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
Abstract
In this thesis several approaches and solutions to the surprise examination are
discussed, in which a teacher tells the students that the exam given next week
will come as a surprise. By a backward induction argument the students are
able to rule out every day of the week as a possibility, reaching a contradiction
and thereby enabling the teacher to surprise them by giving the exam at any
day of the week. Three known logical puzzles, Smaller or Bigger, the Muddy
Children and Number+1, are made self-referential and the several solutions
are in an adjusted form applied to these puzzles to gain knowledge about
the concept of surprise. In this thesis the different solutions to the surprise
paradox are compared and shown to be more similar than one at first sight
would think.
Information
Title: On the Surprise Examination
Author: Ottilia Kasbergen, [email protected], 5939992
Supervisor: dr. Alexandru Baltag
Second reader: dr. Sonja Smets
End date: June 19, 2012
The Institute for Logic, Language and Computation
University of Amsterdam
Science Park 904, 1098 XH Amsterdam
http://www.illc.uva.nl
Contents
1 Introduction
1.1 The Surprise Examination . . . . . .
1.2 Paradox . . . . . . . . . . . . . . . .
1.3 Self-Referentiality . . . . . . . . . . .
1.4 Solution . . . . . . . . . . . . . . . .
1.5 Discussing the Surprise Examination
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2 Logical Preliminaries
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3 Solutions to the surprise paradox
3.1 Quine . . . . . . . . . . . . . . . . . .
3.2 Gerbrandy . . . . . . . . . . . . . . . .
3.3 Baltag and Smets . . . . . . . . . . . .
3.3.1 Surprise in Terms of Knowledge
3.3.2 Surprise in Terms of Belief . . .
3.4 Kritchman and Raz . . . . . . . . . . .
3.5 Discussion . . . . . . . . . . . . . . . .
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5 Conclusion and Relations to Other Work
5.1 Relating to Other Work . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Truth Revision versus Belief Revision . . . . . . . . . . . . . . . . .
50
51
51
4 The Surprise Paradox Elsewhere
4.1 Looking Elsewhere . . . . . . . . . . .
4.2 Smaller or Bigger . . . . . . . . . . . .
4.2.1 The Story . . . . . . . . . . . .
4.2.2 Solutions to ‘Smaller or Bigger’
4.3 Muddy Children . . . . . . . . . . . . .
4.3.1 The Story . . . . . . . . . . . .
4.3.2 Solutions to ‘Muddy Children’ .
4.4 Number+1 . . . . . . . . . . . . . . . .
4.4.1 The Story . . . . . . . . . . . .
4.4.2 Solutions to ‘Number+1’ . . . .
1
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5.2
5.1.2 Non-Monotonic Logic . . . . . . . . . . . . . . . . . . . . . . . . . .
Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
53
Popular Summary
54
Appendix
5.3 A New Theory of Truth Explained . . . . . . . . . . . . . . . . . . . . . .
56
56
2
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 defense
units are properly prepared, no one will know in advance on what day this exercise will take
place.”[18]
A Swedish mathematician, Lennart Ekbom, recognized the contradictory nature of this
announcement. He discussed it with his students and afterwards this puzzling story spread
around the world.
In 1948 the phenomenon first appeared in print, discussed by O’Connor in the British
magazine Mind. After that it appeared in many other publications all over the world, while
taking several forms. From the civil defense exercise to the unexpected egg, in which an
egg was hidden in one of ten boxes and it was announced that you cannot expect where
you will find the egg. From the unexpected hanging, in which a judge tells a convict that
he will be hanged next week but will remain ignorant about this fact until the morning on
which 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 be
able 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 is
commonly understood that an exam comes as a surprise if you do not know the evening
before that it is given the next day.”
The contradictory part in this announcement lies in a backward induction argument
which the students can practise. They reason:
“If the exam is given on Friday, then on Thurday evening we will not have gotten an exam
3
yet and therefore know that it must be given on Friday. But then it would not come as
a surprise, and therefore it cannot be given on Friday. But then, suppose the exam is
given 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 not
on Friday. Thus the exam has to be given on Thursday. Again this would not come as a
surprise, so we can rule out Thursday as well. If we repeat this argument we can also rule
out Wednesday, Tuesday and Monday. There are no days on which we can have a surprise
exam: 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 the
students 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 examination
perhaps a genuine paradox? And, paradox or not, is there a satisfying solution? What will
this solution be?
1.2
Paradox
Let us first have a look at what ‘paradox’ really means. What conditions should a puzzling
story such as the surprise examination meet for us to call it a paradox? Intuitively, we
regard as a paradox a seemingly sound piece of reasoning based on an apparently true
assumption that leads to a contradiction.
The Encyclopedia of Philosophy defines:
“A logical paradox consists of two contrary or even contradictory propositions to which we
are led by apparently sound arguments. The arguments are considered sound because when
used in other contexts they do not seem to create any difficulty.”[24]
Among the paradoxes we can distinguish three main categories: the semantic, settheoretic and epistemic paradoxes. The first categorie speaks of paradoxes based on the
semantic notion of truth, the second on the formal definition of a set and the third on
the concept of knowledge. Well-known paradoxes for these categories are, respectively, the
Liar’s paradox, Russell’s Paradox and the Knower paradox. I will give a brief explanation
of 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”.
1
2
Or on any other day of the week.
Thus, he is a Cretan himself.
4
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 that
it 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 it
is actually false, but that would mean that it is true, and so on ad infinitum. We see that
the sentence, when uttered and assigned a truth value, changes this truth value again and
again ad infinitum.
The Liar’s paradox is an example of a semantic paradox. Let us regard now the settheoretic paradox Russell put forward in the beginning of the previous century. The British
logician 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 R
do 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 not
belong to itself: R ∈ R. All elements of R do not contain themself, thus because R is not
an element of R it does contain itself: R ∈ R. But this would mean it actually does not
contain itself, which in turn would mean it does, and so on ad infinitum. We see that the
following 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, although
it is a bit more complex. By assuming to obtain a contradiction that “This sentence is not
known 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. This
is a contradiction.
The Liar’s paradox, Russell’s paradox and Knower paradox all seem to have a paradoxical nature: to be both true and false at the same time. We can regard them as true
paradoxes: although the argumentation steps seem to be right we get a contradictory outcome.
5
1.3
Self-Referentiality
The paradoxes introduced in the previous section share one common feature: they are all
self-referential: they refer to themself or their own referent. Their cyclic self-referential
nature 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’, which
makes 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 subsequent 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 like
described above. First it can be proved that for all k Sk cannot be true. We assume for
contradiction that Si is true for some i. Thus it holds by definition that for all k > i Sk is
untrue. In particular this holds for k = i + 1. Sentence Si+1 states that for all k > i + 1 Sk
is untrue, thus this sentence being untrue implies that there has to exist a j > i + 1 such
that 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 contradiction
with what we have shown before.
Yablo’s example showed us that (cyclic) self-referentiality is not needed for a paradox
to 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 at
solutions to the paradoxes.
1.4
Solution
Paradoxes are important for the understanding of the foundational concepts involved and
the solutions raised to the several paradoxes therefore have far-reaching consequences. The
different categories of paradoxes had different consequences: semantical paradoxes led to
several theories of truth, set-theoretic paradoxes to new foundations for set-theory and new
definitions for the term set and epistemic paradoxes to new theories of knowability. These
are however somewhat outside the scope of this thesis.
3
Though his example is analogous with a self-referential paradox: the Liar paradox.
6
Other solutions to paradoxes are even more rigourous: the British philosopher Priest
tells us to just accept that some sentences are both true and false. He concludes this because all known solutions fail to indicate which premise or argumentation step is invalid or
to give an illuminating independent not ad hoc reason for rejecting it. The ideal solution
Priest describes resembles Haack’s[15] division in the formal and philosophical solution: the
formal solution should indicate which apparently unexceptionable premises or principle of
inference must be disallowed, whereas the philosophical solution should be an explanation
of 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 assign
it 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 not
hold.
The Liar paradox and Russell’s paradox mentioned before have solutions with the first
as starting point: there seems to be something wrong with the proposition whereupon it
is impossible to assign it a truth value. The Liar paradox is by Revision Theory of Truth
said to be an unstable sentence, and therefore alternates it truth value for ever.5 The set
R 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 paradox, just like the Knower paradox mentioned in the previous sections. For the Knower
paradox it is clear that it is indeed a true paradox, for the surprise examination, even this
remains to be controversial. However, because the reasoning in the surprise examination
seems to be sound and the outcome is contradictory, in this thesis I will take the liberty
of sometimes referring to it as a paradox.
In the teacher’s statement we can distinguish two seperate statements: “next week
there will be an exam” and “the exam will come as a surprise”. Some scientists who raised
4
Ditmarsch and Kooi proposed this to solve epistemic paradoxes, like the surprise examination.
For a further discussion on this I refer to the appendix.
6
Whereas according to Naive Set Theory any definable collection of sets is a set.
7
For a further explanation of this I refer to Devlin[11].
5
7
a solution took the first part as a certainty, some did not. However, in the chapter 3 we
will see that this does not really change their solutions.
But then, how should “surprise” be formalized? In many solutions this is interpreted
as “not knowing beforehand (the evening before) that something is the case”. Others interprete it in terms of belief or even in terms of provability.
There are different variations of the surprise paradox, in which mostly time plays a
role. However, Sorensen proved in 1988 that this is not necessary, with his ‘designated
student’.[21] The solutions discussed in chapter 3 however are all based on variations in
which time does play a role. These solutions are standardized: all deal with the surprise
examination, taking “there will be an exam next week” as a certainty. Some say it is indeed
a paradox, others say it is not: there might be something strange going on that creates
unexpected results. One major question remains whether the teacher’s statement has to
be and stay true.
Before we can have a closer look at the various solutions to the surprise paradox, I will
introduce some logical concepts which will be used in the later chapters.
8
Chapter 2
Logical Preliminaries
To be able to discuss the several solutions to the surprise examination, we first need to have
some logical concepts introduced. In this chapter the definitions and axioms are introduced
which are needed in later chapters.
We are able to translate any real situation into a model of worlds, relations and a
valuation, a possible worlds model. In this model the individuals spoken about are called
agents and the several worlds represent the several scenarios that may be representing the
actual truth. If a world is accessible for a particular agent, agent i, by accessibility relation
Ri , this means that this agent considers this world possible. The relations an agent has is
called it’s propositional attitude. We should now define the language in which the formulae
that hold in the worlds are written.
