woke up and wonders if she has a bad hair day today. She does not have a mirror, soshe is completely ignorant about if she has a bad hair day or not. Let us define b = ‘Alicehas a bad hair day’. If we write 1 and 2 for the possible worldsin which respectively b and¬b holds and a double pointed arrow for the accessibility relation RA of Alice, the processgraph corresponding to this situation, model M0, is then as follows:

1, b↔A 2,¬b

We can see it is the case that M,1 |= ¬KAb, M,1 |= ¬KA¬b, M,2 |= ¬KAb andM,2 |= ¬KA¬b: Alice is completely indifferent regarding to the state of her hair. We mightalso want to formally reason about what happens if an agent gets new information andtherefore change their propositional attitudes. This can be done in Public AnnouncementLogic. After a public announcement is made with information ϕ, all agents get to knowthat ϕ is the case. We denote the action of making ϕ publicly known with [!ϕ] and write[!ϕ]Kiϕ: after the announcement has been made agent i knows that ϕ. Agent i nowhas to update her model with this new information. This is formally defined in the nextdefinitions.

Definition 2.4. [22] (Public Announcement Logic language LPAL). The language of PALis the epistemic language with added action expressions, as well as dynamic modalities forthese, defined by the syntax rules:Formulae P ::= p | ¬ϕ | (ϕ ∧ ψ) | KiϕAction expressions A :!P

The epistemic language is interpreted as before. Note that again ϕ∨ψ and ϕ→ ψ canbe constructed by using negation and conjunction. We require that P has to be true to beannounced, otherwise the knowledge implied by it in the agents’ propositional attitudeswould be untrue, which contradicts Veracity of Knowledge stated in definition 2.10. If wewrite M |P for the updated model after the announcement that P , the new dynamic actionmodality is defined[22] by

M, s |= [!P ]ϕ⇔ if M, s |= P then M |P , s |= ϕ

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We can now write sentences like [!ϕ]Kiϕ. For example, let us suppose now that Alicefinds a mirror in the room of her housemate, and can now assure herself dat she does nothave a bad hair day and can go to the university just like this. Suppose she sees that herhair is perfectly fine. The model corresponding to this situation is then M |b:

1, b.

The world in which ¬b held was deleted, because it was inconsistent with Alice’s newlyreceived information. Suppose now the mirror Alice found was really dirty, so looking in itshe might think that she does not have a bad hair day. Alice acknowledges she cannot besure of this because the mirror is so filthy; she only believes that she does not have a badhair day. When we want to speak about the beliefs someone has, we have to introduce aplausibility relation ‘≤’. This relation gives an order on the worlds of being less or moreplausible, and formulae that hold in all most plausible worlds are believed. We demandthat the plausibility relation is a totally connected preorder. ‘Totally connected’ means thatall worlds have to be comparable: ∀s, t ∈ S it holds that s ≤ t or t ≤ s[3]. A ‘preorder’ isreflexive and transitive: ∀s ∈ S s ≤ s and ∀s, t, u ∈ S s ≤ t ∧ t ≤ u → s ≤ u[3]. If weput this plausibility relation together with the formerly defined possible worlds model, weget the plausibility model, which is just a model (S,≤, ‖ · ‖) having a plausibility structureunderlying it.

Definition 2.5. [6] (Plausibility Model). This model is a quadrupple S = (S,Ri,≤i, ‖ · ‖)such that(i) S is a non-empty set of possible worlds;(ii) Ri is an accessibility relation for agent i between the worlds of S;(iii) ≤i is a plausibility relation for agent i between the worlds of S;(iv) the valution map ‖ · ‖ assigns to each atomic sentence p ∈ AT some set ‖p‖ ⊆ S ofworlds in which p holds.

For shortage, we sometimes write S = (S,≤i, ‖ · ‖) to denote S if we are only speakingabout it’s plausibility relations. If there is a plausibility relation between two worlds thisimplies an accessibility relation between these worlds. We now find yet a more compre-hensive language, the language of doxastic epistemic logic:

Definition 2.6. [5] (Doxastic Epistemic Logic language LDEL). For G a set of agents,i ∈ G and A a basic set of atoms named by p, q, r, . . .,it is inductively defined how to construct further expressions:ϕ ::= p | ¬ϕ | (ϕ ∧ ψ) | Bψ

i ϕ

Again ϕ ∨ ψ and ϕ → ψ can be constructed by using negation and conjunction. Inthis logic, knowledge Kiϕ and ‘unconditional’ belief Biϕ can be derived from conditionalbelief Bψ

i ϕ by putting Kiϕ := B¬ϕi ϕ and Biϕ := B>i ϕ. Belief in ϕ by agent i, Biϕ, andconditional belief in ϕ by agent i given ψ, Bψ

i ϕ, are defined below. We write s(i) for theset of worlds that is reachable for agent i from s, the information cell of i at s.

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Definition 2.7. [6] (Maximal worlds). A world s is a maximal (most plausible) world ifand only if it belongs to the set of maximal worlds Max≤S := {s ∈ S : t ≤ s for all t ∈ S}.

Definition 2.8. [6] (Belief). A sentence ϕ is believed at world s by agent i, Biϕ, if andonly if it holds in all maximal worlds of s’ information cell:Max≤s(i) = {s ∈ s(i) : t ≤ s for all t ∈ s(i)} ⊆ ‖ϕ‖.

Definition 2.9. [6] (Conditional Belief). A sentence ϕ is believed given on ψ at world sby agent i, Bψ

i ϕ, if and only if it holds in all most plausible worlds satisfying ψ:Max≤‖ψ‖ := {s ∈ ‖ψ‖ : t ≤ s for all t ∈ ‖ψ‖} ⊆ ‖ϕ‖.

We regard the following validities connecting knowledge and belief:

Definition 2.10. [6] (Validities for Knowledge and Belief).

• Veracity of Knowledge: Kϕ⇒ ϕ

• Positive introspection of Knowledge: Kϕ⇒ KKϕ

• Negative Introspection of Knowledge: ¬Kϕ⇒ K¬Kϕ

• Consistency of Belief: ¬B(ϕ ∧ ¬ϕ)

• Positive Introspection of Belief: Bϕ⇒ BBϕ

• Negative Introspection of Belief: ¬Bϕ⇒ B¬Bϕ

• Strong Positive Introspection of Belief: Bϕ⇒ KBϕ

• Strong Negative Introspection of Belief: ¬Bϕ⇒ K¬Bϕ

• Knowledge implies Belief: Kϕ⇒ Bϕ

Now we only still want to need to add “Kripke’s axioms”

K(ϕ⇒ ψ)⇒ K(ϕ⇒ Kψ)B(ϕ⇒ ψ)⇒ B(ϕ⇒ Bψ)

and the inference rules

Modus Ponens : From ϕ and ϕ⇒ ψ infer ψand Necessitation: From ϕ infer Kϕ.

Having defined all this we can speak of Alice’s beliefs about the condition of her hair afterlooking in the dirty mirror. If we write a pointed arrow towards the world which she thinks

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is more plausible, it seems reasonable to depict her situation as follows:

1, b→A 2,¬b

But we should make this updating (or better: upgrading) of her model more formal. Todo this, the softer upgrades ’⇑’ and ’↑’ are introduced. They do not induce knowledge, like’ !’ does, but rather beliefs.

Definition 2.11. [3] (Radical - or lexicographic upgrade ⇑). An upgrade with ⇑ ϕ makesall ϕ-worlds are more plausible than all non-ϕ-worlds.

Definition 2.12. [3] (Conservative upgrade ↑). An upgrade with ↑ ϕ makes all maximalϕ-worlds more plausible than all other worlds.

We see that a radical upgrade induces a stronger kind of belief than a conservativeupgrade.These different upgrades and the update correspond to different attitudes:

Definition 2.13. [6] (Corresponding attitudes).Update: The source is infallible; it is guaranteed to be truthful.Radical upgrade: The source is fallible, but highly reliable, or at least very persuasive. Thesource is strongly believed to be truthful.Conservative upgrade: The source is trusted, but only barely. The source is believed to betruthful, but this belief can be easily given up later.

Just like we defined ’!ϕ’ to induce that non-ϕ-worlds are deleted, call this hard infor-mation, we can define a model transformer Tϕ for soft information more formally:

Definition 2.14. [6] (Model Transformer). A belief upgrade with a sentence ϕ is a modeltransformer Tϕ that takes any plausibility model S = (S,≤i, ‖·‖), and returns a new modelTϕ(S) = (S ′,≤′i, ‖ · ‖ ∩ S ′), such that (i) the new non-empty set of possible worlds S ′ ⊆ S;(ii) a new plausibility relation ≤′i, satisfying ‖ϕ‖S ∩ S ′ 6= ∅ =⇒ Max≤′S ′ ⊆ ‖ϕ‖S .(iii) the restriction of the valution map to S ′, ‖ · ‖ ∪ S ′.

An upgrade T is soft if and only if for every model S it is total: there are no worlds deleted,S’=S for all S[3]. We shall denote upgraded models as M |Tϕ for the model obtained byupgrading model M with Tϕ.

Let us consider again Alice and the filthy mirror. Before she looks in it and receives softinformation, her model is M0, complete indifference regarding the condition of her hair.Looking into the mirror provides her with the soft information that ¬b. Her upgradedmodel M1 is then M0|⇑ϕ or M0|↑ϕ; in this case there is no difference between the two:

1, b→A 2,¬b

We can now also indicate an actual world to this model, representing the real situation.In the process graph this is depicted with a box around it. Suppose Alice in reality has a

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bad hair day. Indicating world 1 to be the actual world, the model would be:

1, b →A 2,¬b

This is an example of having a wrong belief.

Only one more concept needs to be introduced. These are the temporal operatorsNEXT and BEFORE. They imply the existence of temporal plausibility models, whichcan be identified with sequences S0,S1, . . . ,Sn, . . . of plausibility models obtained by suc-cessive upgrades T0, T1 · · · such that Sn+1 = Tn(Sn). The future-tense operator NEXTand the past-tense operator BEFORE are defined as follows:

Definition 2.15. [6] (NEXT operator).NEXTϕ holds at Sn if and only if ϕ holds at Sn+1.

Definition 2.16. [6] (BEFORE operator).BEFOREϕ holds at Sn for n ≥ 1 if and only if ϕ holds at Sn−1.

Linking this to the example with Alice and her bad hair day: M0 |= NEXT (BA¬b)and M1 |= BEFORE(¬KAb).

Now we have defined all the concepts that we need in our discussion of the solutions tothe surprise examination in the next chapters.

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Chapter 3

Solutions to the surprise paradox

Many interesting solutions have been put forward the last decades, and still the philoso-phers and logicians do not agree on which solution is the correct one. Part of the problem isthat the surprise examination can be interpreted in several ways. If we look at the teacher’sannouncement ‘next week there will be a surprise exam’, we see this can actually be splitin two announcements: ‘next week there will be an exam’ and ‘this exam will come as asurprise’. I will refer to the first part of the statement as ‘exam’ and to the second partas ’surprise’. In some solutions it is assumed that ‘exam’ is part of the announcement,while in others it is treated as an established fact. For intuition, regard a school in whichthe teacher gives one exam every week. Then ‘exam’ is already known by the studentsbefore the teacher utters his strange announcement. To get a more standardized form forthe solutions and to be able to compare them, we will adjust the solutions in which ‘exam’is part of the announcement. Thus, in all the solutions presented in this chapter, ‘exam’is treated as an established fact, it is known by the students. This adjustment thus madein the solution of Quine and Kritchman and Raz however does not change the idea of thesolutions. This will become clear in the discussion in the last section of this chapter.

Now, let us first look at the reasoning of the students a bit more carefully. In thestudents’ possible worlds model we associate the days (Monday to Friday) with numbers1, 2, 3, 4 and 5, and define worlds 1,2,3,4 and 5 as follows:We call the world in which (only) 1 holds 1, the world in which (only) 2 holds 2, etcetera.Therefore in any model M which consists of these five worlds and the students’ plausi-bility relations, M,1 |= 1 ∧ (¬2 ∧ ¬3 ∧ ¬4 ∧ ¬5), M,2 |= 2 ∧ (¬1 ∧ ¬3 ∧ ¬4 ∧ ¬5), . . . ,M,5 |= 5 ∧ (¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4).

On a kind of school were one exam is given each week, a teacher announces that itwill be a surprise on which day the exam will take place. Let us think of surprise as’not knowing (beforehand) that the exam will be on a certain day while in fact it will begiven on that day’. I will leave the formulation of ‘surprise’ for a moment undetermined.The reasoning of the students below works for both Gerbrandy’s and Baltag and Smet’sformalization. As mentioned before, we treat ‘exam’ as something the students know:

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K(∨

1≤i≤5 i) ∧∧

1≤i<j≤5K(¬(i ∧ j)).

So, before the announcement the students know ‘exam’, and they are completely indifferentregarding to on which day the exam will take place. Their initial model therefore lookslike this:

1↔ 2↔ 3↔ 4↔ 5

Suppose that the teacher’s announcement causes the students to know it: K(surprise).They know from ‘exam’ that if on Thursday evening the exam was not given on the pre-

vious days, it has to be given on Friday, K(

(¬1∧¬2∧¬3∧¬4)→ 5)

. From K(surprise)

they now infer that this cannot be the case: they know that the exam will be a surprise:they should not be able to know on Thursday evening the exam will be given on Friday.Therefore, the exam cannot be on Friday1: K(¬5). This causes an update in their modelM1 = M|¬5, the deletion of Friday:

1↔ 2↔ 3↔ 4

Similarly, from knowing there has to be an exam, the students deduce K(

(¬1∧¬2∧¬3)→

(4 ∨ 5))

; if the exam is not on Monday, Tuesday or Wednesday, the students will know

that it has to be given on Thursday or Friday. But, because the students already ruled outFriday (K¬5) as a candidate, this implies that on Wednesday evening they would knowthat the exam will be given on Thursday. This cannot be the case because the teacher toldthem the exam will be a surprise, thus they conclude that the exam cannot be given onThursday either: K(¬4). This causes another update in their model M1: M2 = M1|¬4,the deletion of Thursday:

1↔ 2↔ 3

Their argument continues in the same way as before, and they will also know that theexam will not be given on Wednesday and Tuesday:Model M3 = M2|¬3 is given by

1↔ 2

and model M4 = M3|¬2 is given by

1

1Note that what we showed was sufficient to eliminate Friday because of the second conjunct of ‘exam’.

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For a last time, following the same argument, the students update their model with thedeletion of Monday: Model M5 = M4|¬1. But now all the worlds are deleted! In thismodel ’exam’ can certainly not hold. This contradicts their background knowledge.

This is the reasoning of the students as many who discussed the surprise examination(or a variant of it) understand it. However, there are numerous different solutions tothis problem. I will present here a number of interesting solution to this paradox. Thesesolutions were not all specified to the surprise examination: Quine discussed the HangmanParadox. I adjusted his solution to the story of the surprise examination, but this doesnot affect the solution.

3.1 Quine

Quine[20] suggests that if the students can conclude after the backward induction that theannouncement will not be fulfilled, they should have been prepared to take this alternativeinto consideration as a possibility from the beginning. This will give the students threeoptions when they look ahead at Thursday evening (as where they first only consideredthe first two options possible):

(a) The exam will have occured at or before that day;

(b) The exam will not have occured at or before that day. The event will (in keeping withthe statement) occur at Friday, and the students will (in violation of the statement)be aware of this promptly after they notice the exam is also not at Thursday;

(c) The exam will not have occured at or before that day. The exam will (in keeping withthe statement) occur at Friday, and the students will (in keeping with the statement)remain ignorant meanwhile of that eventuality;

With the help of a reductio ad absurdum explained in the beginning of this chapter the stu-dents will after hearing ‘surprise’ be entitled to eliminate (b), but not (c). According toQuine, to suppose that the statement impedes (c) is to confuse two things: the hypothesisthat the statement will be fulfilled and the hypothesis that at Thursday the students willknow that the decree will be fulfilled. Thus, Quine reasons, the students will not be ableto exclude Friday by their argument. Therefore they will keep their original beliefs, thecorresponding model and plausibility relations stay the same. In other words: the studentsadopt a ’neutral attitude’ towards the teacher.

Quine for the sake of his argument also brings the surprise paradox back to the form ofone possible day. In his case, this refers to the hangman in which the judge tells the con-vict on Sunday afternoon that he, the convict, will be hanged the following noon and willremain ignorant of the fact until the intervening morning. The convict reasons that only

17

options “I shall not be hanged tomorrow noon, and I don’t know it now.” 2 and “I shallbe hanged tomorrow noon, and do not know it now.” are possibilities, and that only thelast one would fulfil the decree. Therefore, the convict thinks: “Rather than charging thejudge with self-contradiction, therefore, let me suspend judgment and hope for the best.”[20]

This example lively illustrates the neutral attitude the students adopt towards theteacher’s announcement: they do not draw conclusions about the day of the exam basedon the contradiction they inferred, they just hope for the best.

3.2 Gerbrandy

Gerbrandy[13] analyses the surprise paradox in Public Announcement Logic. He tells usthat the surprise statement becoming true after reaching a contradiction seems to implythat ‘knowing a contradiction’ should be part of the formalized statement. To be able todo this, he treats knowledge like a K45-operator. That is, knowledge doesn’t have to befactive ( 6` Kϕ→ ϕ), nor does it have to be consistent( 6` ¬K⊥). Gerbrandy for simplicitybrings the number of days back to three: Wednesday, Thursday and Friday, and his for-malization of the teacher’s announcement then is as follows:

S = (we ∧ ¬Kwe) ∨ (th ∧ [!¬we]¬Kth) ∨ (fr ∧ [!¬we] [!¬th]¬Kfr) ∨K⊥

with ‘we’ standing for ‘the exam is on Wednesday’, ‘th’ for ‘the exam is on Thursday’ and‘fr’ for ‘the exam is on Friday’3.

In his analysis Gerbrandy discusses the principle of success, which states that if youlearn something, you come to know it is true: [!ϕ]Kϕ. Accepting this principle makes thereasoning of the students as follows: after the announcement (S) that there is going to bea surprise exam, the students will know this, KS. This sets the students thinking and likeshown before they infer to know a contradiction. From this, because of the definition ofS and because K does not have to be factive or consistent, they infer that there indeed isgoing to be a surprise. Formally,

[!S]KS ` [!S]K⊥, and[!S]K⊥ ` [!S]S.

Gerbrandy however sees a shortcoming in this: he reasons that the students can excludeFriday, but that their argument should then stop. He thinks the students are right aboutconcluding that after the announcement the exam will not be a surprise if it is given on

2In our version it would be taken as a certainty that the convict will be hanged, and therefore he wouldnot consider this alternative.

3The formalization Gerbrandy suggested is ‘(we∧¬Kwe)∨(th∧[¬we]¬Kth)∨(fr∧[¬we] [¬th]¬Kfr)∨K⊥’. To prevent from ambiguity I decided to add ’!’ here, because we will later also discuss the upgrades’⇑’ and ’↑’.

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Thursday, and they are correct in that the teacher said that it would be, but they are notcorrect to see a contradiction between these two claims. He states that the sentence canbe true before the teacher utters it and becomes false after learning of its truth; which isconfusing, but not paradoxical. Gerbrandy feels that the implicit use of the principle ofsuccess is what makes the surprise examination so puzzling. Just announcing ‘S’ wouldprovide the students with the information that the exam cannot be on Friday, but if wedon’t accept the principle of success, they cannot exclude any more days.

Gerbrandy also suggests that we can interprete the teacher’s statement differently. Forexample, we could understand it as

S ∧ [!S]S

This can be read as: the exam will come as a surprise, and after announcing it will bea surprise, it will still come as a surprise. The first conjunct tells the students that theexam is not on Friday. The second conjunct says that after learning that the exam is noton Friday, it will still be a surprise and therefore it will not be on Thursday either. Thestudents can now exclude the last two days as possibilities.

Carrying Gerbrandy’s suggestion a bit further, we can say that

S ∧ [!S]S ∧ [!S] [!S]S

implies that the students are able to rule out the last three days, that after hearing

S ∧ [!S]S ∧ [!S] [!S]S ∧ [!S] [!S] [!S]S

the students consider only Monday as a possibility, and even that

S ∧ [!S]S ∧ [!S] [!S]S ∧ [!S] [!S] [!S]S ∧ [!S] [!S] [!S] [!S]S

makes the students rule out all days of the week. Any further iteration of this kind leavesus with the same result: the students eliminated all days of the week. This iterated notionin closely related to the self-referential interpretation of the statement. Gerbrandy alsodiscussed such an interpretation of the surprise exam, formulated as ϕ ↔ S ∧ [!ϕ]. Hewrites that such a sentence would be “‘contingently paradoxical’: in certain situations, itis both true and false”[13].

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3.3 Baltag and Smets

Baltag and Smets[3] discuss the surprise examination in detail, and introduce an interest-ing new viewpoint on the matter: if the students reach a contradiction after hearing theteacher’s announcement, they should lower their trust towards him. One could choose tointerpret the teacher’s sentence “you’ll be surprised” as knowledge or as belief. Baltag andSmets discussed both, but prefer to use the belief-version, because they think this is morein line with the meaning of surprise in the natural language.

Baltag and Smets put forward the following interpretation of the teacher’s announce-ment: “The exam will be a surprise, (even) after I’m telling you this.”. This notion makesthe statement self-referential, which can be stated as a non-self-referential ‘future-oriented’doxastic announcement[3] by the use of the temporal operators NEXT and BEFORE.Recall that NEXTϕ in a series of updated/upgraded models (for example at model Sn)means that ϕ holds at the next model (Sn+1), and that BEFOREϕ means exactly theopposite, to hold at the previous model (at model Sn+1 it means that ϕ holds at model Sn).NEXTsurprise means that something will be true after the announcement, but does notsay anything about surprise being true before. Therefore, for the teacher’s statement tobe true before being announced, the complete formalization of the teacher’s announcementis

ϕ := surprise ∧NEXTsurprise.

Baltag and Smets argue that when the students infer a contradiction after hearing theteacher’s statement, they will revise their attitude towards him: he is not infallible afterall! The students will lower their trust grade.In this new light, we might consider it as not so established that the attitude which thestudents have is the highest degree of trust and that the teacher is infallible, as is Ger-brandy’s interpretation. For this reason we now write the yet undetermined update orupgrade ‘T ’ towards the teacher:

Tϕ := T (surprise) ∧ T (NEXTsurprise).

We should now find out which attitude is the right one for the students. In the previouschapter the hard upgrade, the soft radical upgrade and the soft conservative upgrade wereintroduced. Baltag and Smets suggest that the students adopt the most positive attitudepossible towards the teacher; they will always trust him as much as they can. After thestudents lower their trust grade they will keep a positive attitude towards the teacher; theyare willing to come to believe the statement, unless they come to know it is false. For aplausibility model S = (S,≤, ‖ · ‖) and the new model Tϕ(S) = (S ′,≤′, ‖ · ‖ ∩ S ′), this canbe formalized as

‖ϕ‖S ∩ S′ 6= ∅ =⇒ Max≤′S′ ⊆ ‖ϕ‖S

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We can read this as: if ϕ holds in some world in our new model S’ (if it is not known tobe false), then the maximal worlds will be ϕ-worlds (ϕ will be believed). This correspondsto a formulation using the temporal operators:

[Tϕ]¬K¬(BEFOREϕ)⇒ [Tϕ]B(BEFORE)ϕ

which can be read as: unless ϕ is known to be false, it is believed now that ϕ was true.We will call this willingness to revise[3]. With this in mind we look at a formalization ofsurprise, to explore what happens when the students adopt the several attitudes towardstheir teacher.

3.3.1 Surprise in Terms of Knowledge

First, interpreting surprise in terms of knowledge, Baltag and Smets put forward theformalization[3]:

surpriseK =∧

1≤i≤5

(i→ ¬K(i|

∧1≤j<i ¬j)

),

where K(ϕ|ψ) := K(ψ → ϕ) means ‘knowing ϕ given that ψ’, a kind of ‘conditional knowl-edge’.

We should now check which attitude for the students towards their teacher is the rightone. Because we assume that they keep the most positive attitude possible towards him,in the very beginning they will look upon him as an infallible source, and correspondinglyupdate their model with ‘!ϕ’, for ϕ := surpriseK ∧ NEXTsurpriseK . It can be shownthat this will lead to a paradox inmediately.4

We can conclude the students cannot maintain their absolute trust in their teacher;they will adopt the most positive attitude possible now. The students will regard upontheir teacher as a fallible but highly reliable source, and correspondingly upgrade theirmodel radically, with ‘⇑ ϕ’.

For an initial model

1↔ 2↔ 3↔ 4↔ 5

an upgrade with ⇑ ϕ(NEXTsurpriseK) would induce the following:

4This follows from the first part of theorem 3.1, because knowledge implies belief and therefore also[!ϕ]NEXTsurpriseK implies [!ϕ]NEXTsurpriseB , which implies a contradiction.