Definition 2.1. [22] (Epistemic Logic language LEL ). For G a set of agents, i ∈ G and
A a basic set of propositions (atoms) named by p, q, r, . . .,
it is inductively defined how to construct further expressions
ϕ ::= p | ¬ϕ | (ϕ ∧ ψ) | Ki ϕ1
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, Ki ϕ, needs some further introduction.
But first one more definition is needed.
Definition 2.2. [6] (Epistemic Possible Worlds Model). For A the set of propositional
atoms and G the set of agents, a possible worlds model for agents i ∈ G is a triple S =
(S, Ri , k · k) 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 k · k assigns to each atomic sentence p ∈ A some set kpk ⊆ S of
worlds in which p holds.
1
In the original definition common knowledge is defined as well. We don’t need it in this thesis, so it
is left out in all definitions in this chapter.
2
ϕ ∨ ψ ≡ ¬(¬ϕ ∧ ¬ψ) and then ϕ → ψ ≡ ¬ϕ ∨ ψ.
9
Now we can define knowledge for agent i as being true in all worlds in the model
accessible 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 satisfy:
M, s |= p ⇔ s ∈ kpk
M, s |= ¬ϕ ⇔ not M, s |= ϕ
M, s |= ϕ ∧ ψ ⇔ M, s |= ϕ and M, s |= ψ
M, s |= Ki ϕ ⇔ for all t with sRi t: M, t |= ϕ
To make all this a bit more insightful, we shall look at an example: consider Alice, who
just woke up and wonders if she has a bad hair day today. She does not have a mirror, so
she is completely ignorant about if she has a bad hair day or not. Let us define b = ‘Alice
has 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 process
graph 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 |= ¬KA b, M, 1 |= ¬KA ¬b, M, 2 |= ¬KA b and
M, 2 |= ¬KA ¬b: Alice is completely indifferent regarding to the state of her hair. We might
also want to formally reason about what happens if an agent gets new information and
therefore change their propositional attitudes. This can be done in Public Announcement
Logic. After a public announcement is made with information ϕ, all agents get to know
that ϕ is the case. We denote the action of making ϕ publicly known with [!ϕ] and write
[!ϕ] Ki ϕ: after the announcement has been made agent i knows that ϕ. Agent i now
has to update her model with this new information. This is formally defined in the next
definitions.
Definition 2.4. [22] (Public Announcement Logic language LP AL ). The language of PAL
is the epistemic language with added action expressions, as well as dynamic modalities for
these, defined by the syntax rules:
Formulae P ::= p | ¬ϕ | (ϕ ∧ ψ) | Ki ϕ
Action expressions A :!P
The epistemic language is interpreted as before. Note that again ϕ ∨ ψ and ϕ → ψ can
be constructed by using negation and conjunction. We require that P has to be true to be
announced, otherwise the knowledge implied by it in the agents’ propositional attitudes
would be untrue, which contradicts Veracity of Knowledge stated in definition 2.10. If we
write M |P for the updated model after the announcement that P , the new dynamic action
modality is defined[22] by
M, s |= [!P ] ϕ ⇔ if M, s |= P then M |P , s |= ϕ
10
We can now write sentences like [!ϕ] Ki ϕ. For example, let us suppose now that Alice
finds a mirror in the room of her housemate, and can now assure herself dat she does not
have a bad hair day and can go to the university just like this. Suppose she sees that her
hair 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 newly
received information. Suppose now the mirror Alice found was really dirty, so looking in it
she might think that she does not have a bad hair day. Alice acknowledges she cannot be
sure of this because the mirror is so filthy; she only believes that she does not have a bad
hair day. When we want to speak about the beliefs someone has, we have to introduce a
plausibility relation ‘≤’. This relation gives an order on the worlds of being less or more
plausible, and formulae that hold in all most plausible worlds are believed. We demand
that the plausibility relation is a totally connected preorder. ‘Totally connected’ means that
all worlds have to be comparable: ∀s, t ∈ S it holds that s ≤ t or t ≤ s[3]. A ‘preorder’ is
reflexive and transitive: ∀s ∈ S s ≤ s and ∀s, t, u ∈ S s ≤ t ∧ t ≤ u → s ≤ u[3]. If we
put this plausibility relation together with the formerly defined possible worlds model, we
get the plausibility model, which is just a model (S, ≤, k · k) having a plausibility structure
underlying it.
Definition 2.5. [6] (Plausibility Model). This model is a quadrupple S = (S, Ri , ≤i , k · k)
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 k · k assigns to each atomic sentence p ∈ AT some set kpk ⊆ S of
worlds in which p holds.
For shortage, we sometimes write S = (S, ≤i , k · k) to denote S if we are only speaking
about it’s plausibility relations. If there is a plausibility relation between two worlds this
implies an accessibility relation between these worlds. We now find yet a more comprehensive 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) | Biψ ϕ
Again ϕ ∨ ψ and ϕ → ψ can be constructed by using negation and conjunction. In
this logic, knowledge Ki ϕ and ‘unconditional’ belief Bi ϕ can be derived from conditional
belief Biψ ϕ by putting Ki ϕ := Bi¬ϕ ϕ and Bi ϕ := Bi> ϕ. Belief in ϕ by agent i, Bi ϕ, and
conditional belief in ϕ by agent i given ψ, Biψ ϕ, are defined below. We write s(i) for the
set of worlds that is reachable for agent i from s, the information cell of i at s.
11
Definition 2.7. [6] (Maximal worlds). A world s is a maximal (most plausible) world if
and only if it belongs to the set of maximal worlds M ax≤ 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, Bi ϕ, if and
only if it holds in all maximal worlds of s’ information cell:
M ax≤ s(i) = {s ∈ s(i) : t ≤ s for all t ∈ s(i)} ⊆ kϕk.
Definition 2.9. [6] (Conditional Belief ). A sentence ϕ is believed given on ψ at world s
by agent i, Biψ ϕ, if and only if it holds in all most plausible worlds satisfying ψ:
M ax≤ kψk := {s ∈ kψk : t ≤ s for all t ∈ kψk} ⊆ kϕk.
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 ψ
and Necessitation: From ϕ infer Kϕ.
Having defined all this we can speak of Alice’s beliefs about the condition of her hair after
looking in the dirty mirror. If we write a pointed arrow towards the world which she thinks
12
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. To
do 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 makes
all ϕ-worlds are more plausible than all non-ϕ-worlds.
Definition 2.12. [3] (Conservative upgrade ↑). An upgrade with ↑ ϕ makes all maximal
ϕ-worlds more plausible than all other worlds.
We see that a radical upgrade induces a stronger kind of belief than a conservative
upgrade.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. The
source is strongly believed to be truthful.
Conservative upgrade: The source is trusted, but only barely. The source is believed to be
truthful, 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 information, we can define a model transformer T ϕ for soft information more formally:
Definition 2.14. [6] (Model Transformer). A belief upgrade with a sentence ϕ is a model
transformer T ϕ that takes any plausibility model S = (S, ≤i , k·k), and returns a new model
T ϕ(S) = (S 0 , ≤0i , k · k ∩ S 0 ), such that (i) the new non-empty set of possible worlds S 0 ⊆ S;
(ii) a new plausibility relation ≤0i , satisfying kϕkS ∩ S 0 6= ∅ =⇒ M ax≤0 S 0 ⊆ kϕkS .
(iii) the restriction of the valution map to S 0 , k · k ∪ S 0 .
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 |T ϕ for the model obtained by
upgrading model M with T ϕ.
Let us consider again Alice and the filthy mirror. Before she looks in it and receives soft
information, 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 upgraded
model 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
13
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 operators
N EXT and BEF ORE. They imply the existence of temporal plausibility models, which
can be identified with sequences S0 , S1 , . . . , Sn , . . . of plausibility models obtained by successive upgrades T0 , T1 · · · such that Sn+1 = Tn (Sn ). The future-tense operator N EXT
and the past-tense operator BEF ORE are defined as follows:
Definition 2.15. [6] (NEXT operator).
N EXT ϕ holds at Sn if and only if ϕ holds at Sn+1 .
Definition 2.16. [6] (BEFORE operator).
BEF OREϕ 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 |= N EXT (BA ¬b)
and M1 |= BEF ORE(¬KA b).
Now we have defined all the concepts that we need in our discussion of the solutions to
the surprise examination in the next chapters.
14
Chapter 3
Solutions to the surprise paradox
Many interesting solutions have been put forward the last decades, and still the philosophers and logicians do not agree on which solution is the correct one. Part of the problem is
that the surprise examination can be interpreted in several ways. If we look at the teacher’s
announcement ‘next week there will be a surprise exam’, we see this can actually be split
in two announcements: ‘next week there will be an exam’ and ‘this exam will come as a
surprise’. I will refer to the first part of the statement as ‘exam’ and to the second part
as ’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 which
the teacher gives one exam every week. Then ‘exam’ is already known by the students
before the teacher utters his strange announcement. To get a more standardized form for
the 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 made
in the solution of Quine and Kritchman and Raz however does not change the idea of the
solutions. 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 the
students’ possible worlds model we associate the days (Monday to Friday) with numbers
1, 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’ plausibility 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 it
will 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 be
given 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’s
formalization. As mentioned before, we treat ‘exam’ as something the students know:
15
W
V
K( 1≤i≤5 i) ∧ 1≤i<j≤5 K(¬(i ∧ j)).
So, before the announcement the students know ‘exam’, and they are completely indifferent
regarding to on which day the exam will take place. Their initial model therefore looks
like 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 previous 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 model
M1 = 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 out
Friday (K¬5) as a candidate, this implies that on Wednesday evening they would know
that the exam will be given on Thursday. This cannot be the case because the teacher told
them the exam will be a surprise, thus they conclude that the exam cannot be given on
Thursday 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 the
exam 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
1
Note that what we showed was sufficient to eliminate Friday because of the second conjunct of ‘exam’.
16
For a last time, following the same argument, the students update their model with the
deletion of Monday: Model M5 = M4 |¬1 . But now all the worlds are deleted! In this
model ’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 to
this problem. I will present here a number of interesting solution to this paradox. These
solutions were not all specified to the surprise examination: Quine discussed the Hangman
Paradox. I adjusted his solution to the story of the surprise examination, but this does
not affect the solution.
3.1
Quine
Quine[20] suggests that if the students can conclude after the backward induction that the
announcement will not be fulfilled, they should have been prepared to take this alternative
into consideration as a possibility from the beginning. This will give the students three
options when they look ahead at Thursday evening (as where they first only considered
the 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 with
the 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 with
the 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 students will after hearing ‘surprise’ be entitled to eliminate (b), but not (c). According to
Quine, to suppose that the statement impedes (c) is to confuse two things: the hypothesis
that the statement will be fulfilled and the hypothesis that at Thursday the students will
know that the decree will be fulfilled. Thus, Quine reasons, the students will not be able
to exclude Friday by their argument. Therefore they will keep their original beliefs, the
corresponding model and plausibility relations stay the same. In other words: the students
adopt a ’neutral attitude’ towards the teacher.
Quine for the sake of his argument also brings the surprise paradox back to the form of
one possible day. In his case, this refers to the hangman in which the judge tells the convict on Sunday afternoon that he, the convict, will be hanged the following noon and will
remain 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 shall
be hanged tomorrow noon, and do not know it now.” are possibilities, and that only the
last one would fulfil the decree. Therefore, the convict thinks: “Rather than charging the
judge 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 the
teacher’s announcement: they do not draw conclusions about the day of the exam based
on 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 us
that the surprise statement becoming true after reaching a contradiction seems to imply
that ‘knowing a contradiction’ should be part of the formalized statement. To be able to
do this, he treats knowledge like a K45-operator. That is, knowledge doesn’t have to be
factive (6` Kϕ → ϕ), nor does it have to be consistent(6` ¬K⊥). Gerbrandy for simplicity
brings the number of days back to three: Wednesday, Thursday and Friday, and his formalization of the teacher’s announcement then is as follows:
S = (we ∧ ¬Kwe) ∨ (th ∧ [!¬we] ¬Kth) ∨ (f r ∧ [!¬we] [!¬th] ¬Kf r) ∨ 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 you
learn something, you come to know it is true: [!ϕ] Kϕ. Accepting this principle makes the
reasoning of the students as follows: after the announcement (S) that there is going to be
a surprise exam, the students will know this, KS. This sets the students thinking and like
shown before they infer to know a contradiction. From this, because of the definition of
S and because K does not have to be factive or consistent, they infer that there indeed is
going 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 exclude
Friday, but that their argument should then stop. He thinks the students are right about
concluding that after the announcement the exam will not be a surprise if it is given on
2
In our version it would be taken as a certainty that the convict will be hanged, and therefore he would
not consider this alternative.
3
The formalization Gerbrandy suggested is ‘(we∧¬Kwe)∨(th∧[¬we] ¬Kth)∨(f r∧[¬we] [¬th] ¬Kf r)∨
K⊥’. To prevent from ambiguity I decided to add ’!’ here, because we will later also discuss the upgrades
’⇑’ and ’↑’.
18
Thursday, and they are correct in that the teacher said that it would be, but they are not
correct to see a contradiction between these two claims. He states that the sentence can
be true before the teacher utters it and becomes false after learning of its truth; which is
confusing, but not paradoxical. Gerbrandy feels that the implicit use of the principle of
success is what makes the surprise examination so puzzling. Just announcing ‘S’ would
provide the students with the information that the exam cannot be on Friday, but if we
don’t accept the principle of success, they cannot exclude any more days.
Gerbrandy also suggests that we can interprete the teacher’s statement differently. For
example, 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 be
a surprise, it will still come as a surprise. The first conjunct tells the students that the
exam is not on Friday. The second conjunct says that after learning that the exam is not
on Friday, it will still be a surprise and therefore it will not be on Thursday either. The
students 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 leaves
us with the same result: the students eliminated all days of the week. This iterated notion
in closely related to the self-referential interpretation of the statement. Gerbrandy also
discussed such an interpretation of the surprise exam, formulated as ϕ ↔ S ∧ [!ϕ]. He
writes that such a sentence would be “‘contingently paradoxical’: in certain situations, it
is both true and false”[13].
19
3.3
Baltag and Smets
Baltag and Smets[3] discuss the surprise examination in detail, and introduce an interesting new viewpoint on the matter: if the students reach a contradiction after hearing the
teacher’s announcement, they should lower their trust towards him. One could choose to
interpret the teacher’s sentence “you’ll be surprised” as knowledge or as belief. Baltag and
Smets discussed both, but prefer to use the belief-version, because they think this is more
in line with the meaning of surprise in the natural language.
Baltag and Smets put forward the following interpretation of the teacher’s announcement: “The exam will be a surprise, (even) after I’m telling you this.”. This notion makes
the statement self-referential, which can be stated as a non-self-referential ‘future-oriented’
doxastic announcement[3] by the use of the temporal operators N EXT and BEF ORE.
Recall that N EXT ϕ in a series of updated/upgraded models (for example at model Sn )
means that ϕ holds at the next model (Sn+1 ), and that BEF OREϕ means exactly the
opposite, to hold at the previous model (at model Sn+1 it means that ϕ holds at model Sn ).
N EXT surprise means that something will be true after the announcement, but does not
say anything about surprise being true before. Therefore, for the teacher’s statement to
be true before being announced, the complete formalization of the teacher’s announcement
is
ϕ := surprise ∧ N EXT surprise.
Baltag and Smets argue that when the students infer a contradiction after hearing the
teacher’s statement, they will revise their attitude towards him: he is not infallible after
all! The students will lower their trust grade.
In this new light, we might consider it as not so established that the attitude which the
students have is the highest degree of trust and that the teacher is infallible, as is Gerbrandy’s interpretation. For this reason we now write the yet undetermined update or
upgrade ‘T ’ towards the teacher:
T ϕ := T (surprise) ∧ T (N EXT surprise).
We should now find out which attitude is the right one for the students. In the previous
chapter the hard upgrade, the soft radical upgrade and the soft conservative upgrade were
introduced. Baltag and Smets suggest that the students adopt the most positive attitude
possible towards the teacher; they will always trust him as much as they can. After the
students lower their trust grade they will keep a positive attitude towards the teacher; they
are willing to come to believe the statement, unless they come to know it is false. For a
plausibility model S = (S, ≤, k · k) and the new model T ϕ(S) = (S 0 , ≤0 , k · k ∩ S 0 ), this can
be formalized as
kϕkS ∩ S0 6= ∅ =⇒ Max≤0 S0 ⊆ kϕkS
20
We can read this as: if ϕ holds in some world in our new model S’ (if it is not known to
be false), then the maximal worlds will be ϕ-worlds (ϕ will be believed). This corresponds
to a formulation using the temporal operators:
[T ϕ] ¬K¬(BEF OREϕ) ⇒ [T ϕ] B(BEF ORE)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 of
surprise, to explore what happens when the students adopt the several attitudes towards
their teacher.
3.3.1
Surprise in Terms of Knowledge
First, interpreting surprise in terms of knowledge, Baltag and Smets put forward the
formalization[3]:
V
V
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 knowledge’.
We should now check which attitude for the students towards their teacher is the right
one. 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 correspondingly
update their model with ‘!ϕ’, for ϕ := surpriseK ∧ N EXT surpriseK . It can be shown
that 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 upon
their teacher as a fallible but highly reliable source, and correspondingly upgrade their
model radically, with ‘⇑ ϕ’.
For an initial model
1↔2↔3↔4↔5
an upgrade with ⇑ ϕ(N EXT surpriseK ) would induce the following:
4
This follows from the first part of theorem 3.1, because knowledge implies belief and therefore also
[!ϕ] N EXT surpriseK implies [!ϕ] N EXT surpriseB , which implies a contradiction.
21
1 ↔ 2 ↔ 3 ↔ 4 ← 5.
The students highly trust their teacher, so the worlds in which N EXT surpriseK hold
will be more plausible than the worlds in which it does not. The sentence N EXT surpriseK
only does not hold in world 5, Friday, and therefore after the upgrade this world will become less plausible than all the others. Upgrading conservatively with ↑ would lead to the
exact same outcome.
Baltag and Smets argue that this outcome is somewhat unsatisfactory, because it fails
to 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 for
the soft upgrades, which only bring about new beliefs. It therefore seems fair to them to
interprete 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:
V
V
surpriseB = 1≤i≤5 i → ¬B(i| 1≤j<i ¬j) ,
where B(ϕ|ψ) := B ψ ϕ means ‘belief in ϕ given ψ’. The sentence surpriseB thus can be
read as: for every day of the week, if the exam is on day i you will not believe this given
that it didn’t happen the previous days.
Let us look at the students’ reasoning for this interpretation
V a bit more carefully. The
W
students already know there will be one exam: K( 1≤i≤5 i) ∧ 1≤i<j≤5 K(¬(i ∧ j)).
of ‘surpriseB ’ causes the students to believe it:
Supposethat the teacher’s utterance
V
V
B
. We will show that this leads to a contradiction.
1≤i≤5 i → ¬B(i| 1≤j<i ¬j)
Because knowledge implies belief, also knowing surpriseB will lead to a contradiction.
calculus, from the ‘exam’ the students deduce that
Now with some easy propositional
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 an
exam on Monday, Tuesday, Wednesday or Thursday, the students believe that the exam
will be given on Friday. But if the exam would indeed
on Friday, this would
be given
V
V
be in contradiction with believing the statement: B
1≤i≤5 i → ¬B(i| 1≤j<i ¬j)
22
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).
W
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 the
students already ‘ruled out’ Friday as a candidate (B¬5), this implies B ¬1∧¬2∧¬3 (4), which
means that given that there will not be an exam given on Monday, Tuesday or Wednesday
the students believe that the exam will be given on Thursday. Their argument continues
in the same way as before, and they will also believe that the exam will not be given on
Thursday.
By a similar argument they ‘exclude’ also Wednesday, Tuesday and Monday as possible
days on which the exam can be given, which contradicts their knowledge that there has to
be given an exam next week. We may conclude that the students cannot believe nor can
they know surpriseB .
We see that interpreting surprise in terms of beliefs shows the same problem in the
students’ reasoning as when interpreting it in terms of knowledge. Now Baltag and Smets
provide us with two solutions.6 One for the real self-referential form of the teacher’s
statement by the use of temporal operators, and one for an approximation of this selfreferential 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 ∧ N EXT surpriseB , which
we update or upgrade with T a still undetermined positive update or upgrade. It can be
shown that the update, and both the radical and conservative upgrade lead to a contradiction.