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1↔ 2↔ 3↔ 4← 5.

The students highly trust their teacher, so the worlds in which NEXTsurpriseK holdwill be more plausible than the worlds in which it does not. The sentence NEXTsurpriseKonly does not hold in world 5, Friday, and therefore after the upgrade this world will be-come less plausible than all the others. Upgrading conservatively with ↑ would lead to theexact same outcome.

Baltag and Smets argue that this outcome is somewhat unsatisfactory, because it failsto accound for the feeling that there is a process of iterated upgrading in the ‘paradoxical’reasoning. Secondly, they think the formalization in terms of knowledge is too weak forthe soft upgrades, which only bring about new beliefs. It therefore seems fair to them tointerprete surprise in terms of beliefs.

3.3.2 Surprise in Terms of Belief

The formalization in terms of belief[3] is analogous to the knowledge-variant:

surpriseB =∧

1≤i≤5

(i→ ¬B(i|

∧1≤j<i ¬j)

),

where B(ϕ|ψ) := Bψϕ means ‘belief in ϕ given ψ’. The sentence surpriseB thus can beread as: for every day of the week, if the exam is on day i you will not believe this giventhat it didn’t happen the previous days.

Let us look at the students’ reasoning for this interpretation a bit more carefully. Thestudents already know there will be one exam: K(

∨1≤i≤5 i) ∧

∧1≤i<j≤5K(¬(i ∧ j)).

Suppose that the teacher’s utterance of ‘surpriseB’ causes the students to believe it:

B

(∧1≤i≤5

(i → ¬B(i|

∧1≤j<i ¬j)

)). We will show that this leads to a contradiction.

Because knowledge implies belief, also knowing surpriseB will lead to a contradiction.

Now with some easy propositional calculus, from the ‘exam’ the students deduce that

K(

(¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4)→ 5)

. Because of the easily verifiable validity in the logic of con-

ditional beliefs ‘K(ϕ → ψ) ⇒ B((ϕ → ψ)|ϕ

)’ it follows that B

(((¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4) →

5)|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4

), and because of the validity ‘B

((ψ → ϕ)|ψ

)⇔ B(ϕ|ψ)’ this is

equal to B(5|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4). This means that, conditionally on there not being anexam on Monday, Tuesday, Wednesday or Thursday, the students believe that the examwill be given on Friday. But if the exam would indeed be given on Friday, this would

be in contradiction with believing the statement: B

(∧1≤i≤5

(i → ¬B(i|

∧1≤j<i ¬j)

))

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implies B(

5 → ¬B(5|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4))

. Combining this with the formerly deduced

B¬1∧¬2∧¬3∧¬4(5) implies the students to believe the exam will not be given on Friday (B¬5).

Similarly, from knowing there has to be an exam (K(∨

1≤i≤5 i))5, the students deduce

K(

(¬1 ∧ ¬2 ∧ ¬3) → (4 ∨ 5))

; if the exam is not on Monday, Tuesday or Wednesday,

the students will know that it has to be given on Thursday or Friday. But, because thestudents already ‘ruled out’ Friday as a candidate (B¬5), this implies B¬1∧¬2∧¬3(4), whichmeans that given that there will not be an exam given on Monday, Tuesday or Wednesdaythe students believe that the exam will be given on Thursday. Their argument continuesin the same way as before, and they will also believe that the exam will not be given onThursday.By a similar argument they ‘exclude’ also Wednesday, Tuesday and Monday as possibledays on which the exam can be given, which contradicts their knowledge that there has tobe given an exam next week. We may conclude that the students cannot believe nor canthey know surpriseB.

We see that interpreting surprise in terms of beliefs shows the same problem in thestudents’ reasoning as when interpreting it in terms of knowledge. Now Baltag and Smetsprovide us with two solutions.6 One for the real self-referential form of the teacher’sstatement by the use of temporal operators, and one for an approximation of this self-referential statement with an infinite sequence of updates or upgrades. I’ll give them both.

A Solution Using Temporal Operators

We are trying to find a solution for the sentence ϕ := surpriseB∧NEXTsurpriseB, whichwe update or upgrade with T a still undetermined positive update or upgrade. It can beshown that the update, and both the radical and conservative upgrade lead to a contra-diction.

Theorem 3.1. !φ,⇑ φ and ↑ φ all lead to a contradiction for φ = NEXTsurpriseB, in

which surprise =∧

1≤i≤5

(i→ ¬B(i|

∧1≤j<i ¬j)

).

Proof. BEFORE(NEXTsurpriseB) ⇔ surpriseB is a validity. Because φ has to betrue before it is announced, we can also state that [!φ]K(BEFOREφ). Combining thesetwo gives us the validity [!(NEXTsurpriseB)]KsurpriseB. We already concluded thatknowing that there has to be an exam next week implies the sentence ‘surpriseB’ cannotbe known: K(

∨1≤i≤5 i) ⇒ ¬KsurpriseB. Because this knowledge about the fact that

5And using the first Kripke modality.6They did the same for the interpretation in terms of knowledge, but for conciseness I only show them

both for the interpretation in terms of belief. The idea is the same. Baltag and Smets spoke about this in[3].

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there has to be an exam in one of these five days is persistant under updates, this impliesK(∨

1≤i≤5 i) ⇒ [!(NEXTsurpriseB)]¬KsurpriseB. This contradicts what we proved be-fore, so we have K(

∨1≤i≤5 i)⇒ [!(NEXTsurpriseB)]⊥. We may conclude that T cannot

be an update.

An analogous proof for the radical and conservative upgrade gives us an impossibilityfor the two soft upgrades as well. I’ll give the proof for the radical upgrade. Note thatdoing a soft upgrade can cause the listener to believe something, and not to know it, likethe hard update does.

Proof. It is again a validity that BEFORE(NEXTsurpriseB)⇔ surprise. By definitionof the radical upgrade we know that [⇑ φ]B(BEFOREφ). Combining these two givesus the validity [⇑ (NEXTsurpriseB)]BsurpriseB. We already concluded that knowingthat there has to be an exam next week implies that the sentence ‘surpriseB’ cannot bebelieved: K(

∨1≤i≤5 i)⇒ ¬BsurpriseB. Because this knowledge about the fact that there

has to be an exam in one of these five days is persistant under upgrades, this impliesK(∨

1≤i≤5 i) ⇒ [⇑ (NEXTsurpriseB)]¬BsurpriseB. This contradicts what we provedbefore, so we have K(

∨1≤i≤5 i) ⇒ [⇑ (NEXTsurpriseB)]⊥. We may conclude that T

cannot be a radical upgrade either. The proof for the conservative upgrade is similar

Because of this, assuming willingness to revise, we can prove that [Tϕ]¬K¬surpriseB ⇒[Tϕ]BsurpriseB ⇒ [Tϕ]⊥ for each positive update or upgrade T . So we should find amodel in which K¬surpriseB holds everywhere after the announcement of ϕ.

It is easy to check that the only possible way to satisfy this condition is

1← 2← 3← 4← 5

but I will show this more carefully. By definition of a positive attitude we know that[Tϕ]K(¬BEFOREϕ) ∨ [Tϕ]B(BEFOREϕ), and therefore[Tϕ]K(¬surpriseB) ∨ [Tϕ]B(surpriseB). Because we’ve already proven that the seconddisjunct leads to a contradiction with the students’ background knowledge (that therehas to be an exam on one of the five days) and thus the first disjunct has to be true.So it has to be the case that |= [Tϕ]K(¬surpriseB), meaning that after this yet unde-termined upgrade with ϕ, ¬surpriseB holds everywhere in the model. We can rewrite

this as |=∨

1≤i≤5

(i ∧ B(i|

∧1≤j<i ¬j)

). Specially, ¬surpriseB has to hold in world

5 (M,5 |=∨

1≤i≤5

(i ∧ B(i|

∧1≤j<i ¬j)

)) and by definition of this world it holds that

M,5 |= ¬1∧¬2∧¬3∧¬4 (meaning that in the world where the exam is given on Friday, theexam is not given on Monday, Tuesday, Wednesday or Thursday). From the last it follows

that M,5 |=∧

1≤i≤4 ¬(i∧B(i|

∧1≤j<i ¬j)

), and therefore M,5 |= 5∧B(5|¬1∧¬2∧¬3∧¬4)

has to be the case. This of course implies that B(5|¬1∧¬2∧¬3∧¬4) has to hold in world

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5, and by strong introspection of belief this must hold in the whole model.¬surpriseB has to hold in world 4 as well, and here, by the definition of this world, M,4 |=¬1∧¬2∧¬3∧¬5. Just like before M,4 |=

∧i∈{1,2,3,5} ¬

(i∧B(i|

∧1≤j<i ¬j)

)is the case, from

which we now can infer that B(4|¬1∧¬2∧¬3) holds here and in the rest of the model. Sim-ilarly we are able to deduce M |= B(3|¬1∧¬2), M |= B(2|¬1) and M |= B(1). Altogetherthis implies that B(1)∧B(2|¬1)∧B(3|¬1∧¬2)∧B(4|¬1∧¬2∧¬3)∧B(5|¬1∧¬2∧¬3∧¬4)has to hold everywhere in the model7, which uniquely corresponds to the model

1← 2← 3← 4← 5

So we see that according to Baltag and Smets indeed the only possible solution to the self-referential version of this paradox is the model in which an earlier day is more plausible tothe students than a later day.

An Approximation to the Surprise Paradox

Alexandru Baltag and Sonja Smets[6] also approximated the self-referential version of thisstatement by identifying it with the limit of an infinite sequence of non-self-referential an-nouncements. With surpriseB as before, we thus identifyϕ := surpriseB ∧ [!ϕ] surpriseBwith!(surpriseB); !(surpriseB); !(surpriseB); · · ·

You could think of this infinite sequence as if the teacher writes “The exam will be asurprise” on a paper and locks it in a drawer, after which the students secretly break intothe drawer and read it. The teacher than notices this, writes another message “The examwill still be a surprise” which he locks in the drawer, the students again break into thedrawer, read it, etcetera.

But, just as before, Baltag and Smets argue that this announcement should not be aninfallible update, because the students will deduce that such an infinite sequential compo-sition of updates is an impossible event. Again the students can reason that the exam willnot be on Friday because they would know this on Thursday evening, and by a similar ar-gument they again exclude all the other days. This contradicts the background knowledgeof the students that there will be an exam.

This essentially means that the teacher cannot be an infallible source: he might tellthe truth, but this is not garanteed. Of course it could be the case that the teacher isdeliberately lying to confuse the students, but we assume this is not the case. Because thestudents deduce the teacher is not an infallible source they will lower their degree of trust.The most positive attitude beneath the update is the radical upgrade, corresponding to a

7Or, if one prefers: B(1) ∧B¬1(2) ∧B¬1∧¬2(3) ∧B¬1∧¬2∧¬3(4) ∧B¬1∧¬2∧¬3∧¬4(5).

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highly reliable source. If we then identify the self-referential statementϕ := surpriseB ∧ [⇑ ϕ] surpriseBwith the limit of the infinite sequence of iterated updates,⇑ (surpriseB);⇑ (surpriseB);⇑ (surpriseB); · · · ,we will see the following happening:In the original model (the students only know that there will be an exam next week) thestudents are completely indifferent regarding to the day on which the exam will occur:

1↔ 2↔ 3↔ 4↔ 5

But after one radical upgrade with the sentence surprise the students will think of Fridayas less plausible than the others, because the sentence is true on Monday till Thursday,but fails on Friday.

1↔ 2↔ 3↔ 4← 5

But after the next radical upgrade with the sentence surpriseB the students will also thinkof Thursday less plausible than Monday, Tuesday and Wednesday, because now ϕ also doesnot hold on Thursday.

1↔ 2↔ 3← 4← 5

Continuing this, after two more successive upgrades, the model will look like this:

1← 2← 3← 4← 5

Now any further iteration will leave the model unchanged; we reached a fixed point in thissequence of upgrades. But this fixed point is a negative one: after the fourth iteration thesentence is known to be false.

We can generalize this: starting with any initial plausibility relation in which thestudent has no hard information about the day of the exam, and iteratively applying softupgrades T (surpriseB) of any kind, we always reach the same fixed point.

The Uniquely Determined Solution

Doing a positive upgrade with ϕ means that new plausibility relation satisfies‖ϕ‖S ∩ S′ 6= ∅ =⇒ Max≤′S′ ⊆ ‖ϕ‖S for a plausibility model S = (S,≤, ‖ ·‖) and the newmodel Tϕ(S) = (S ′,≤′, ‖ · ‖ ∩ S ′).This condition means that, unless the new information was already known to be untruebefore the upgrade (it is, there are no worlds accessible in which the new informationholds), it will be believed after the upgrade (the maximal worlds will be ϕ-worlds).