Theorem 3.1. !φ, ⇑ φ and
to a contradiction
for φ = N EXT surpriseB , in
↑ φ all lead
V
V
which surprise = 1≤i≤5 i → ¬B(i| 1≤j<i ¬j) .
Proof. BEF ORE(N EXT surpriseB ) ⇔ surpriseB is a validity. Because φ has to be
true before it is announced, we can also state that [!φ] K(BEF OREφ). Combining these
two gives us the validity [!(N EXT surpriseB )] KsurpriseB . We already concluded that
knowing that there
W has to be an exam next week implies the sentence ‘surpriseB ’ cannot
be known: K( 1≤i≤5 i) ⇒ ¬KsurpriseB . Because this knowledge about the fact that
5
And using the first Kripke modality.
They 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].
6
23
there
W has to be an exam in one of these five days is persistant under updates, this implies
K( 1≤i≤5 i) ⇒ [!(NW
EXT surpriseB )] ¬KsurpriseB . This contradicts what we proved before, so we have K( 1≤i≤5 i) ⇒ [!(N EXT surpriseB )] ⊥. We may conclude that T cannot
be an update.
An analogous proof for the radical and conservative upgrade gives us an impossibility
for the two soft upgrades as well. I’ll give the proof for the radical upgrade. Note that
doing a soft upgrade can cause the listener to believe something, and not to know it, like
the hard update does.
Proof. It is again a validity that BEF ORE(N EXT surpriseB ) ⇔ surprise. By definition
of the radical upgrade we know that [⇑ φ] B(BEF OREφ). Combining these two gives
us the validity [⇑ (N EXT surpriseB )] BsurpriseB . We already concluded that knowing
that there has
W to be an exam next week implies that the sentence ‘surpriseB ’ cannot be
believed: K( 1≤i≤5 i) ⇒ ¬BsurpriseB . Because this knowledge about the fact that there
hasWto be an exam in one of these five days is persistant under upgrades, this implies
K( 1≤i≤5 i) ⇒ [⇑ (N EXT
W surpriseB )] ¬BsurpriseB . This contradicts what we proved
before, so we have K( 1≤i≤5 i) ⇒ [⇑ (N EXT surpriseB )] ⊥. 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 [T ϕ] ¬K¬surpriseB ⇒
[T ϕ] BsurpriseB ⇒ [T ϕ] ⊥ for each positive update or upgrade T . So we should find a
model 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
[T ϕ] K(¬BEF OREϕ) ∨ [T ϕ] B(BEF OREϕ), and therefore
[T ϕ] K(¬surpriseB ) ∨ [T ϕ] B(surpriseB ). Because we’ve already proven that the second
disjunct leads to a contradiction with the students’ background knowledge (that there
has 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 |= [T ϕ] K(¬surpriseB ), meaning that after this yet undetermined upgrade with
ϕ, ¬surpriseB holds everywhere in the model. We can rewrite
V
W
this as |= 1≤i≤5 i ∧ B(i| 1≤j<i ¬j) . Specially, ¬surpriseB has to hold in world
W
V
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, the
exam is not given on Monday,
Tuesday, Wednesday
or Thursday). From the last it follows
V
V
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
24
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
of this
the definition
world, M, 4 |=
V
V
¬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. Similarly we are able to deduce M |= B(3|¬1 ∧ ¬2), M |= B(2|¬1) and M |= B(1). Altogether
this 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 selfreferential version of this paradox is the model in which an earlier day is more plausible to
the students than a later day.
An Approximation to the Surprise Paradox
Alexandru Baltag and Sonja Smets[6] also approximated the self-referential version of this
statement by identifying it with the limit of an infinite sequence of non-self-referential announcements. With surpriseB as before, we thus identify
ϕ := surpriseB ∧ [!ϕ] surpriseB
with
!(surpriseB ); !(surpriseB ); !(surpriseB ); · · ·
You could think of this infinite sequence as if the teacher writes “The exam will be a
surprise” on a paper and locks it in a drawer, after which the students secretly break into
the drawer and read it. The teacher than notices this, writes another message “The exam
will still be a surprise” which he locks in the drawer, the students again break into the
drawer, read it, etcetera.
But, just as before, Baltag and Smets argue that this announcement should not be an
infallible update, because the students will deduce that such an infinite sequential composition of updates is an impossible event. Again the students can reason that the exam will
not be on Friday because they would know this on Thursday evening, and by a similar argument they again exclude all the other days. This contradicts the background knowledge
of the students that there will be an exam.
This essentially means that the teacher cannot be an infallible source: he might tell
the truth, but this is not garanteed. Of course it could be the case that the teacher is
deliberately lying to confuse the students, but we assume this is not the case. Because the
students 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
7
Or, if one prefers: B(1) ∧ B ¬1 (2) ∧ B ¬1∧¬2 (3) ∧ B ¬1∧¬2∧¬3 (4) ∧ B ¬1∧¬2∧¬3∧¬4 (5).
25
highly reliable source. If we then identify the self-referential statement
ϕ := surpriseB ∧ [⇑ ϕ] surpriseB
with 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) the
students 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 Friday
as 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 think
of Thursday less plausible than Monday, Tuesday and Wednesday, because now ϕ also does
not 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 this
sequence of upgrades. But this fixed point is a negative one: after the fourth iteration the
sentence is known to be false.
We can generalize this: starting with any initial plausibility relation in which the
student has no hard information about the day of the exam, and iteratively applying soft
upgrades 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
kϕkS ∩ S0 6= ∅ =⇒ Max≤0 S0 ⊆ kϕkS for a plausibility model S = (S, ≤, k · k) and the new
model T ϕ(S) = (S 0 , ≤0 , k · k ∩ S 0 ).
This condition means that, unless the new information was already known to be untrue
before the upgrade (it is, there are no worlds accessible in which the new information
holds), it will be believed after the upgrade (the maximal worlds will be ϕ-worlds).
26
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 ksurpriseB kS ∩S’ =
∅ ⇔ the model S is given by (1 ← 2 ← 3 ← 4 ← 5).
Proof. ksurpriseB kS ∩ S0 = ∅ ⇔ S ⊆k¬surpriseB kS , in other words: ¬surpriseB holds
everywhere in this model. We have proven before that the only corresponding model in
which ¬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
‘ksurpriseB kS ∩S0 6= ∅ =⇒ Max≤0 S0 ⊆ksurpriseB kS ’ is true and therefore the consequent
as well. So the soft upgrade will have effect: the maximal worlds will be surpriseB -worlds
after the upgrade.
Before the iteration of T(surpriseB ), or as action modality [T surpriseB ], the following
is the case:
Worlds 1,2,3 and 4 can be surpriseB -worlds, but do not have to be. But we know for
sure that 5 is not a surpriseB -world, because M, 5 |= 5 ∧ B(5|¬1 ∧ ¬2 ∧ ¬3 ∧ ¬4). Hence
M, 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 all
the surprise-worlds. But not only that: it is also less plausible than all the other nonsurpriseB -worlds! This is because, for a world i to be a non-surpriseB -world, in i it has to
hold that i and not believe iVgiven that the ’earlier’ worlds (lower in number) j are believed
to 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 otherwise
it would not be among the maximal worlds, and it would not be believed). So, we can
conclude 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 plausible 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 nonsurpriseB -world, by definition of surpriseB . World 1 is not a non-surpriseB -world anymore, because worlds 2,3 and 4 are now more plausible worlds and therefore 1 is not longer
8
On page 24 of this thesis.
27
believed to be true. Another thing to notice: because 5 is now less plausible than all the
others, it is also less plausible than 4, which inmediately makes this a non-surpriseB -world:
M, 4 |= [T surpriseB ] ¬surpriseB . So we now know for sure that after the second upgrade
with surpriseB , both 4 and 5 do not satisfy surpriseB . And therefore, after the second
iteration 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 even
less plausible than 4 (by the argument explained before). The non-surpriseB -worlds are
again in their plausibility ordered just as by their number; the non-surpriseB -worlds which
denote a smaller number, are more plausible than the non-surpriseB -worlds that denote
a bigger number. Now notice that 4 stays a non-surpriseB -world: because it is more
plausible than 5, it satisfies both 4 and B ¬1∧¬2∧¬3 (4). So before the third upgrade we find
that 4,5 and 3 have to be non-surpriseB -worlds. By similar arguments as before, after
the third upgrade we find:
2↔1←3←4←5
And again, after the fourth upgrade, the shift in plausibility relations goes the same, which
results in:
1←2←3←4←5
Now we find that, if we upgrade any further, nothing will happen anymore. This is because
of the lemma proven above, because the antecedent is not true anymore the consequent
does not have to be either. We do not change our plausibility relations anymore by upgrading, because we concluded that surpriseB contradicts our knowledge at this point:
M |= K¬surpriseB . We reached a fixed point. Mark that this solution for approximated
self-referentiality is similar to Baltag and Smets’ solution for the self-referential version of
the statement.
3.4
Kritchman and Raz
Kritchman and Raz[17] connect the surprise paradox to Gödel’s second incompleteness
theorem[14]. They introduce a new proof for this theorem, and use a similar proof to give
a solution to the surprise paradox. For the surprise statement S, the following formalization is given:
V
S := 1≤i≤5 P rT,S (pm ≥ i → m = iq) → (m 6= i)
28
This tells us that for every day i, if one can prove (using this statement) that if the exam
did not occur on the previous days then the exam will be on day i, then the exam will
not 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 a
theory T as formulated in Gödels second incompleteness theorem, surprise statement S
gives us a system T + S of which the students cannot prove its consistency in itself. Their
solution thus implies that if the students believe in the consistency of T + S, they believe
the exam cannot be held on Friday, because on Thursday night the students will know
that if T + S is consistent the exam will be held on Friday. But as they cannot prove this
consistency, the exam can be held on any other day of the week. We can visualize this as
follows:
1↔2↔3↔4←5
3.5
Discussion
In this chapter we have seen that there are several interpretations possible for the teacher’s
statement, and several ways of ‘solving’ this paradox. Things in which they can differ are
among other things knowledge or belief, how strongly the teacher is trusted, the degree of
self-referentiality and whether “there will be an exam next week” can be treated as certain
or not.
Adjustments done for standardization
Let us start discussing the last. We should explore how big an influence taking ‘exam’ as
certain or not has on the solution. Remember that we adjusted Quine’s and Kritchman
and Raz’ versions, in the other three cases it was already taken as a certainty that there
will be one exam next week. If we would follow Quine more closely, “there is an exam next
week” would be a part of the statement, and therefore the student’s can also doubt the
existence of the exam. Therefore, Quine expressed one more option10 : “The exam will (in
violation 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 the
same, because the students can eliminate this option in the same way as they eliminate
option (b). Thus in the end the solutions are the same.
I also adjusted Kritchman and Raz’ solution to our standardized form: originally the
formalization of surprise statement S 0 is as follows:
9
Kritchman and Raz originally took this as a part of the announcement. This does not change the
solution, which will be discussed in the next chapter.
10
The 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.
29
0
V
S := [m ∈ {1, . . . , 5}]
!
m = jq)
1≤i≤5
P rT,S (pm ≥ i → m = iq)
V
1≤j≤5,j6=i
¬P rT,S (pm ≥ i →
→ (m 6= i)
This formalization consists of two parts. The first says that the exam will be given in
one of the five days. The second tells us that for every day i, if one can prove (using this
statement) that if the exam did not occur on the previous days then the exam will be on
day i and one cannot prove in the same manner that then the exam will also be on another
day j, then the exam will not be given on day i. So, instead of already knowing that there
is no such j for which m = j, it is part of the provability antecedent: if they can prove
such i exists and they cannot prove another such j exists then the exam will not be given
on day i. The reasoning stays exactly the same: because the students cannot prove the
consistency of T + S 0 in itself, they can only believe that the exam is not being held on
Friday if they believe in the consistency of the system. But they cannot exclude any other
day 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 Gerbrandy’s formulization, this would not change the solution either. The formalization then
would
be S 0 = (we ∧ ¬Kwe) ∨ (th ∧ [!¬we] ¬Kth) ∨ (f r ∧ [!¬we] [!¬th] ¬Kf r) ∨ K ⊥
∧ (we ∨ th ∨ f r) ∧ ¬(we ∧ th) ∧ ¬(we ∧ f r) ∧ ¬(th ∧ f r) . Announcing S 0 once would
again induce the students to eliminate Friday as a possible candidate, because Friday is
still the only day not satisfying S 0 . Announcing S 0 ∧ [S 0 ] S 0 would induce them to eliminate
both Friday and Thursday. The solution stays the same.
The fact that Baltag and Smets take “there will be an exam next week” as a certainty
does not really change the outcome either, it just makes it easier. We could have just as
well regarded “there will not be an exam next week” as an extra world in the model (to
make it possible for this part of the announcement to fail as well), making their initial
model:
1 ↔ 2 ↔ 3 ↔ 4 ↔ 5 ↔ 6,
with world 6 representing the situation in which no exam
W is given at
V all. We would have
then formalized the whole sentence as ‘surprise0B := K( 1≤i≤5 (i))∧ 1≤i<j≤5 K(¬(i∧j))∧
V
V
’. Because the first conjunct of surprise0B is untrue in
1≤i≤5 i → ¬B(i| 1≤j<i ¬j)
this newly created world 6, it will always be a non-surprise0B -world, and therefore not
be believed nor known. Using temporal operators or the iteration of non-self-referential
upgrades, the solution will be
30
1 ← 2 ← 3 ← 4 ← 5 ← 6,
corresponding to the unique solution of Baltag and Smets for surpriseB when ‘exam’ is
treated as an established fact.
We may conclude that taking ‘exam’ as a part of the announcement or as something
certain (known) does not affect the core of the solutions discussed. Therefore our adjustments 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 cannot eliminate upfront the possibility that the exam will be on Friday but that they do not
expect this on Thursday evening. Therefore they cannot exclude any of the days: they are
stuck with their original knowledge and beliefs, the teacher’s statement does not impose a
change of their model. We can link this to Gerbrandy’s approach. Quine sees a fallacy in
the reasoning of the students: if they can conclude in the end that the teacher’s statement
was untrue, they should have considered this as a possibility from the beginning. One
could say following Gerbrandy that if the students in the end conclude that they were
wrong to accept the principle of success after all, they should consider this principle to fail
as a possibility from the beginning. Thus, Quine speaks of the teacher’s statement being
untrue, and Gerbrandy speaks of the teacher’s statement as being true before announcing
and becoming untrue inmediately after. This difference is the cause of Gerbrandy’s students eliminating (only) Friday as a possibility and Quine’s students not changing their
propositional attitudes at all.
Quine did not speak about the degree of self-referentiality of the problem he discusses
in ‘on a so-called paradox’. Where he speakes of The Hangman11 , we can see that he
interpretes the sentence announced by the judge as “you will be hanged and you will not
know on which day, until the fatal day.” This looks like self-referentiality: you will not
know until judgement day, even after I’m telling you this. Quine also treats the statement
this way. He does not consider the possibility that the judge’s statement was true before
announcing it, and becomes false inmediately after. In other words: Quine inherently accepts the principle of success.
Although Gerbrandy uses his formalization to show that maybe the principle of success
must not be accepted and points out that the teacher’s statement might be a self-refuting12
11
The story around the same paradox in which a judge tells a convict that he shall be hanged one of the
following seven noons, but that he will be kept in ignorance until the morning of the fatal day as to just
which noon it would be.
12
Statements which become untrue because of uttering it. For example the well-known Moore sentences
‘p ∧ ¬Bp’.
31
one, one can have some comments on the way S is being interpretated. He suggests we
should also consider ‘knowing a contradiction’ as ‘being surprised’. You could side with
Gerbrandy, because you can argue that someone who is a little acquainted with the rules of
logic would be a bit surprised to know a contradiction. But besides this argument, it feels
somewhat too easy, maybe ad hoc even, to - if you reach a contradiction - just add a contradiction to the formalization of surprise. Besides this it has to be said that Gerbrandy’s
non-factive, non-consistent notion of knowledge bears more resemblance to the common
notion of belief than it does to the common notion of knowledge, although it is usually
assumed that beliefs are consistent. Something that stands out in Gerbrandy’s iterated
non-selfreferential statement is its similarity to the students’ reasoning in the introduction
of this chapter. Looking at the model changes: they are completely the same! In this way
this 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 simply
became false and the students get to know this.
Baltag and Smets take a completely different road in their discussion of the surprise
examination paradox. They take the perspective of the student and argue that, if the
teacher says something that leads to a contradiction, the students naturally will not see
him as an infallible source anymore. Repeatedly lowering their (still positive) degree of
trust, the students will in the end have an attitude equal to the limit of a sequense iterated
positive soft upgrades. Baltag argues that this can be linked to the inductive reasoning
process of an ‘ideal’ student13 by means of trial and error.
This iterated upgrading of the non-self-referential surpriseB is something Baltag and Smets
also do for the knowledge-variant of surprise, surpriseK . For conciseness I did not also discuss this in section 3.3.1, but identifying the teacher’s self-referential announcement with
the limit of an infinite sequence of announcements T (surpriseK ); T (surpriseK ); . . ., just
like done in the belief-version, is another way to reach the same conclusion for all positive
updates and upgrades T. For ’ !’ this coincides exactly with Gerbrandy’s iterated solution.
Something more to notice: Baltag and Smets’ iterated non-selfreferential statement surpriseB
is similar to the students’ reasoning for the belief-version of surprise. Looking at the model
changes: they are also completely the same! In this way this solution does really account
for the intuitive feeling of the reasoning of the students, just like the their treatment of
surpriseK 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 contradiction.
One could also encode the belief-version of the surprise examination as a game14 , and
this gives a result similar to the solution raised by Baltag and Smets. Let player 1 be the
students and player 2 the teacher. After the teacher tells the students that there will be
a surprise exam, they (player 1) form a belief about whether the exam will be tomorrow
13
14
A student who is a perfect logician and willing to trust the teacher.
Ken Binmore first did so in [9], page 45-46.
32
(t) or not (t). On Monday morning the teacher (player 2) chooses an action, giving the
exam that day (e) or not (e). The teacher does not know which beliefs the students had
formed the evening before. If player 2 gives the exam the game will end and depending on
the 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 the
student: ttttt. Now we know not only that the statement of the teacher is false, but also
that the students can get to know that it is false. This winning strategy encodes a plausibility 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 fact
that, to win, the student has to be prepared every day in case the teacher decides to give
the exam. The solution found with the use of game theory thus tallies with that of Baltag
and Smets.
Just as before, als for the solution in game theory adding ‘exam’ to the formalization
would not change the solution. It would enable the teacher (player 2) to choose not to
give the exam in one of the five days: he can choose eeeee. Then we should say that the
game ends at the minimum of the day at which player 2 chooses e and after day five. We
should specify who wins the game when it ends after day five with the teacher choosing
e. It seems reasonable to designate the students (player 1) as winner: there was no exam
that they did not expect. Defining the game like this, player 1 still will have his winning
strategy: ttttt. The solution is again not changed by taking ’there will be an exam’ as part
of the teacher’s announcement.
The last discussed solution in this chapter comes from an unexpected quarter: taking an alternative proof for Gödel’s second incompleteness theorem as fundament for a
new way to look at the surprise examination. This interesting new perspective, in which
they take the self-referential notion of surprise into account, gives yet another solution to
the paradox, Friday being less plausible and indifference about the other days. Curiously
enough, this is the solution which many people who are not introduced in the formal rules
of 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 argue
that the students can indeed not rule out Thursday.
Something already named that we might want to discuss a bit further, is the similarity
between Gerbrandy’s iterated solution and Baltag and Smets’ iteration of surpriseK with
the highest possible positive attitude, ‘!’. To get a similar notation to Gerbrandy’s, I wrote
S ∧ [!S] S ∧ [!S] [!S] S ∧ [!S] [!S] [!S] S ∧ [!S] [!S] [!S] [!S] S
for the four times iterated formalization of surprise. This could, somewhat suggestively,
33
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 the
update ‘!’ (almost) fully coincides with that of Gerbrandy - also in an intuitive way- , their
solution considering [!ϕ] N EXT surprise is the same as well.
But there is a difference: Gerbrandy’s formalization of surprise differs from that of Baltag
and Smets. Gerbrandy added K⊥ to his formalization, for the rest it can easily be seen
that the two formalizations coincide.