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To show that the solution described by Baltag and Smet I will use the following lemma:

Lemma: If S is a model with S = {1,2,3,4,5} the set of worlds, then ‖surpriseB‖S∩S’ =∅ ⇔ the model S is given by (1← 2← 3← 4← 5).Proof. ‖surpriseB‖S ∩ S′ = ∅ ⇔ S ⊆‖¬surpriseB‖S , in other words: ¬surpriseB holdseverywhere in this model. We have proven before that the only corresponding model inwhich ¬surpriseB holds everywhere is (1← 2← 3← 4← 5)8.

Thus, as long as the model is not given by (1← 2← 3← 4← 5), the antecedent of‘‖surpriseB‖S∩S′ 6= ∅ =⇒ Max≤′S′ ⊆‖surpriseB‖S ’ is true and therefore the consequentas well. So the soft upgrade will have effect: the maximal worlds will be surpriseB-worldsafter the upgrade.

Before the iteration of T(surpriseB), or as action modality [TsurpriseB], the followingis the case:Worlds 1,2,3 and 4 can be surpriseB-worlds, but do not have to be. But we know forsure that 5 is not a surpriseB-world, because M,5 |= 5 ∧B(5|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4). HenceM,5 6|= 5→ ¬B(5|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4), and therefore M,5 |= ¬surpriseB.

Because 5 is a non-surpriseB-world, after the upgrade it is less plausible than allthe surprise-worlds. But not only that: it is also less plausible than all the other non-surpriseB-worlds! This is because, for a world i to be a non-surpriseB-world, in i it has tohold that i and not believe i given that the ’earlier’ worlds (lower in number) j are believedto be false: M, i |= i∧B(i|

∧1≤j<i ¬j). The first conjunct is true by definition, but for the

second to be true, i has to be more plausible than all the later worlds (because otherwiseit would not be among the maximal worlds, and it would not be believed). So, we canconclude that 5 is the least plausible world.

We can make this visual by taking an arbitrary plausibility relation on the five worlds.Let us suppose the initial plausibility relation is as follows:

1← 5← 2↔ 3↔ 4

Both 5 and 1 are non-surpriseB-worlds, so after the first iteration they become less plau-sible than the rest while the plausibility between these two remains the same:

2↔ 3↔ 4← 1← 5

But now we notice that after the shifting in the plausibility relations, 5 is still a non-surpriseB-world, by definition of surpriseB. World 1 is not a non-surpriseB-world any-more, because worlds 2,3 and 4 are now more plausible worlds and therefore 1 is not longer

8On page 24 of this thesis.

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believed to be true. Another thing to notice: because 5 is now less plausible than all theothers, it is also less plausible than 4, which inmediately makes this a non-surpriseB-world:M,4 |= [TsurpriseB]¬surpriseB. So we now know for sure that after the second upgradewith surpriseB, both 4 and 5 do not satisfy surpriseB. And therefore, after the seconditeration the following is the case:

2↔ 3← 1← 4← 5

Since 4 and 5 are non-surpriseB-worlds, they became less plausible than all the others,that is, 4 became less plausible than all the worlds with a smaller number, and 5 evenless plausible than 4 (by the argument explained before). The non-surpriseB-worlds areagain in their plausibility ordered just as by their number; the non-surpriseB-worlds whichdenote a smaller number, are more plausible than the non-surpriseB-worlds that denotea bigger number. Now notice that 4 stays a non-surpriseB-world: because it is moreplausible than 5, it satisfies both 4 and B¬1∧¬2∧¬3(4). So before the third upgrade we findthat 4,5 and 3 have to be non-surpriseB-worlds. By similar arguments as before, afterthe third upgrade we find:

2↔ 1← 3← 4← 5

And again, after the fourth upgrade, the shift in plausibility relations goes the same, whichresults in:

1← 2← 3← 4← 5

Now we find that, if we upgrade any further, nothing will happen anymore. This is becauseof the lemma proven above, because the antecedent is not true anymore the consequentdoes not have to be either. We do not change our plausibility relations anymore by up-grading, because we concluded that surpriseB contradicts our knowledge at this point:M |= K¬surpriseB. We reached a fixed point. Mark that this solution for approximatedself-referentiality is similar to Baltag and Smets’ solution for the self-referential version ofthe statement.

3.4 Kritchman and Raz

Kritchman and Raz[17] connect the surprise paradox to Godel’s second incompletenesstheorem[14]. They introduce a new proof for this theorem, and use a similar proof to givea solution to the surprise paradox. For the surprise statement S, the following formaliza-tion is given:

S :=∧

1≤i≤5

(PrT,S(pm ≥ i→ m = iq)→ (m 6= i)

)

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This tells us that for every day i, if one can prove (using this statement) that if the examdid not occur on the previous days then the exam will be on day i, then the exam willnot be given on day i. Again it is taken as a certainty that there will be one exam:‘[m ∈ {1, . . . , 5}] ∧ [(m = i ∧m = j)→ i = j]’.9 Kritchman and Raz argue that, using atheory T as formulated in Godels second incompleteness theorem, surprise statement Sgives us a system T + S of which the students cannot prove its consistency in itself. Theirsolution thus implies that if the students believe in the consistency of T + S, they believethe exam cannot be held on Friday, because on Thursday night the students will knowthat if T + S is consistent the exam will be held on Friday. But as they cannot prove thisconsistency, the exam can be held on any other day of the week. We can visualize this asfollows:

1↔ 2↔ 3↔ 4← 5

3.5 Discussion

In this chapter we have seen that there are several interpretations possible for the teacher’sstatement, and several ways of ‘solving’ this paradox. Things in which they can differ areamong other things knowledge or belief, how strongly the teacher is trusted, the degree ofself-referentiality and whether “there will be an exam next week” can be treated as certainor not.

Adjustments done for standardization

Let us start discussing the last. We should explore how big an influence taking ‘exam’ ascertain or not has on the solution. Remember that we adjusted Quine’s and Kritchmanand Raz’ versions, in the other three cases it was already taken as a certainty that therewill be one exam next week. If we would follow Quine more closely, “there is an exam nextweek” would be a part of the statement, and therefore the student’s can also doubt theexistence of the exam. Therefore, Quine expressed one more option10: “The exam will (inviolation of the statement) fail to occur at Friday (and it did not occur the days before)”.This is not taken as an alternative in our version. His argumentation, however, stays thesame, because the students can eliminate this option in the same way as they eliminateoption (b). Thus in the end the solutions are the same.

I also adjusted Kritchman and Raz’ solution to our standardized form: originally theformalization of surprise statement S ′ is as follows:

9Kritchman and Raz originally took this as a part of the announcement. This does not change thesolution, which will be discussed in the next chapter.

10The option described here is the option Quine formulated translated to the surprise examination.Quine discussed the ‘Hangman Paradox’, which differs only in the story around the paradox.

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S ′ := [m ∈ {1, . . . , 5}]∧

1≤i≤5

((PrT,S(pm ≥ i → m = iq)

∧1≤j≤5,j 6=i ¬PrT,S(pm ≥ i →

m = jq)

)→ (m 6= i)

)This formalization consists of two parts. The first says that the exam will be given inone of the five days. The second tells us that for every day i, if one can prove (using thisstatement) that if the exam did not occur on the previous days then the exam will be onday i and one cannot prove in the same manner that then the exam will also be on anotherday j, then the exam will not be given on day i. So, instead of already knowing that thereis no such j for which m = j, it is part of the provability antecedent: if they can provesuch i exists and they cannot prove another such j exists then the exam will not be givenon day i. The reasoning stays exactly the same: because the students cannot prove theconsistency of T + S ′ in itself, they can only believe that the exam is not being held onFriday if they believe in the consistency of the system. But they cannot exclude any otherday of the week. The solution is still

1↔ 2↔ 3↔ 4← 5

If we would, the other way around, add “there will be one surprise next week” to Ger-brandy’s formulization, this would not change the solution either. The formalization thenwould be S ′ =

((we ∧ ¬Kwe) ∨ (th ∧ [!¬we]¬Kth) ∨ (fr ∧ [!¬we] [!¬th]¬Kfr) ∨ K ⊥)

∧((we ∨ th ∨ fr) ∧ ¬(we ∧ th) ∧ ¬(we ∧ fr) ∧ ¬(th ∧ fr)

). Announcing S ′ once would

again induce the students to eliminate Friday as a possible candidate, because Friday isstill the only day not satisfying S ′. Announcing S ′∧ [S ′]S ′ would induce them to eliminateboth Friday and Thursday. The solution stays the same.

The fact that Baltag and Smets take “there will be an exam next week” as a certaintydoes not really change the outcome either, it just makes it easier. We could have just aswell regarded “there will not be an exam next week” as an extra world in the model (tomake it possible for this part of the announcement to fail as well), making their initialmodel:

1↔ 2↔ 3↔ 4↔ 5↔ 6,

with world 6 representing the situation in which no exam is given at all. We would havethen formalized the whole sentence as ‘surprise′B := K(

∨1≤i≤5(i))∧

∧1≤i<j≤5K(¬(i∧j))∧(∧

1≤i≤5

(i → ¬B(i|

∧1≤j<i ¬j)

))’. Because the first conjunct of surprise′B is untrue in

this newly created world 6, it will always be a non-surprise′B-world, and therefore notbe believed nor known. Using temporal operators or the iteration of non-self-referentialupgrades, the solution will be

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1← 2← 3← 4← 5← 6,

corresponding to the unique solution of Baltag and Smets for surpriseB when ‘exam’ istreated as an established fact.

We may conclude that taking ‘exam’ as a part of the announcement or as somethingcertain (known) does not affect the core of the solutions discussed. Therefore our adjust-ments are all right, and we are able to compare the several solutions.

How the several solutions relate to eachother

Quine’s solution is a rigourous one: the students cannot exclude Friday, because they can-not eliminate upfront the possibility that the exam will be on Friday but that they do notexpect this on Thursday evening. Therefore they cannot exclude any of the days: they arestuck with their original knowledge and beliefs, the teacher’s statement does not impose achange of their model. We can link this to Gerbrandy’s approach. Quine sees a fallacy inthe reasoning of the students: if they can conclude in the end that the teacher’s statementwas untrue, they should have considered this as a possibility from the beginning. Onecould say following Gerbrandy that if the students in the end conclude that they werewrong to accept the principle of success after all, they should consider this principle to failas a possibility from the beginning. Thus, Quine speaks of the teacher’s statement beinguntrue, and Gerbrandy speaks of the teacher’s statement as being true before announcingand becoming untrue inmediately after. This difference is the cause of Gerbrandy’s stu-dents eliminating (only) Friday as a possibility and Quine’s students not changing theirpropositional attitudes at all.

Quine did not speak about the degree of self-referentiality of the problem he discussesin ‘on a so-called paradox’. Where he speakes of The Hangman11, we can see that heinterpretes the sentence announced by the judge as “you will be hanged and you will notknow on which day, until the fatal day.” This looks like self-referentiality: you will notknow until judgement day, even after I’m telling you this. Quine also treats the statementthis way. He does not consider the possibility that the judge’s statement was true beforeannouncing it, and becomes false inmediately after. In other words: Quine inherently ac-cepts the principle of success.

Although Gerbrandy uses his formalization to show that maybe the principle of successmust not be accepted and points out that the teacher’s statement might be a self-refuting12

11The story around the same paradox in which a judge tells a convict that he shall be hanged one of thefollowing seven noons, but that he will be kept in ignorance until the morning of the fatal day as to justwhich noon it would be.

12Statements which become untrue because of uttering it. For example the well-known Moore sentences‘p ∧ ¬Bp’.

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one, one can have some comments on the way S is being interpretated. He suggests weshould also consider ‘knowing a contradiction’ as ‘being surprised’. You could side withGerbrandy, because you can argue that someone who is a little acquainted with the rules oflogic would be a bit surprised to know a contradiction. But besides this argument, it feelssomewhat too easy, maybe ad hoc even, to - if you reach a contradiction - just add a con-tradiction to the formalization of surprise. Besides this it has to be said that Gerbrandy’snon-factive, non-consistent notion of knowledge bears more resemblance to the commonnotion of belief than it does to the common notion of knowledge, although it is usuallyassumed that beliefs are consistent. Something that stands out in Gerbrandy’s iteratednon-selfreferential statement is its similarity to the students’ reasoning in the introductionof this chapter. Looking at the model changes: they are completely the same! In this waythis solution does really account for the intuitive feeling of the reasoning of the students,but does it really solve anything? Since all the worlds are deleted, the statement simplybecame false and the students get to know this.