If we would bring
back Baltag and Smets formal
V
V
W
ization to three days: 1≤i≤3 i → ¬K(i| 1≤j<i ¬j) , we can rewrite this as 1≤i≤3 i ∧
V
¬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 f r with 3, this is Gerbrandy’s formalization without K⊥. So if their findings are exactly the same, we can ask ourselves what will be the
added value of K⊥. I think this has a purely semantical value: it enables the students to
still be surprised after hearing the teacher’s announcement. Whereas Baltag and Smets’
students conclude that the teacher’s announcement became untrue, Gerbrandy’s students
are surprised.
We see another way in which Gerbandy’s and Baltag and Smets’ solution coincide if
we look at the discussion of the latter of N EXT surpriseK with a radical or conservatice
upgrade. This looks like Gerbrandy’s discussion of S (not the iteration of it). Doing an
upgrade ⇑ (N EXT surpriseK ) causes the students to make Friday less plausible than the
other 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 the
students, for example highly trusting the teacher, the two solutions would have been the
same (mark that, would we have been using the iteration of ⇑ surpriseK , we would have
reached a fixed point here. Any further upgrading with ⇑ surpriseK will leave the model
unchanged).
Although maybe in first sight Kritchman and Raz appear to have raised a completely
different solution to the surprise examination, there are striking similarities between their
approach and that of Gerbrandy and Baltag and Smets. First, a comparance with Gerbrandy’s solution. We can see a parallel between knowledge in his solution with the provability used in Kritchman and Raz’: knowing that there will be an exam on a specific day
versus being able to prove that there will be an exam on a specific day implies that the
exam will not be given that day. Furthermore, we can associate the principle of success
with the consistency of arithmetic: if Gerbrandy’s students accept the principle of success
they 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 on
their argument, but they cannot. Consequently, both the reasoning of Gerbrandy’s and
Kritchman and Raz’ students stops after excluding Friday as a possibility; in the first case
they 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 Smets
discussed as well, namely surpriseK with high trust (⇑)15 . In this case, we identify provability 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 teacher
to tell the truth. This is, Kritchman and Raz’ students reason that if they can prove the
consistency, Friday could be deleted. They cannot do this, but believe in the consistency
and therefore for them Friday is the least plausible day. Baltag and Smets’ students on
the other hand reason that if they could know that the sentence is true, Friday could be
deleted. They cannot do this (updating with ‘!ϕ’ implies knowledge that leads to a contradiction), but believe the teacher to tell the truth and therefore for them Friday is the
least plausible day.
Conclusion
While the several solutions explained in this chapter first looked quite different, we have
now seen that there are a lot similarities between them. We standardized the solutions
to versions in which “there will be one exam next week” is taken as a certainty, and saw
that this did not affect the idea of the solution. Doing this we made it easier to see
where the differences lie. Quine’s solution is unique in seeking the fault totally in the
students’ reasoning and leaving them with their initial knowledge and beliefs. Gerbrandy
adds ‘knowing a contradiction’ to the formalization of surprise, and defies the principle
of 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 towards
the teacher and to interprete surprise in terms of beliefs. The degree of self-referentiality
in which the teacher’s statement is understood differs from solution to solution, and for
each understanding there may be a case.
How should this paradox be formalized? The ambiguity of the formulation is the lion’s
share of it’s difficulty. All yet raised solutions have their qualities, and give interesting
sights to the matter, but yet no all about adopted satisfying solution has been put forward. Conversance with some of the new solutions might be relevant to this, but also the
fact that there are so many explanations of the teacher’s words. Which solution is the
best yet remains to be something individual, a matter of taste; we wait for a universally
accepted solution.
15
Upgrading with ↑, ’simply’ believing the teacher, would lead to a similar result.
35
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 bit
further. We can do this by looking elsewhere and examining in which way the solutions
can be applied. In this manner we learn to look from a different perspective, which might
bring us closer to a completely satisfying solution. The other way around, we could then
solve more problems than the original surprise paradox only and broaden our view on the
meaning of ’surprise’. In the next three sections I will discuss some known puzzles which
I put in a more ’surprise’-like form. These puzzles are made self-referential, and a similar
approach to them is taken as the several solutions raised to the surprise examination.
4.2
4.2.1
Smaller or Bigger
The Story
Story: Alice (A), Bob (B) and Claire (C) are in a cafe. Claire asks the former two to
think of a natural number bigger than zero and to whisper it to her so that only she can
hear 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 numbers
Claire tells them: “You both thought of a different number, but you’ll never be able to
deduce who choose the bigger number.”1 . Alice and Bob are not allowed to communicate
with eachother, all they can do is reasoning. 2
After the announcement A will think: “If B has number 1, he will know that his
number is smaller than mine. But C just told us that we cannot know that, so B cannot
have number 1. But then, if he cannot have number 1, he also cannot have number 2: he
1
We can see the similarity with the surprise examination here: if your number is smaller or bigger, this
will come as a surprise.
2
The original puzzle is discussed in [18], p. 113.
36
knows by a similar argument that I cannot have number 1, we both know our numbers are
different 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 2
either, etcetera”. In this way A continues, and is able to rule out for B every number smaller
than her own3 . When she ruled out all numbers smaller than n as possible candidates for
m she will know that her number is the smaller one, which is in contradiction with C’s
announcement that she will never know. Similarly, B reasons about A’s possibilities and
infers 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 , and
the sentence “B’s number is smaller than A’s number” we will name D2 . We write KA Di ∧
KB Di simpler as KA,B Di , 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 smaller
than A’s. Recall that A choose number n and B number m. We take it as a certainty that
A 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 B
would not even be able to exclude number 1. We can look at the options they have in their
reasoning, 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 (in
violation of the statement) be aware of his number being smaller than mine promptly
after the statement;
(c) The other’s number (in keeping with the statement) is equal to 1, and he will (in
keeping 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 attitude
towards person C, and not change their beliefs. The outcome is that both A and B do
not get to know whether their number is smaller/bigger than the other one’s number. So
C’s statement became true after all. A special case is of course that n or m is actually
equal to 1, but then for this person the case would be similar to Quine’s Hangman paradox
with one day[20]. The person inmediately reaches a contradiction but is not able to gain
knowledge from it.
3
In fact, she is able to rule out every natural number to be m. In this way she also reaches a contradiction, but without using induction she would have to make infinite think steps to infer this contradiction.
37
Gerbrandy
When we apply an adjusted form of the solution of Gerbrandy, who actually speaks about
the 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,B D1 ) ∨ (D2 ∧ ¬KA,B D2 ) ∨ K ⊥.
The last disjunct makes it possible for C’s statement to be succesful, even after person
A and B reach a contradiction. With this non-self-referential formulation only 1 can be
excluded by A and B to be the other one’s number. Therefore A and B do not gain
knowledge by hearing C’s statement, unless their number is 1 or 2. If one of them has
number 1 he inmediately knows this is the smallest number, which makes C’s statement
incorrect. If one of them has number 2 he can exclude for the other one to have number 1
and thus deduces that he himself has to be the one with the smaller number. This makes
C’s statement incorrect.
Baltag and Smets
Remember that Baltag and Smets also interpreted surprise as a statement V
which induces
beliefs, we will do the same in this puzzle. If we write notbelieve for i∈{1,2} (Di →
¬BA,B Di ), then ϕ := [T ϕ] notbelieve ∧ [T ϕ] N EXT notbelieve would be the self-referential
statement C makes.
We can approximate this statement (analogous to Baltag and Smets’ iterated solution
explained in the previous chapter) by introducing rounds of this game, in which C repeatedly 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 and
Bob will each round respond with yes or no. In case they do know they of course specify
if 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 as
follows: they begin with total indifference. Of course A and B know their own number, so
combinations like (n, m) for n 6= 3 and m 6= 5 are ruled out inmediately. Furthermore, just
like in the surprise examination, the part of the announcement “you have different natural
numbers 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 cannot
come 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 upgrade
T. All soft upgrades will lead to the same result, just as we have seen in the previous chapter. After the first upgrade with notbelieve A and B come to think that number 1 is less
plausible for the other to be her or his number ([T notbelieve] KA ¬BB D2 ⇒ BA (m 6= 1)
and [T notbelieve] KB ¬BA D1 ⇒ 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, they
still 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 plausible
for 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 = 1
and m = 2 are less plausible than all the others. So A comes to correctly believe that her
number is smaller, and announces this. She responds to C’s question with “Yes, I believe
my number is smaller”, ‘BA D1 ’, and C’s statement becomes untrue.5 For a countably
infinite 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 B
would have come to believe that his number is smaller, which is incorrect. But still, they
will 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 and
adjust it to this puzzle, we could say that if person A and B believe in the consistency of
the system, they would be able to rule out number 1 as a candidate. But as they will never
4
If there are countably infinitely many rounds.
Now it seems reasonable to assume that B adopts the conclusion: BB D1 , 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 is
however not needed to falsify C’s announcement.
5
39
be able to prove the consistency of this system within this system, they will not be able
to continue this argument. In the end if they believe in the consistency of the system they
believe the other doesn’t have number 1, but they are not able to deduce whether their
number is smaller or bigger than the other one’s. Using the same example as before, the
situation 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. Whereas
Quine’s solution adjusted to this puzzle leaves Alice and Bob completely ignorant and indifferent, Gerbrandy’s, Kritchman and Raz’ and Baltag and Smets’ adjusted solutions give
the two an opportunity to learn from the statement C makes. In different gradations of
course: the solution in the line of Gerbrandy let’s A and B only eliminate 1 for the other
to be his number and in Kritchman and Raz’ adjusted solution A and B only consider 1
to 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 this
puzzle person A and B take the risk of being wrong to come to the conclusion that C’s
statement 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 understood. Does she say
V that “A and B both individually will not correctly believe which
number is smaller”,’ i∈{1,2} (¬(Di ∧ BA Di ) ∧ ¬(Di ∧ BB Di ))’, which can be written as
V
‘ i∈{1,2} ¬(Di ∧ (BA Di ∨ BB Di ))’, or does she mean that they will not “both correctly
V
V
believe it”, ’ i=1,2 ¬(Di ∧ BA,B Di )’, which can be written as i=1,2 ¬(Di ∧ BA Di ∧ BB Di ).
The latter, I think, would be the natural way to understand this sentence, and this is what
I 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 better one, A and B would maybe not reach a contradiction. In this case, C’s announcement
can be succesful. We do not see a like-wise ambiguity in understanding in the surprise
examination, because the students act as one: they are all in the same situation, whereas
A and B in this puzzle are not.
Something in which this puzzle differs from the surprise paradox is the way in which
wrong beliefs have an effect on persons A and B. In the surprise paradox, it is clear, the
students will prefer to be prepared for an exam which is not given than to not be prepared
for 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 not
prefer one above the other. There are no ’good’ or ’bad’ wrong beliefs. This makes the
solution different as well.
You could argue about whether this puzzle is more paradoxical than the surprise paradox or less. You could think less, because always one of persons A and B will be right
about his number being bigger or smaller. But it might also be more paradoxical: they
individually do not have a ‘satisfying’ solution because there are no ’good’ wrong beliefs
in this case. On the other hand, one could say that this puzzle could be solved rather
easily: by just taking opposite beliefs for A and B, this would always make C’s statement
false. This however would be ad hoc and because A and B are not allowed to communicate
solving this so pragmatically is impossible. It is interesting to notice that in the solution
of Baltag and Smets, the first one to get beliefs about her number being bigger or smaller,
is right about this.6
4.3
4.3.1
Muddy Children
The Story
Story: There are n children who played outside together. Their father calls them in and
announces: “At least one of you is dirty and you will never know if you are dirty.”7 . No
communication is allowed, looking in the mirror is prohibited; for the children there is no
other way to deduce whether they’re dirty or not than just by reasoning. And their father
just told them even this is impossible.8
Firstly I’ll introduce the general reasoning of the children: this strange announcement
sets them thinking about it individually. We take it as a certainty that one of them is
dirty, and therefore the option ‘all children are clean’ is ruled out. The father asks them
again and again (we call this rounds) if they know if they’re dirty. The children respond
with “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 he
has to be the one who is dirty. This is in violation of the statement, and thus ‘one of the
children 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 one
child is dirty’ is in violation with the statement. So the option that exactly two children
are dirty is ruled out as well, because this contradicts the statement that a dirty child will
never know that he’s dirty. Continuing this line of thought, every possibility can be ruled
6
When the first one does not announce his beliefs, the other will deduce the same: thinking his number
is smaller as well. Thus the same result as the pragmatic one is reached, without being ad hoc.
7
We can see the similarity with the surprise examination here: if you’re dirty it will come as a surprise.
8
For 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 is
of course the general case. In the real situation the children can see the rest (which we
can 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 to
before 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 their
argument are able to conclude that the announcement will not be fulfilled, they should
have been prepared to consider this alternative as a possibility from the beginning. This
gives each child not only two, but three possible options, and he is not able to eliminate
the case in which he (in keeping with the statement) is dirty and he (in keeping with the
statement) does not know this after round n − 19 (and also not in the rounds after). Unlike
in the surprise examination, the passing of rounds does not give the children any hard
information; the strange announcement made this impossible. Just as before, the children
will keep their original beliefs, which means that they are indifferent about their own state
of dirtiness.
Gerbrandy
Trying to apply the solution raised by Gerbrandy adjusted to this puzzle and trying to solve
the non-selfreferential form of this puzzle, we will use the notion of knowledge such that it
does not have to be factive, nor consistent. For the children it is now possible to ‘know’ a
contradiction, 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 ∧ ¬Ki di ) ∨ 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 one
child 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 correctly
believe if you are dirty”. Instead
V of updating their models, the children softly upgrade
them. We define notbelieve := 1≤i≤n (di → ¬Bi di ) 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
9
In the original puzzle this is the round in which the dirty children are able to deduce that they’re dirty.
42
V
dirty.”. The case that 1≤i≤n ¬di is already ruled out. Let us perceive what happens if we
take the infinite sequense of (not yet determined) upgrades
T notbelieve; T notbelieve; T notbelieve; · · ·
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 least
one of them is dirty:
7 ddd
O g
1
w
cdd
O k
2
ccd
3
2
'
3 ddc
O
dcd
< b
3
1
s
"
1
3
|
+
cdc
2
dcc
Then, after the first upgrade the children think of all the worlds in which one child is dirty
as less plausible than the cases in which more children are dirty (because now ‘di → ¬Bi di ’
does not hold in worlds ccd, dcc and cdc):
7 ddd
O g
1
w
cdd
O k
3
2
'
3 ddc
O
dcd
< b
3
1
2
ccd
1
3
2
cdc
dcc
After the second upgrade also the worlds in which two children are dirty become less
plausible than the world in which all children are dirty, so that the model is as follows:
7 ddd
O g
1
cdd
O k
2
ccd
3
2
3 ddc
O
dcd
< b
3
1
1
3
2
cdc
dcc
We reached a fixed point. Remember that this was the general case. If we choose an
actual world arbitrarily, for instance ddc the initial model would be:
ddd g
3
'
3 ddc
O
1
|
cdc
43
2
dcc
and after the second upgrade this would change into:
ddd g
3
'
3 ddc
O
1
2
cdc
dcc
Mark that if we continue upgrading and no one says anything, the third child would have
a wrong belief:
ddd g
3
3 ddc
O
1
2
cdc
dcc
Child 3 thinks he is dirty, while actually he is clean. But children 1 and 2 one round before
already correctly believed that they are dirty, and they would announce this as a response
to 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 it
more plausible that there are two or more children dirty than only one if they believe in
the consistency of the system. But they are not able to continue this argument unless they
can prove the consistency of their system in this system itself. Because of Gödel’s second
incompleteness theorem they cannot do this, and therefore the resulting model for each
child i will be, for the same example as before with 3 children and actual world ddc:
7 ddd
O g
1
w
cdd
O k
2
ccd
3
2
'
3 ddc
O
dcd
< b
3
1
1
3
cdc
44
2
dcc
Discussion
If we try to apply the solutions discussed in the previous chapter to the puzzle with the
muddy children, we can again doubt how to formalize the father’s statement. For the
children there are only two possible options: being dirty and being clean. But you could
also see this as one option: to be dirty or not. We might want to see what happens if we
add ∨(¬i ∧ ¬Ki ¬i)) in Gerbrandy’s formalization, ‘child i is clean and he does not know
this’. But adding this does not change the solution! After the update, still ccd, cdc and dcc
are the worlds not satisfying this sentence and are therefore deleted. Again the children
cannot 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 knowing
being 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: by
upgrading with the total sentence for the first time, again the worlds with only one dirty
child is less plausible than the rest (the option ‘all children are clean’ is already ruled out
or least plausible) because they do not satisfy this sentence and the rest still does. If we
add ‘if you are clean, you don’t know it’ using game theory however, the solution would
change: now there is no winning strategy possible for the students anymore: they cannot
both always expect to be clean and always expect to be dirty, so one of these may come
as a surprise. This means that the children cannot falsify their father’s statement, which
they would be able to do if he only asks about their dirtiness. This gives us the insight
that this solution only works if the children will not be asked about both sides: about
being dirty and about not being dirty. Linking this to the surprise paradox: the solution
in 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
4.4.1
Number+1
The Story
Story: Claire assigns Alice and Bob two numbers, n ∈ N∗ and m ∈ N∗ respectively, and
writes it on their forehead so they can only see the other one’s number. Then she tells
them: “You’re number is one less or one more than the other one’s, but you will never
both know your number.”10 . Again no communication or mirrors are allowed, and the only
thing A and B can do to get to know their number is reasoning. If one of them get’s to
know 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 is
assigned number 2 and Alice is assigned number 3, m = 2 and n = 3. Now person
A can see that m = 2, so he infers from the statement that she has number 1 or 3:
10
We can see the similarity with the surprise examination here: if your number is the smaller or bigger
one, 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 if
I would have number 1, Bob would inmediately know that he has number 2. But this is
not possible by Claire’s statement, so number 1 can be ruled out” and Alice knows her
number has to be equal to 3. And so Alice is able to deduce her number, which is again
in 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 end
consider it possible that C’s statement is incorrect, they should do so in the beginning
as well. Returning to the previous example, A cannot even rule out n = 1 because she
cannot exclude the option that her number is 1 (in keeping with the statement) and that B
remains ignorant of his number being 2 (in keeping with the statement). Therefore the one
with the biggest number (A) cannot exclude that her number is the smaller one and B on
the 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 formalization 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 only
give them information if the other has number 2. Thus, in the previous example, applying
this solution, A rightly concludes to know her number is 3, which makes C’s announcement
untrue. In all other cases11 A and B do not get any more information out of it, which makes
C’s announcement true.
Baltag and Smets
When we adjust Baltag and Smets’ solution to this puzzle, we speak of the belief-version
of C’s announcement (“You’ll never correctly believe your number”) and approximate the
self-referentiality with an iterated upgrading with the non-self-referential statement. We
introduce 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 case
and than visualize this by picking a specific (but arbitrary) case. Assuming A and B start
with total indifference, the initial possible worlds model corresponding to their situation is:
(1, 2) ↔A (3, 2) ↔B (3, 4) ↔A (5, 4) ↔B . . .
11
Except 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 and
B inmediately decide that their number cannot be equal to one, otherwise the other one
would have came to (correctly) believe her or his number, and then C’s statement would
have 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 round
after round that they do not correctly believe whether their number is the other one’s
minus one, or the other one’s number plus one. To illustrate what happens, without loss
of generality, we assume that A has the higher number (n = m + 1) and that this number
is 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 from
confusing the worlds (because A and B would enumerate them differently - they see only
the other one’s number). The plausibility relations of A and B in the corresponding initial
model 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
12
In the case that n is 1 A would inmediately believe her number to be the smaller one. The roles of A
and B are completely interchangeable.
47
w1 →A w2 ↔B w3
And therefore, already A comes to correctly believe her number, which makes C’s statement false. Note that if A had the number m − 1 B would have came to correctly believe
his number after m − 2 rounds. So we see that by iterating the statement always the one
with the bigger one comes to correctly believe his number, making the statement false. If
we carried on, after m rounds B would have came to incorrectly believe his number to be
the 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 that
if person A and B believe in the consistency of the system, they would be able to rule out
number 1 as a candidate. But as they will never be able to prove the consistency of this
system within this system, they will not be able to continue this argument. In the end they
believe they don’t have number 1, but they are not able to deduce whether their number
is the other one’s number plus one or the one’s minus one, unless of course the other has
number 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 (which
is hard information) and therefore only distinguishing two cases. For example, if A has
number 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. They
do if one of them has number 213 . For example if A has number 2 and B number 3, the
model 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 as
easily had a wrong belief, if he had number 1 and A still number 2 (but then A would have
13
Or number 1, but then the other would clearly have number 2. Requiring that one of them has number
2 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 said
it is a special variant of this former puzzle, the case in which the numbers of A and B differ
only 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, and
they 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 the
other’s number to be equal to one. The adjusted solutions of Gerbrandy and Kritchman
and 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 to
correctly believe her or his number, but this can be explained by the changed roles of A
an 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 one
of A and B will guess it right. And again one could argue that there is always an easy
pragmatic (ad hoc) solution to this puzzle, because there are only two options: if they both
believe to have the smaller number (or if they both believe to have the bigger number) C’s
statement will be falsified.