Baltag and Smets take a completely different road in their discussion of the surpriseexamination paradox. They take the perspective of the student and argue that, if theteacher says something that leads to a contradiction, the students naturally will not seehim as an infallible source anymore. Repeatedly lowering their (still positive) degree oftrust, the students will in the end have an attitude equal to the limit of a sequense iteratedpositive soft upgrades. Baltag argues that this can be linked to the inductive reasoningprocess of an ‘ideal’ student13 by means of trial and error.This iterated upgrading of the non-self-referential surpriseB is something Baltag and Smetsalso do for the knowledge-variant of surprise, surpriseK . For conciseness I did not also dis-cuss this in section 3.3.1, but identifying the teacher’s self-referential announcement withthe limit of an infinite sequence of announcements T (surpriseK);T (surpriseK); . . ., justlike done in the belief-version, is another way to reach the same conclusion for all positiveupdates and upgrades T. For ’ !’ this coincides exactly with Gerbrandy’s iterated solution.Something more to notice: Baltag and Smets’ iterated non-selfreferential statement surpriseBis similar to the students’ reasoning for the belief-version of surprise. Looking at the modelchanges: they are also completely the same! In this way this solution does really accountfor the intuitive feeling of the reasoning of the students, just like the their treatment ofsurpriseK did. Different from Gerbrandy, or Baltag and Smets’ solution in surpriseK ,their belief-version offers more of a solution to the problem, rather than just a contradic-tion.

One could also encode the belief-version of the surprise examination as a game14, andthis gives a result similar to the solution raised by Baltag and Smets. Let player 1 be thestudents and player 2 the teacher. After the teacher tells the students that there will bea surprise exam, they (player 1) form a belief about whether the exam will be tomorrow

13A student who is a perfect logician and willing to trust the teacher.14Ken Binmore first did so in [9], page 45-46.

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(t) or not (t). On Monday morning the teacher (player 2) chooses an action, giving theexam that day (e) or not (e). The teacher does not know which beliefs the students hadformed the evening before. If player 2 gives the exam the game will end and depending onthe beliefs of player 1 one of them will win. If player 1 chooses t when player 2 chooses e,player 1 wins. If player 1 chooses t while player 2 chooses e, player 2 wins.The announced sentence is true if and only if there is a winning strategy for the teacher.But if we draw a game tree we can easily see that their exists a winning strategy for thestudent: ttttt. Now we know not only that the statement of the teacher is false, but alsothat the students can get to know that it is false. This winning strategy encodes a plausi-bility relation in the model of the students, every day of the week they expect the exam:

1← 2← 3← 4← 5

This is the only winning strategy for the student, which can easily be seen by the factthat, to win, the student has to be prepared every day in case the teacher decides to givethe exam. The solution found with the use of game theory thus tallies with that of Baltagand Smets.

Just as before, als for the solution in game theory adding ‘exam’ to the formalizationwould not change the solution. It would enable the teacher (player 2) to choose not togive the exam in one of the five days: he can choose eeeee. Then we should say that thegame ends at the minimum of the day at which player 2 chooses e and after day five. Weshould specify who wins the game when it ends after day five with the teacher choosinge. It seems reasonable to designate the students (player 1) as winner: there was no examthat they did not expect. Defining the game like this, player 1 still will have his winningstrategy: ttttt. The solution is again not changed by taking ’there will be an exam’ as partof the teacher’s announcement.

The last discussed solution in this chapter comes from an unexpected quarter: tak-ing an alternative proof for Godel’s second incompleteness theorem as fundament for anew way to look at the surprise examination. This interesting new perspective, in whichthey take the self-referential notion of surprise into account, gives yet another solution tothe paradox, Friday being less plausible and indifference about the other days. Curiouslyenough, this is the solution which many people who are not introduced in the formal rulesof logic, think of as the most plausible solution. “But why can they rule out Thursday too,on Wednesdaynight there are still two possibilities left for the exam to be held, isn’t it?”-was a question I heard more than once. Interesting, because Kritchman and Raz arguethat the students can indeed not rule out Thursday.

Something already named that we might want to discuss a bit further, is the similaritybetween Gerbrandy’s iterated solution and Baltag and Smets’ iteration of surpriseK withthe highest possible positive attitude, ‘!’. To get a similar notation to Gerbrandy’s, I wroteS ∧ [!S]S ∧ [!S] [!S]S ∧ [!S] [!S] [!S]S ∧ [!S] [!S] [!S] [!S]Sfor the four times iterated formalization of surprise. This could, somewhat suggestively,

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also be written as!S; !S; !S; !S.Now the link to Baltag and Smets’ iteration can easily be seen: they wrote:!(surpriseK); !(surpriseK); !(surpriseK); !(surpriseK).It should be said that, while Baltag and Smets’ iterated treatment of surpriseK with theupdate ‘!’ (almost) fully coincides with that of Gerbrandy - also in an intuitive way- , theirsolution considering [!ϕ]NEXTsurprise is the same as well.But there is a difference: Gerbrandy’s formalization of surprise differs from that of Baltagand Smets. Gerbrandy added K⊥ to his formalization, for the rest it can easily be seenthat the two formalizations coincide. If we would bring back Baltag and Smets formal-

ization to three days:∧

1≤i≤3

(i → ¬K(i|

∧1≤j<i ¬j)

), we can rewrite this as

∨1≤i≤3

(i ∧

¬K(i|∧

1≤j<i ¬j))

, which is the same as (1∧¬2)∨ (2∧ [!¬1]¬K2)∨ (3∧ [!¬1] [!¬2]¬K3).

Associating we with 1, th with 2 and fr with 3, this is Gerbrandy’s formalization with-out K⊥. So if their findings are exactly the same, we can ask ourselves what will be theadded value of K⊥. I think this has a purely semantical value: it enables the students tostill be surprised after hearing the teacher’s announcement. Whereas Baltag and Smets’students conclude that the teacher’s announcement became untrue, Gerbrandy’s studentsare surprised.

We see another way in which Gerbandy’s and Baltag and Smets’ solution coincide ifwe look at the discussion of the latter of NEXTsurpriseK with a radical or conservaticeupgrade. This looks like Gerbrandy’s discussion of S (not the iteration of it). Doing anupgrade ⇑ (NEXTsurpriseK) causes the students to make Friday less plausible than theother days:1↔ 2↔ 3↔ 4← 5,where doing an update !S causes the students to exclude Friday:1↔ 2↔ 3↔ 4.We can see that if Gerbrandy had spoken of another, lower but positive, attitude of thestudents, for example highly trusting the teacher, the two solutions would have been thesame (mark that, would we have been using the iteration of ⇑ surpriseK , we would havereached a fixed point here. Any further upgrading with ⇑ surpriseK will leave the modelunchanged).

Although maybe in first sight Kritchman and Raz appear to have raised a completelydifferent solution to the surprise examination, there are striking similarities between theirapproach and that of Gerbrandy and Baltag and Smets. First, a comparance with Ger-brandy’s solution. We can see a parallel between knowledge in his solution with the prov-ability used in Kritchman and Raz’: knowing that there will be an exam on a specific dayversus being able to prove that there will be an exam on a specific day implies that theexam will not be given that day. Furthermore, we can associate the principle of successwith the consistency of arithmetic: if Gerbrandy’s students accept the principle of successthey would be able to carry on their argument and if Kritchman and Raz’ students are

34

able to prove consistency of the system in the system itself they would be able to carry ontheir argument, but they cannot. Consequently, both the reasoning of Gerbrandy’s andKritchman and Raz’ students stops after excluding Friday as a possibility; in the first casethey eliminate Friday, in the second they think of it as less plausible than the rest.

We can compare Kritchman and Raz’ approach to one of the versions Baltag and Smetsdiscussed as well, namely surpriseK with high trust (⇑)15. In this case, we identify prov-ability with knowledge again. But instead of seeking the parallel with the axiom of succes,we associate believing in the consistency of the system T + S with believing the teacherto tell the truth. This is, Kritchman and Raz’ students reason that if they can prove theconsistency, Friday could be deleted. They cannot do this, but believe in the consistencyand therefore for them Friday is the least plausible day. Baltag and Smets’ students onthe other hand reason that if they could know that the sentence is true, Friday could bedeleted. They cannot do this (updating with ‘!ϕ’ implies knowledge that leads to a con-tradiction), but believe the teacher to tell the truth and therefore for them Friday is theleast plausible day.

Conclusion

While the several solutions explained in this chapter first looked quite different, we havenow seen that there are a lot similarities between them. We standardized the solutionsto versions in which “there will be one exam next week” is taken as a certainty, and sawthat this did not affect the idea of the solution. Doing this we made it easier to seewhere the differences lie. Quine’s solution is unique in seeking the fault totally in thestudents’ reasoning and leaving them with their initial knowledge and beliefs. Gerbrandyadds ‘knowing a contradiction’ to the formalization of surprise, and defies the principleof success. Kritchman and Raz are the first to interprete surprise in terms of provability,Baltag and Smets are the first and only to introduce different possible attitudes towardsthe teacher and to interprete surprise in terms of beliefs. The degree of self-referentialityin which the teacher’s statement is understood differs from solution to solution, and foreach understanding there may be a case.How should this paradox be formalized? The ambiguity of the formulation is the lion’sshare of it’s difficulty. All yet raised solutions have their qualities, and give interestingsights to the matter, but yet no all about adopted satisfying solution has been put for-ward. Conversance with some of the new solutions might be relevant to this, but also thefact that there are so many explanations of the teacher’s words. Which solution is thebest yet remains to be something individual, a matter of taste; we wait for a universallyaccepted solution.

15Upgrading with ↑, ’simply’ believing the teacher, would lead to a similar result.

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Chapter 4

The Surprise Paradox Elsewhere

4.1 Looking Elsewhere

We might want to explore the variety of solutions described in the previous chapter a bitfurther. We can do this by looking elsewhere and examining in which way the solutionscan be applied. In this manner we learn to look from a different perspective, which mightbring us closer to a completely satisfying solution. The other way around, we could thensolve more problems than the original surprise paradox only and broaden our view on themeaning of ’surprise’. In the next three sections I will discuss some known puzzles whichI put in a more ’surprise’-like form. These puzzles are made self-referential, and a similarapproach to them is taken as the several solutions raised to the surprise examination.

4.2 Smaller or Bigger

4.2.1 The Story

Story : Alice (A), Bob (B) and Claire (C) are in a cafe. Claire asks the former two tothink of a natural number bigger than zero and to whisper it to her so that only she canhear it. Let’s say Alice chooses the number n ∈ N∗ and Bob the number m ∈ N∗, with N∗representing the set of natural numbers without zero. After hearing the chosen numbersClaire tells them: “You both thought of a different number, but you’ll never be able todeduce who choose the bigger number.”1. Alice and Bob are not allowed to communicatewith eachother, all they can do is reasoning. 2

After the announcement A will think: “If B has number 1, he will know that hisnumber is smaller than mine. But C just told us that we cannot know that, so B cannothave number 1. But then, if he cannot have number 1, he also cannot have number 2: he

1We can see the similarity with the surprise examination here: if your number is smaller or bigger, thiswill come as a surprise.

2The original puzzle is discussed in [18], p. 113.

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knows by a similar argument that I cannot have number 1, we both know our numbers aredifferent and so if B would have number 2 he would know his number is smaller than mine.This is not possible, because of the announcement C made, so B cannot have number 2either, etcetera”. In this way A continues, and is able to rule out for B every number smallerthan her own3. When she ruled out all numbers smaller than n as possible candidates form she will know that her number is the smaller one, which is in contradiction with C’sannouncement that she will never know. Similarly, B reasons about A’s possibilities andinfers a contradiction as well.

4.2.2 Solutions to ‘Smaller or Bigger’

For simplicity, the sentence “A’s number is smaller than B’s number” we will name D1, andthe sentence “B’s number is smaller than A’s number” we will name D2. We write KADi∧KBDi simpler as KA,BDi, which may be part of the formalization of C’s announcement:you both do not know whether A’s number is smaller than B’s or B’s number is smallerthan A’s. Recall that A choose number n and B number m. We take it as a certainty thatA and B chose a different number: ‘D1 ∨D2’.

Quine

If we apply Quine’s solution discussed in the previous chapter to this puzzle, A and Bwould not even be able to exclude number 1. We can look at the options they have in theirreasoning, similar to the ones presented in Quine’s solution for the surprise examination:

(a) The other’s number will be bigger than 1;

(b) The other’s number (in keeping with the statement) will be equal to 1, and he will (inviolation of the statement) be aware of his number being smaller than mine promptlyafter the statement;

(c) The other’s number (in keeping with the statement) is equal to 1, and he will (inkeeping with the statement) remain ignorant of his number being smaller than mine;

Because person A and B cannot exclude option (c), they will adopt a neutral attitudetowards person C, and not change their beliefs. The outcome is that both A and B donot get to know whether their number is smaller/bigger than the other one’s number. SoC’s statement became true after all. A special case is of course that n or m is actuallyequal to 1, but then for this person the case would be similar to Quine’s Hangman paradoxwith one day[20]. The person inmediately reaches a contradiction but is not able to gainknowledge from it.

3In fact, she is able to rule out every natural number to be m. In this way she also reaches a contradic-tion, but without using induction she would have to make infinite think steps to infer this contradiction.

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Gerbrandy

When we apply an adjusted form of the solution of Gerbrandy, who actually speaks aboutthe non-referential notion of surprise, person A and B come to (K45-)know a contradiction.The statement C makes could be formalized as follows:(D1 ∧ ¬KA,BD1) ∨ (D2 ∧ ¬KA,BD2) ∨K ⊥.The last disjunct makes it possible for C’s statement to be succesful, even after personA and B reach a contradiction. With this non-self-referential formulation only 1 can beexcluded by A and B to be the other one’s number. Therefore A and B do not gainknowledge by hearing C’s statement, unless their number is 1 or 2. If one of them hasnumber 1 he inmediately knows this is the smallest number, which makes C’s statementincorrect. If one of them has number 2 he can exclude for the other one to have number 1and thus deduces that he himself has to be the one with the smaller number. This makesC’s statement incorrect.

Baltag and Smets

Remember that Baltag and Smets also interpreted surprise as a statement which inducesbeliefs, we will do the same in this puzzle. If we write notbelieve for

∧i∈{1,2}(Di →

¬BA,BDi), then ϕ := [Tϕ]notbelieve∧ [Tϕ]NEXTnotbelieve would be the self-referentialstatement C makes.

We can approximate this statement (analogous to Baltag and Smets’ iterated solutionexplained in the previous chapter) by introducing rounds of this game, in which C repeat-edly tells person A and B: “You’ll both never correctly believe who’s number is smaller.”and asks them every round again: “Do you know if your number is smaller?”. Alice andBob will each round respond with yes or no. In case they do know they of course specifyif their number is smaller or bigger than the other’s.

Now for example, if n = 3 for A and m = 5 for B, the reasoning for A and B goes asfollows: they begin with total indifference. Of course A and B know their own number, socombinations like (n,m) for n 6= 3 and m 6= 5 are ruled out inmediately. Furthermore, justlike in the surprise examination, the part of the announcement “you have different naturalnumbers bigger than zero” is taken as a certainty. Their initial model is thus as follows,for (n,m) representing the situation in which A chose number n and B number m:

(3,1)↔A (3,2)↔A (3,4)↔A (3,5) ↔A (3,6)↔A . . .

(1,5)↔B (2,5)↔B (3,5) ↔B (4,5)↔B (6,5)↔B . . .

Then C tells them that they have different natural numbers bigger than zero, and cannotcome to correctly believe who’s number is bigger. Because, similarly to the surprise para-

38

dox, a hard update (!) would lead to a contradiction4, we use an unspecified soft upgradeT. All soft upgrades will lead to the same result, just as we have seen in the previous chap-ter. After the first upgrade with notbelieve A and B come to think that number 1 is lessplausible for the other to be her or his number ([Tnotbelieve]KA¬BBD2 ⇒ BA(m 6= 1)and [Tnotbelieve]KB¬BAD1 ⇒ BB(n 6= 1)). So the model changes :

(3,1)→A (3,2)↔A (3,4)↔A (3,5) ↔A (3,6)↔A . . .

(1,5)→B (2,5)↔B (3,5) ↔B (4,5)↔B (6,5)↔B . . .

In the next round, if person C asks A and B if they know which number is smaller, theystill say ‘no’. A believes that B cannot have number 2 after C’s second announcement,because otherwise he would have announced this. Similarly B thinks of 2 as less plausiblefor person A:

(3,1)→A (3,2)→A (3,4)↔A (3,5) ↔A (3,6)↔A . . .

(1,5)→B (2,5)→B (3,5) ↔B (4,5)↔B (6,5)↔B . . .

Now in the next round A reaches a crucial point: m = 3 cannot be the case and m = 1and m = 2 are less plausible than all the others. So A comes to correctly believe that hernumber is smaller, and announces this. She responds to C’s question with “Yes, I believemy number is smaller”, ‘BAD1’, and C’s statement becomes untrue.5 For a countablyinfinite iteration of C’s announcement the situation would be as follows:

(3,1)→A (3,2)→A (3,4)→A (3,5) →A (3,6)→A . . .

(1,5)→B (2,5)→B (3,5) →B (4,5)→B (6,5)→B . . .

If A had not announced that she believed her number to be smaller, after n rounds Bwould have come to believe that his number is smaller, which is incorrect. But still, theywill falsify C’s statement: always one of them will obtain the right beliefs. In this way,they learn from the incorrect statement C makes.

Kritchman and Raz

When we use the solution of Kritchman and Raz raised to the surprise examination andadjust it to this puzzle, we could say that if person A and B believe in the consistency ofthe system, they would be able to rule out number 1 as a candidate. But as they will never

4If there are countably infinitely many rounds.5Now it seems reasonable to assume that B adopts the conclusion: BBD1, because he believes that

person A is right to conclude this (were he in the role of A he would have concluded this as well). This ishowever not needed to falsify C’s announcement.

39

be able to prove the consistency of this system within this system, they will not be ableto continue this argument. In the end if they believe in the consistency of the system theybelieve the other doesn’t have number 1, but they are not able to deduce whether theirnumber is smaller or bigger than the other one’s. Using the same example as before, thesituation for person A and B is as follows:

(3,1)→A (3,2)↔A (3,4)↔A (3,5) ↔A (3,6)↔A . . .

(1,5)→B (2,5)↔B (3,5) ↔B (4,5)↔B (6,5)↔B . . .

Discussion

We see that the several approaches produce various solutions to this puzzle. WhereasQuine’s solution adjusted to this puzzle leaves Alice and Bob completely ignorant and in-different, Gerbrandy’s, Kritchman and Raz’ and Baltag and Smets’ adjusted solutions givethe two an opportunity to learn from the statement C makes. In different gradations ofcourse: the solution in the line of Gerbrandy let’s A and B only eliminate 1 for the otherto be his number and in Kritchman and Raz’ adjusted solution A and B only consider 1to be less plausible than other numbers. This gives them no information about D1 or D2

if their own number is bigger than 2. In the Baltag and Smets’ solution adjusted to thispuzzle person A and B take the risk of being wrong to come to the conclusion that C’sstatement was false and to learn from it, just like in the surprise paradox.

Staying in this belief-variant, we can doubt in which way C’s sentence has to be un-derstood. Does she say that “A and B both individually will not correctly believe whichnumber is smaller”,’

∧i∈{1,2}(¬(Di ∧ BADi) ∧ ¬(Di ∧ BBDi))’, which can be written as

‘∧i∈{1,2} ¬(Di ∧ (BADi ∨ BBDi))’, or does she mean that they will not “both correctly

believe it”, ’∧i=1,2 ¬(Di ∧BA,BDi)’, which can be written as

∧i=1,2 ¬(Di ∧BADi ∧BBDi).

The latter, I think, would be the natural way to understand this sentence, and this is whatI used in the previously. In this way, if only one of them correctly believes either D1 or D2,C’s statement becomes false. But if we would consider the first understanding as the bet-ter one, A and B would maybe not reach a contradiction. In this case, C’s announcementcan be succesful. We do not see a like-wise ambiguity in understanding in the surpriseexamination, because the students act as one: they are all in the same situation, whereasA and B in this puzzle are not.

Something in which this puzzle differs from the surprise paradox is the way in whichwrong beliefs have an effect on persons A and B. In the surprise paradox, it is clear, thestudents will prefer to be prepared for an exam which is not given than to not be preparedfor an exam which is being given. In this puzzle there are two kind of wrong beliefs:

• Believing that your number is smaller, but it isn’t.

• Believing that your number is bigger, but it isn’t.

40

Person A and B are completely indifferent about these two wrong beliefs, they do notprefer one above the other. There are no ’good’ or ’bad’ wrong beliefs. This makes thesolution different as well.

You could argue about whether this puzzle is more paradoxical than the surprise para-dox or less. You could think less, because always one of persons A and B will be rightabout his number being bigger or smaller. But it might also be more paradoxical: theyindividually do not have a ‘satisfying’ solution because there are no ’good’ wrong beliefsin this case. On the other hand, one could say that this puzzle could be solved rathereasily: by just taking opposite beliefs for A and B, this would always make C’s statementfalse. This however would be ad hoc and because A and B are not allowed to communicatesolving this so pragmatically is impossible. It is interesting to notice that in the solutionof Baltag and Smets, the first one to get beliefs about her number being bigger or smaller,is right about this.6

4.3 Muddy Children

4.3.1 The Story

Story : There are n children who played outside together. Their father calls them in andannounces: “At least one of you is dirty and you will never know if you are dirty.”7. Nocommunication is allowed, looking in the mirror is prohibited; for the children there is noother way to deduce whether they’re dirty or not than just by reasoning. And their fatherjust told them even this is impossible.8

Firstly I’ll introduce the general reasoning of the children: this strange announcementsets them thinking about it individually. We take it as a certainty that one of them isdirty, and therefore the option ‘all children are clean’ is ruled out. The father asks themagain and again (we call this rounds) if they know if they’re dirty. The children respondwith “no”, “Yes, I am dirty” or “Yes, I am not dirty”.

The children, each for themselves, will inmediately (even without seeing the other)deduce that if one child sees that all the others are clean, he inmediately deduces that hehas to be the one who is dirty. This is in violation of the statement, and thus ‘one of thechildren is dirty’ cannot be the case. But also ‘two children are dirty’ cannot be the case,because then a dirty child would be able to deduce he’s dirty after concluding ‘only onechild is dirty’ is in violation with the statement. So the option that exactly two childrenare dirty is ruled out as well, because this contradicts the statement that a dirty child willnever know that he’s dirty. Continuing this line of thought, every possibility can be ruled

6When the first one does not announce his beliefs, the other will deduce the same: thinking his numberis smaller as well. Thus the same result as the pragmatic one is reached, without being ad hoc.

7We can see the similarity with the surprise examination here: if you’re dirty it will come as a surprise.8For an explanation of the original puzzle I refer to [23].

41

out by the children and they infer that their father’s statement cannot be true. This isof course the general case. In the real situation the children can see the rest (which wecan treat as a hard update inducing knowledge about the dirtiness of the other children)leaving each child with only two possibilities: “I am dirty” and “I am clean”. Similar tobefore we take “at least one of you is dirty” as a certainty.

4.3.2 Solutions to ‘Muddy Children’

Quine

Applying Quine’s solution to this puzzle we should think that, if the children by theirargument are able to conclude that the announcement will not be fulfilled, they shouldhave been prepared to consider this alternative as a possibility from the beginning. Thisgives each child not only two, but three possible options, and he is not able to eliminatethe case in which he (in keeping with the statement) is dirty and he (in keeping with thestatement) does not know this after round n−19 (and also not in the rounds after). Unlikein the surprise examination, the passing of rounds does not give the children any hardinformation; the strange announcement made this impossible. Just as before, the childrenwill keep their original beliefs, which means that they are indifferent about their own stateof dirtiness.

Gerbrandy

Trying to apply the solution raised by Gerbrandy adjusted to this puzzle and trying to solvethe non-selfreferential form of this puzzle, we will use the notion of knowledge such that itdoes not have to be factive, nor consistent. For the children it is now possible to ‘know’ acontradiction, and including this possibility in the formalization of the father’s statement,makes it possible to be true. For each child i we thus formalize this announcement as(di ∧ ¬Kidi) ∨Ki⊥,which we can read as “i is dirty and he does not know this, or he knows a contradiction”.Not accepting the principle of success unduces the children to only eliminate ’exactly onechild is dirty’ as an option. They do not get more information out of it.