49
Chapter 5
Conclusion and Relations to Other
Work
In this thesis I discussed the paradoxicality of the surprise examination and five main
approaches to them1 . Then I tried to apply these solutions in an adjusted way to three
puzzles known in the logical field, but then made self-referential. Using this method I tried
to find out what exactly makes the surprise examination so puzzling, and which ingredients
are needed for a similar paradox to occur.
We saw that the ideas of the solutions raised to (variants of) the surprise examination differ
in idea from blaming the students’ reasoning, interpreting the statement non-selfreferential
and 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 of
trust (while still keeping a possible attitude). The results of this approaches differ from
remaining with the original propositional attitude to only ruling out Friday (and therefore
in both cases enabling the teacher’s statement to come true) to making the teacher’s statement false by everyday believing the exam to come. We have seen that taking “there will
be 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 say
at first sight.
From applying these solutions to the self-referential puzzles we saw that the students
act 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 something
as 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 were
not 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 formulation 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.
1
Although 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 the
importance of this.
Besides the many things I discussed in this thesis there are many things which I would
have wanted to do, but I could not do because of lack of time or because it was outside the
scope of this thesis. For instance, there were many other solutions proposed to the surprise
examination or variations on it, and many papers written discussing the phenomenon. I
simply could not read them all. Something else which I had to leave undiscussed is if
the surprise examination fits in what Priest[19] calls ’Russell’s Schema’ which considers
the paradoxes of self-reference to have a common underlying structure, irrespective of its
category. Maybe in this way or another we could find similar paradoxes tot the surprise
examination in a semantic or set-theoretic setting. It would also have been interesting to
look at the more general solutions to all paradoxes and the consequences they have for the
study 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 own
truth predicate is impossible. To model the kind of paradoxical reasoning involved in the
Liar paradox within a two-valued context, Gupta and Herzberger in the early eighties independently developed a new theory of truth, the Revision Theory of Truth[16]. The core of
this theory is the revision process by which hypothesis about the truth values of sentences
are revised2 .
Now how can this new theory of truth be linked to the surprise paradox? Let us look at
on one hand the iterated version of Gerbrandy’s non-self-referential formalization (equal to
Baltag and Smets’ iterated updating for surpriseK for an infallible source) and on the other
hand the iterated soft upgrading of Baltag and Smets’ surpriseB for a trusted but fallible source, and how these methods compare to the formerly discussed truth revision theory.
Let us suggestively call the initial plausibility model representing the students’ propositional attitude h0 . Then for Gerbrandy the first update leads to deleting world 5 and
for Baltag and Smets the first upgrading leads to making (amongst others) these worlds
less plausible. This can be viewed as the first hypothesis which is revised into the new
hypothesis h1 and h01 for Gerbrandy and Baltag an Smets respectively. Another updating
2
For a more comprehensive introduction on this theory of truth I refer to the Appendix.
51
or upgrading then brings us to hypothesis h2 and h02 , and so on until h4 and h04 .
We can see the similarities between the knowledge or belief revision and the truth
revision process. However, the similarity between belief revision and truth revision is more
apt: just like in the truth revision process there is nothing definite in the belief revision
process. An agent can always come back to his beliefs about the several worlds, also once
changed. This is different in Gerbrandy’s knowledge revision process: once deleted a world
cannot ever become knowledge again. That is, not in classical logic. This would however
be 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 conclusions drawn previously. This is very useful in everyday life and in fields like computer
science: inferring things which are then known until other new information might change
this. An important feature of a non-monotic logic is that the size of a knowledge base does
not 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 lose
your knowledge on having a bad hairday.
We can also link this logic to solutions to the surprise examination discussed in this
thesis. For example, if we look at the solution raised by Kritchman and Raz, we recognise
the ‘strong’ classical inference used in their solution. Write ConsistentT +S for consistency
of the system T + S 3 , with S again referring to the surprise sentence. It is provable that
T + S `strong ConsistentT +S → ¬5, but because T + S `strong ConsistentT +S is not provable 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 believe
in the consistency of the system, say ConsistentT +S holds in all maximal worlds. Then
you would maybe be able to infer T + S `weak ¬5 from it, in the weak sense. But now
your induction argument is indeed possible, it does not stop! It can then be inferred that
T + S, ¬5 `weak ¬4, that T + S, ¬5 ∧ ¬4 `weak ¬3 and so on. We could identify this with
a plausibility relation in which Friday is less plausible than Thursday less plausible than
Wednesday and so on:
1←2←3←4←5
This is exactly the solution raised by Baltag and Smets for the belief-variant of surprise
and a fallible but trusted source. We may conclude that if we take Kritchman and Raz’
3
See solution Kritchman and Raz in chapter 3.
52
solution for the provability-variant of surprise as a starting point and use this kind of weak
logical entailment in non-monotonic logic, we end up in the solution raised by Baltag and
Smets.
5.2
Conclusion
Having said all this leads me to the end of my thesis, and the conclusions I draw from
the work done. I did not think beforehand that in this thesis I was going to find THE
solution to the paradoxicality of the surprise examination. What I did expect was to get a
better understanding of ‘surprise’ and of paradoxes as a whole. I think I succeeded in this
respect, however I remain with many questions for which answers I will keep looking after
finishing this thesis.
The several solutions raised to the surprise examination all have their charms. I am
especially attracted to on one hand Kritchman and Raz’ proof using Gödels theorem and
on the other hand Baltag and Smets’ idea to look at a softer notion of surprise and a
lower trust in the teacher. For the knowledge-version of surprise I also think Quine has a
good point when he sees the fallacy in the students’ reasoning, but lowering their trust towards 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 this
story. Writing this thesis I did not only get one answer: I got several! Another reason for
me to do this research is that I think that paradoxes matter. A view which is widely shared
is that paradoxes may point out that there is something wrong with our understanding of
the fundamental concepts involved. For example, take Zeno’s paradox of Achilles and the
Tortoise. In this ‘paradox’ Achilles runs 10 times faster than the tortoise and therefore
gives him a head start of 100 meters. Now the main reasoning in this story is that whenever
Achilles reaches a point where the tortoise has been, the tortoise went a little bit further
and therefore Achilles never can overtake him.
It turned out that the paradox rested on an inadequate understanding of infinity. The
paradox was solved.
Paradoxes matter and can be solved. And although I think we should not forget the
normal cases in which there is no paradoxicality at all, it is interesting to examine the
paradoxes closely to gain in knowledge (or belief) of the fundamental concepts involved.
The different interpretations of the surprise-sentence raised in this thesis actually create
different paradoxes. We can always reformulate them more closely to get different stories
with 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 solved
because we reach a better understanding of its fundamental concepts, just like it was the
case for Achilles and the Tortoise.
53
Popular Summary
The surprise examination is a paradox in which a teacher tells his students that the exam
that will be given next week will come as a surprise. Suppose it is commonly understood
that the exam will be a surprise if the students cannot know the evening before that the
exam wil be tomorrow. The smart students then reason that the exam cannot be given
on Friday, because they would know this at Thursday evening (because the exam did not
occur on the previous days only one option is left) and thus at Friday the exam would
not come as a surprise. If we represent this reasoning in a model with the double pointed
arrows 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 they
would know this at Wednesday evening, because they just now derived that the exam cannot possibly be given on Friday:
Monday ↔ Tuesday ↔ Wednesday
With this backward induction argument the students are able to eliminate any day of the
week as a possible day for the exam to occur. They reach a contradiction. But now the
teacher 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. A
few 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’ reasoning to eliminate Friday even fails and others say that reaching a contradiction will make
the students lower their degree of trust in the teacher. For instance, if the students do not
completely trust the teacher after the announcement, they may think it is less plausible
for the exam to occur on Friday but do not completely rule this out as a possibility. In a
54
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 not
deleting worlds but just making them less plausible. However the backward induction
argument 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 from
not gaining in knowledge or beliefs at all to falsifying the teacher’s statement and expecting
the 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 as
referring to itself (stating its own truth even after being uttered) or not, interprete surprise
in terms of knowledge or belief and choose from possible (positive or neutral) attitudes
towards the teacher. The ambiguity of the formulation of the surprise sentence uttered by
the 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 made
self-referential. In this way we are able to examine more closely what is exactly so puzzling
about the surprise examination and why the solutions seem to work. The goal of this
thesis is getting a better understanding of the concept of surprise and of paradoxes more
generally.
55
Appendix
In this chapter the Revision Theory of Truth is explained by taking the Liar paradox as
an example.
5.3
A New Theory of Truth Explained
The Revision Theory of Truth[8] is one of the main rivals to the threevalued semantics
which state that sentences can be true, untrue or neither (or both). To give an idea of
how the truth revision works, we return to the example of the Liar[16]. We write the Liar
sentence as
(1)
(1) is not true
We can revise the Liar’s truth value as follows: let us first assume that sentence (1) is not
true; hypothesis h0 = (1) is not true. Then by the its T-biconditional the name ‘(1) is not
true’ 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 therefore
we 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 with
hypothesis h1 = (1) is true. The T-biconditional is again ‘(1) is not true’ is true if and
only 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 our
new new revised hypothesis, h2 . We see that we get an alternating sequence of hypotheses
with hi is (1) is not true for all odd i and hi is (1) is true for all even i. If we would have
started 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 reach
an infinetely alternating sequence. Gupta and Belnap call this unstable. Being unstable
means not being stably true nor stably untrue: the revision sequence of revised hypothesis
does not stabilize on one of the two truth-values.
What explained formerly however is only the informal idea of the reasoning proposed
by Gupta and Belnap. For further work I refer to their book The Revision Theory of
Truth[8].
56
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