Baltag and Smets

When we adjust the solution raised by Baltag and Smets, the father rather tells the children“At least one of you is dirty and, (even) after I’m telling you this, you will never correctlybelieve if you are dirty”. Instead of updating their models, the children softly upgradethem. We define notbelieve :=

∧1≤i≤n(di → ¬Bidi) for children i ∈ {1, . . . , n} with n the

number of children.If we would try to apply the approximated solution, this would correspond to the story

where the father each round tells the children “You do not correctly believe if you are

9In the original puzzle this is the round in which the dirty children are able to deduce that they’re dirty.

42

dirty.”. The case that∧

1≤i≤n ¬di is already ruled out. Let us perceive what happens if wetake the infinite sequense of (not yet determined) upgradesTnotbelieve;Tnotbelieve;Tnotbelieve; · · ·For a father with 3 children, we see the following happening: before the first round ‘ccc’is deleted, because this is not in keeping with the childrens’ hard information that at leastone of them is dirty:

ddd771

ww

OO

2��

gg3

''cddOO2��

kk3

""

dcd<<

1ss

bb

3++

ddcOO2��

331

||ccd cdc dcc

Then, after the first upgrade the children think of all the worlds in which one child is dirtyas less plausible than the cases in which more children are dirty (because now ‘di → ¬Bidi’does not hold in worlds ccd, dcc and cdc):

ddd771

ww

OO

2��

gg3

''cddOO2

kk3

dcd<<

1

bb

3

ddcOO2

331

ccd cdc dcc

After the second upgrade also the worlds in which two children are dirty become lessplausible than the world in which all children are dirty, so that the model is as follows:

ddd771

OO

2

gg3

cddOO2

kk3

dcd<<

1

bb

3

ddcOO2

331

ccd cdc dcc

We reached a fixed point. Remember that this was the general case. If we choose anactual world arbitrarily, for instance ddc the initial model would be:

ddd gg3

''ddcOO

2��

331

||cdc dcc

43

and after the second upgrade this would change into:

ddd gg3

''ddcOO

2

331

cdc dcc

Mark that if we continue upgrading and no one says anything, the third child would havea wrong belief:

ddd gg3

ddcOO2

331

cdc dcc

Child 3 thinks he is dirty, while actually he is clean. But children 1 and 2 one round beforealready correctly believed that they are dirty, and they would announce this as a responseto their father’s question. The father’s announcement is falsified.

Kritchman and Raz

Applying an adjusted version of Kritchman and Raz’ solution, the children might think itmore plausible that there are two or more children dirty than only one if they believe inthe consistency of the system. But they are not able to continue this argument unless theycan prove the consistency of their system in this system itself. Because of Godel’s secondincompleteness theorem they cannot do this, and therefore the resulting model for eachchild i will be, for the same example as before with 3 children and actual world ddc:

ddd771

ww

OO

2��

gg3

''cddOO2

kk3

dcd<<

1

bb

3

ddcOO2

331

ccd cdc dcc

44

Discussion

If we try to apply the solutions discussed in the previous chapter to the puzzle with themuddy children, we can again doubt how to formalize the father’s statement. For thechildren there are only two possible options: being dirty and being clean. But you couldalso see this as one option: to be dirty or not. We might want to see what happens if weadd ∨(¬i ∧ ¬Ki¬i)) in Gerbrandy’s formalization, ‘child i is clean and he does not knowthis’. But adding this does not change the solution! After the update, still ccd, cdc and dccare the worlds not satisfying this sentence and are therefore deleted. Again the childrencannot continue this argument. The same holds for the solution of Kritchman and Raz.Similarly the outcome following Quine’s reasoning also does not change adding not knowingbeing clean. If we follow Baltag and Smets, adding “if you are clean you do not know it,(even) after I’m telling you this” would again not change the (approximated) solution: byupgrading with the total sentence for the first time, again the worlds with only one dirtychild is less plausible than the rest (the option ‘all children are clean’ is already ruled outor least plausible) because they do not satisfy this sentence and the rest still does. If weadd ‘if you are clean, you don’t know it’ using game theory however, the solution wouldchange: now there is no winning strategy possible for the students anymore: they cannotboth always expect to be clean and always expect to be dirty, so one of these may comeas a surprise. This means that the children cannot falsify their father’s statement, whichthey would be able to do if he only asks about their dirtiness. This gives us the insightthat this solution only works if the children will not be asked about both sides: aboutbeing dirty and about not being dirty. Linking this to the surprise paradox: the solutionin game theory would not work if the teacher also says “if the exam is not at a certain day,you will not know this in advance”.

4.4 Number+1

4.4.1 The Story

Story : Claire assigns Alice and Bob two numbers, n ∈ N∗ and m ∈ N∗ respectively, andwrites it on their forehead so they can only see the other one’s number. Then she tellsthem: “You’re number is one less or one more than the other one’s, but you will neverboth know your number.”10. Again no communication or mirrors are allowed, and the onlything A and B can do to get to know their number is reasoning. If one of them get’s toknow his number anyway, he will shout “I know my number!”.

We can illustrate the general case by taking an example: let us assume that Bob isassigned number 2 and Alice is assigned number 3, m = 2 and n = 3. Now personA can see that m = 2, so he infers from the statement that she has number 1 or 3:

10We can see the similarity with the surprise examination here: if your number is the smaller or biggerone, this will come as a surprise (for both of you).

45

‘KA(m = 2) ⇒ KA((n = 1) ∨ (n = 3))’. But, Alice reasons, “Bob sees my number, so ifI would have number 1, Bob would inmediately know that he has number 2. But this isnot possible by Claire’s statement, so number 1 can be ruled out” and Alice knows hernumber has to be equal to 3. And so Alice is able to deduce her number, which is againin violation with the statement.

4.4.2 Solutions to ‘Number+1’

Quine

When we adjust Quine’s solution to this puzzle we can argue that if A and B in the endconsider it possible that C’s statement is incorrect, they should do so in the beginningas well. Returning to the previous example, A cannot even rule out n = 1 because shecannot exclude the option that her number is 1 (in keeping with the statement) and that Bremains ignorant of his number being 2 (in keeping with the statement). Therefore the onewith the biggest number (A) cannot exclude that her number is the smaller one and B onthe other hand can also not conclude from C’s statement that he has the smaller number.Therefore both A and B remain with their initial propositional attitudes.

Gerbrandy

Applying the adjustment of Gerbrandy’s solution again a ‘∨K⊥ is added to the formaliza-tion of the sentence. We cannot for no reason accept the principle of success. Therefore,after the announcement A and B can only exclude 1 as a possible number. This will onlygive them information if the other has number 2. Thus, in the previous example, applyingthis solution, A rightly concludes to know her number is 3, which makes C’s announcementuntrue. In all other cases11 A and B do not get any more information out of it, which makesC’s announcement true.

Baltag and Smets

When we adjust Baltag and Smets’ solution to this puzzle, we speak of the belief-versionof C’s announcement (“You’ll never correctly believe your number”) and approximate theself-referentiality with an iterated upgrading with the non-self-referential statement. Weintroduce rounds in which C asks the two if they already have beliefs about their numbers.Just like we did with the Muddy Children puzzle, we will first look at the general caseand than visualize this by picking a specific (but arbitrary) case. Assuming A and B startwith total indifference, the initial possible worlds model corresponding to their situation is:

(1,2)↔A (3,2)↔B (3,4)↔A (5,4)↔B . . .

11Except of course for the completely interchangeable case in which A has number 2 and B number 3.

46

(2,1)↔B (2,3)↔A (4,3)↔B (4,5)↔A . . .

When C tells them for the first time “You do not correctly believe your number”, A andB inmediately decide that their number cannot be equal to one, otherwise the other onewould have came to (correctly) believe her or his number, and then C’s statement wouldhave been a lie. After the first round the model would thus be given by

(1,2)→A (3,2)↔B (3,4)↔A (5,4)↔B . . .

(2,1)→B (2,3)↔A (4,3)↔B (4,5)↔A . . .

and one can inmediately see what would happen if we would take the limit of the iteration.The resulting model then would be:

(1,2)→A (3,2)→B (3,4)→A (5,4)→B . . .

(2,1)→B (2,3)→A (4,3)→B (4,5)→A . . .

Now we return to a more specific, but still arbitrary example. C tells A and B roundafter round that they do not correctly believe whether their number is the other one’sminus one, or the other one’s number plus one. To illustrate what happens, without lossof generality, we assume that A has the higher number (n = m+ 1) and that this numberis bigger than 112. Because A only sees two possibilities (n is equal to m − 1 or m + 1),and so does B (m is equal to n− 1 or n+ 1) we are left with only three worlds:

• w1, which denotes the actual world (n,m);

• w2, which denotes the situation in which n = m− 1, or rather world (n− 2,m);

• w3, which denotes the situation in which m = n+ 1, or rather world (n,m + 2).

It seems a bit complex to define these worlds w1,w2 and w3, but this prevents us fromconfusing the worlds (because A and B would enumerate them differently - they see onlythe other one’s number). The plausibility relations of A and B in the corresponding initialmodel M are given by:

w1 ↔A w2 ↔B w3

Just like before, after m− 1 rounds, w1 will become a less plausible world for A than w3,and therefore the resulting model is

12In the case that n is 1 A would inmediately believe her number to be the smaller one. The roles of Aand B are completely interchangeable.

47

w1 →A w2 ↔B w3

And therefore, already A comes to correctly believe her number, which makes C’s state-ment false. Note that if A had the number m− 1 B would have came to correctly believehis number after m− 2 rounds. So we see that by iterating the statement always the onewith the bigger one comes to correctly believe his number, making the statement false. Ifwe carried on, after m rounds B would have came to incorrectly believe his number to bethe bigger one, which is represented in this model:

w1 →A w2 →B w3

Kritchman and Raz

Again when we use the adjusted version of Kritchman and Raz’ solution, we could say thatif person A and B believe in the consistency of the system, they would be able to rule outnumber 1 as a candidate. But as they will never be able to prove the consistency of thissystem within this system, they will not be able to continue this argument. In the end theybelieve they don’t have number 1, but they are not able to deduce whether their numberis the other one’s number plus one or the one’s minus one, unless of course the other hasnumber two, which makes C’s statement inmediately incorrect.The plausibility relation for both person A and B for n and m, respectively, is as follows:

(1,2)→A (3,2)↔B (3,4)↔A (5,4)↔B . . .

(2,1)→B (2,3)↔A (4,3)↔B (4,5)↔A . . .

This is the general case, in real person A and B would see the other’s number (whichis hard information) and therefore only distinguishing two cases. For example, if A hasnumber 3 an B number 4, the model would be as follows:

(3,2)↔B (3,4) ↔A (5,4)

In this case both A and B did not gain in knowledge or beliefs by the announcement. Theydo if one of them has number 213. For example if A has number 2 and B number 3, themodel would be as follows:

(2,1)→B (2,3) ↔A (4,3)

Now B believes to have number 3, which is correct in this case. He could have just aseasily had a wrong belief, if he had number 1 and A still number 2 (but then A would have

13Or number 1, but then the other would clearly have number 2. Requiring that one of them has number2 is enough.

48

known her number to be 2):

(2,1) →B (2,3)

Discussion

This interesting puzzle looks a bit like the first puzzle, ‘Smaller or Bigger’. It can be saidit is a special variant of this former puzzle, the case in which the numbers of A and B differonly one and in which A and B have to guess their own number instead of the other’s.That there are only two possibilities for their numbers makes the solutions a lot easier, andthey are similar to the solutions in section 2. The adjustment of Quine’s solution is similar;they cannot rule out to have number one, where first they were not able to rule out for theother’s number to be equal to one. The adjusted solutions of Gerbrandy and Kritchmanand Raz are also similar to the solutions in section 1 of this chapter. Baltag and Smets’adjusted solutions differs in the way that the one with the bigger number now first gets tocorrectly believe her or his number, but this can be explained by the changed roles of Aan B (because they have to guess their own number and know the other person’s number,instead of the other way around). Again this is somehow ‘half paradoxical’: always oneof A and B will guess it right. And again one could argue that there is always an easypragmatic (ad hoc) solution to this puzzle, because there are only two options: if they bothbelieve to have the smaller number (or if they both believe to have the bigger number) C’sstatement will be falsified.

49

Chapter 5

Conclusion and Relations to OtherWork

In this thesis I discussed the paradoxicality of the surprise examination and five mainapproaches to them1. Then I tried to apply these solutions in an adjusted way to threepuzzles known in the logical field, but then made self-referential. Using this method I triedto find out what exactly makes the surprise examination so puzzling, and which ingredientsare needed for a similar paradox to occur.We saw that the ideas of the solutions raised to (variants of) the surprise examination differin idea from blaming the students’ reasoning, interpreting the statement non-selfreferentialand therefore making the induction argument impossible, defying the principle of success,to interpreting a surprise in terms of belief and letting the students lower their degree oftrust (while still keeping a possible attitude). The results of this approaches differ fromremaining with the original propositional attitude to only ruling out Friday (and thereforein both cases enabling the teacher’s statement to come true) to making the teacher’s state-ment false by everyday believing the exam to come. We have seen that taking “there willbe an exam” as part of the announcement or not does not affect the idea of the solution,and that the different solutions raised to the problem are more similar than one would sayat first sight.

From applying these solutions to the self-referential puzzles we saw that the studentsact as ‘one student’, and that otherwise the reasoning might change. But more importantly,we saw from the puzzles that in the surprise examination there exists actually somethingas a ‘good’ or ‘bad’ wrong belief. A ‘good’ wrong belief would be “not be given an exam,while you were expecting it” and a ‘bad’ wrong belief “be given an exam, while you werenot expecting it”. But then, would “not be given an exam, while you were expecting it”maybe be surprising as well? Or would “a not expected exam” perhaps be a better for-mulation for the teacher’s announcement? Although, “the exam will come as a surprise”seems to speak only of the exam being a surprise, not about “no exam” being a surprise.

1Although you could reason that Baltag and Smets actually had more than one approach.

50

Trying to apply the solutions raised to the surprise examination to the puzzles stressed theimportance of this.

Besides the many things I discussed in this thesis there are many things which I wouldhave wanted to do, but I could not do because of lack of time or because it was outside thescope of this thesis. For instance, there were many other solutions proposed to the surpriseexamination or variations on it, and many papers written discussing the phenomenon. Isimply could not read them all. Something else which I had to leave undiscussed is ifthe surprise examination fits in what Priest[19] calls ’Russell’s Schema’ which considersthe paradoxes of self-reference to have a common underlying structure, irrespective of itscategory. Maybe in this way or another we could find similar paradoxes tot the surpriseexamination in a semantic or set-theoretic setting. It would also have been interesting tolook at the more general solutions to all paradoxes and the consequences they have for thestudy of logic. This leads me to the connections that can be made to other work.

5.1 Relating to Other Work

In this section connections are made between this thesis and two other logical concepts:the revision theory of truth and non-monotonic logic.

5.1.1 Truth Revision versus Belief Revision

Tarski showed that giving a consistent theory of truth for languages containing their owntruth predicate is impossible. To model the kind of paradoxical reasoning involved in theLiar paradox within a two-valued context, Gupta and Herzberger in the early eighties inde-pendently developed a new theory of truth, the Revision Theory of Truth[16]. The core ofthis theory is the revision process by which hypothesis about the truth values of sentencesare revised2.

Now how can this new theory of truth be linked to the surprise paradox? Let us look aton one hand the iterated version of Gerbrandy’s non-self-referential formalization (equal toBaltag and Smets’ iterated updating for surpriseK for an infallible source) and on the otherhand the iterated soft upgrading of Baltag and Smets’ surpriseB for a trusted but falli-ble source, and how these methods compare to the formerly discussed truth revision theory.

Let us suggestively call the initial plausibility model representing the students’ propo-sitional attitude h0. Then for Gerbrandy the first update leads to deleting world 5 andfor Baltag and Smets the first upgrading leads to making (amongst others) these worldsless plausible. This can be viewed as the first hypothesis which is revised into the newhypothesis h1 and h′1 for Gerbrandy and Baltag an Smets respectively. Another updating

2For a more comprehensive introduction on this theory of truth I refer to the Appendix.

51

or upgrading then brings us to hypothesis h2 and h′2, and so on until h4 and h′4.

We can see the similarities between the knowledge or belief revision and the truthrevision process. However, the similarity between belief revision and truth revision is moreapt: just like in the truth revision process there is nothing definite in the belief revisionprocess. An agent can always come back to his beliefs about the several worlds, also oncechanged. This is different in Gerbrandy’s knowledge revision process: once deleted a worldcannot ever become knowledge again. That is, not in classical logic. This would howeverbe possible in non-monotonic logic, discussed in the next section.

5.1.2 Non-Monotonic Logic

Non-monotonic[2] logic covers a family of formal frameworks in which ‘defeasible inference’is central: a more tentative kind of inference in which one is able to retract from conclu-sions drawn previously. This is very useful in everyday life and in fields like computerscience: inferring things which are then known until other new information might changethis. An important feature of a non-monotic logic is that the size of a knowledge base doesnot always increase by adding new information, but can also stay the same or even shrink.An example: if you inferred you have a bad hairday, but later infer otherwise, you will loseyour knowledge on having a bad hairday.

We can also link this logic to solutions to the surprise examination discussed in thisthesis. For example, if we look at the solution raised by Kritchman and Raz, we recognisethe ‘strong’ classical inference used in their solution. Write ConsistentT+S for consistencyof the system T + S3, with S again referring to the surprise sentence. It is provable thatT + S `strong ConsistentT+S → ¬5, but because T + S `strong ConsistentT+S is not prov-able also T + S `strong ¬5 cannot be inferred.

Now suppose we are in a non-monotonic logic, with a ‘weaker’ kind of inference, `weak.You could again say that T + S `weak ConsistentT+S → ¬5. But suppose you also believein the consistency of the system, say ConsistentT+S holds in all maximal worlds. Thenyou would maybe be able to infer T + S `weak ¬5 from it, in the weak sense. But nowyour induction argument is indeed possible, it does not stop! It can then be inferred thatT + S,¬5 `weak ¬4, that T + S,¬5 ∧ ¬4 `weak ¬3 and so on. We could identify this witha plausibility relation in which Friday is less plausible than Thursday less plausible thanWednesday and so on:

1← 2← 3← 4← 5

This is exactly the solution raised by Baltag and Smets for the belief-variant of surpriseand a fallible but trusted source. We may conclude that if we take Kritchman and Raz’

3 See solution Kritchman and Raz in chapter 3.

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solution for the provability-variant of surprise as a starting point and use this kind of weaklogical entailment in non-monotonic logic, we end up in the solution raised by Baltag andSmets.

5.2 Conclusion

Having said all this leads me to the end of my thesis, and the conclusions I draw fromthe work done. I did not think beforehand that in this thesis I was going to find THEsolution to the paradoxicality of the surprise examination. What I did expect was to get abetter understanding of ‘surprise’ and of paradoxes as a whole. I think I succeeded in thisrespect, however I remain with many questions for which answers I will keep looking afterfinishing this thesis.

The several solutions raised to the surprise examination all have their charms. I amespecially attracted to on one hand Kritchman and Raz’ proof using Godels theorem andon the other hand Baltag and Smets’ idea to look at a softer notion of surprise and alower trust in the teacher. For the knowledge-version of surprise I also think Quine has agood point when he sees the fallacy in the students’ reasoning, but lowering their trust to-wards the teacher after reaching a contradiction seems a very natural reaction to me as well.

As told before, one main goal was to find out what exactly is so puzzling about thisstory. Writing this thesis I did not only get one answer: I got several! Another reason forme to do this research is that I think that paradoxes matter. A view which is widely sharedis that paradoxes may point out that there is something wrong with our understanding ofthe fundamental concepts involved. For example, take Zeno’s paradox of Achilles and theTortoise. In this ‘paradox’ Achilles runs 10 times faster than the tortoise and thereforegives him a head start of 100 meters. Now the main reasoning in this story is that wheneverAchilles reaches a point where the tortoise has been, the tortoise went a little bit furtherand therefore Achilles never can overtake him.It turned out that the paradox rested on an inadequate understanding of infinity. Theparadox was solved.

Paradoxes matter and can be solved. And although I think we should not forget thenormal cases in which there is no paradoxicality at all, it is interesting to examine theparadoxes closely to gain in knowledge (or belief) of the fundamental concepts involved.The different interpretations of the surprise-sentence raised in this thesis actually createdifferent paradoxes. We can always reformulate them more closely to get different storieswith different solutions. But I trust that also the ’real paradoxes’ will be solved in time.And maybe in some years also the paradox of the surprise examination will be solvedbecause we reach a better understanding of its fundamental concepts, just like it was thecase for Achilles and the Tortoise.

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Popular Summary

The surprise examination is a paradox in which a teacher tells his students that the examthat will be given next week will come as a surprise. Suppose it is commonly understoodthat the exam will be a surprise if the students cannot know the evening before that theexam wil be tomorrow. The smart students then reason that the exam cannot be givenon Friday, because they would know this at Thursday evening (because the exam did notoccur on the previous days only one option is left) and thus at Friday the exam wouldnot come as a surprise. If we represent this reasoning in a model with the double pointedarrows standing for regarding two days as equally plausible:

Monday↔ Tuesday↔Wednesday↔ Thursday↔ Friday

which becomes after the update

Monday↔ Tuesday↔Wednesday↔ Thursday

The students argue that the exam cannot be given on Thursday either because then theywould know this at Wednesday evening, because they just now derived that the exam can-not possibly be given on Friday:

Monday↔ Tuesday↔Wednesday

With this backward induction argument the students are able to eliminate any day of theweek as a possible day for the exam to occur. They reach a contradiction. But now theteacher can give the exam at any day of the week, and it will still come as a surprise:because of eliminating all days as possibilities they do not expect the exam to be given.

To solve the mystery which arises in this paradox, many solutions have been raised. Afew of them have been discussed in this thesis, taking different approaches to the problem.Some say the backward induction argument does not hold, one says that the students’ rea-soning to eliminate Friday even fails and others say that reaching a contradiction will makethe students lower their degree of trust in the teacher. For instance, if the students do notcompletely trust the teacher after the announcement, they may think it is less plausiblefor the exam to occur on Friday but do not completely rule this out as a possibility. In a

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model with arrows pointed towards the day which the students regard as more plausible,this can be represented as follows:

Monday↔ Tuesday↔Wednesday↔ Thursday← Friday

They could then perhaps again exert the backward induction shown before, while notdeleting worlds but just making them less plausible. However the backward inductionargument stops here, unless you would interprete surprise as “not believing beforehand”.This is comprehensively discussed in this thesis.

The different approaches raised to the problem have different outcomes, varying fromnot gaining in knowledge or beliefs at all to falsifying the teacher’s statement and expectingthe exam to come every day. There are of course different interpretations of this paradox,which are widely discussed in this thesis. One could for instance interprete the statement asreferring to itself (stating its own truth even after being uttered) or not, interprete surprisein terms of knowledge or belief and choose from possible (positive or neutral) attitudestowards the teacher. The ambiguity of the formulation of the surprise sentence uttered bythe teacher is the lion’s share of the difficulty and mystery of the surprise examination.

In this thesis I also adapted the several solutions to known logical puzzles which I madeself-referential. In this way we are able to examine more closely what is exactly so puzzlingabout the surprise examination and why the solutions seem to work. The goal of thisthesis is getting a better understanding of the concept of surprise and of paradoxes moregenerally.

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Appendix

In this chapter the Revision Theory of Truth is explained by taking the Liar paradox asan example.

5.3 A New Theory of Truth Explained

The Revision Theory of Truth[8] is one of the main rivals to the threevalued semanticswhich state that sentences can be true, untrue or neither (or both). To give an idea ofhow the truth revision works, we return to the example of the Liar[16]. We write the Liarsentence as

(1) (1) is not true

We can revise the Liar’s truth value as follows: let us first assume that sentence (1) is nottrue; hypothesis h0 = (1) is not true. Then by the its T-biconditional the name ‘(1) is nottrue’ is true if and only if (1) is not true. From these two by modus ponens we know that‘(1) is not true’ is indeed true. The identity (1) = ‘(1) is not true’ holds, and thereforewe can conclude by replacement that (1) is true. This is our new revised hypothesis, h1.Now we can again revise this hypothesis by using the T-biconditional. Now we start withhypothesis h1 = (1) is true. The T-biconditional is again ‘(1) is not true’ is true if andonly if (1) is not true, but now we can infer from this that ‘(1) is not true’ is not true.Because of the identity we can use replacement again and infer (1) is not true. This is ournew new revised hypothesis, h2. We see that we get an alternating sequence of hypotheseswith hi is (1) is not true for all odd i and hi is (1) is true for all even i. If we would havestarted with h0= (1) is true, this would be the other way around.

We see that how we start does not really effect the outcome: in any way we reachan infinetely alternating sequence. Gupta and Belnap call this unstable. Being unstablemeans not being stably true nor stably untrue: the revision sequence of revised hypothesisdoes not stabilize on one of the two truth-values.

What explained formerly however is only the informal idea of the reasoning proposedby Gupta and Belnap. For further work I refer to their book The Revision Theory ofTruth[8].

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