Phil 2511: Paradoxes

Lecture 3:  Paradoxes of Rationality and some Epistemic Paradoxes

 

The paradoxes we shall consider today are paradoxes about what it is rational to do or to believe, and paradoxes concerning what you can or cannot know.

 

The Prisoner's Dilemma

The police present two suspects with the following alternatives: if either confesses while the other does not, he will receive two years in prison while the other will receive twelve; if neither confesses, each will receive four years in prison; if both confess, each will receive ten years in prison.

(Source: Robert Nozick, Philosophical Explanations (Cambridge MA, Harvard University  Press. 1981), p.452).   This problem was devised in 1950 by Albert W. Tucker, Department of Mathematics, Princeton University.  The following matrix represents the prisoners' choices:

 

                                    B confesses        B does not confess

 

A confesses                    <10,10>                     <2,12>

 

A does not confess          <12,2>                      <4,4>

 

[The left number in each pair represents years in prison for A; the right number, years in prison for B]

(Good discussion in Sainsbury, p.64 ff.)

 

Suspect A reasons that it is in his best interest to confess.  For, if he stays silent and B confesses, then he will get 12 years (by confessing he would only have got 10).  And if he stays silent and B stays silent, then he will get 4 years (whereas had he confessed with B staying silent he would only have got 2).  B, going through similar reasoning, figures that it is in his best interest also to confess.  Result: between them they serve 20 years in prison instead of the 8 they would have got had both remained silent.  In other words, the `rational’ choice – to confess – predictably leads to a worse result for both!

 

This problem has been discussed by psychologists and social scientists, as well as by philosophers.  Experiments have been done in which a pair of subjects have been subjected to a series of such trials, and informed, after each trial, what the other decided.  It was found that, after a sufficiently long series, subjects tended to cooperate, i.e. take the `remain silent' option.

 

The Prisoner’s Dilemma is related, in an interesting way, to Newcomb’s Puzzle, which we met in the last lecture.

 

The Two-envelope Paradox

You are presented with two sealed envelopes, and are informed that one of them contains twice as much money as the other.  You select one at random.  You are then offered the chance to swap and take the other instead.  (Compare the Monty Hall non-paradox.)  If your selected envelope contains x, and your swap is lucky, you get 2x, but, if you are unlucky, you get ½ x.  So it seems that your expected utility if you swap is

(½ . 2x) + (½ . ½ x),

which is 1¼ x.  For example, if you have $20 in your chosen envelope, the other envelope must have either $10 or $40, average $25.  So it looks as if you should swap.  However, exactly the same argument would have been available if you had picked the other argument in the first place.

 

For a thorough (and difficult) discussion of this paradox, see Michael Clark and Nicholas Shakel, `The Two-envelope Paradox’, Mind 109/435 (July, 2000), pp.415-442.  But there is also a paper, not yet published, which to my mind, solves this paradox satisfactorily.  It is Graham Priest and Greg Restall,  `Envelopes and Indifference’ (2003), which you can find at http://consequently.org/papers/envelopes.pdf

 

One of the funny things the Two Envelope Paradox, is that, whichever envelope you were first given, it seems that you would be better off with the other one, and I suggested the motto `The grass is always greener on the other side of the fence’, the Chinese counterpart to which is 隔離飯香.

                         

 

The Bottle Imp

There is a bottle containing an imp, and someone offers to sell it to you for $1000.  The imp will grant you any wish, except for those that undermine the condition of the sale.  That condition is that you sell the bottle before you die for less than you paid for it, otherwise you will burn in hell for eternity.  The minimum unit of currency is one cent.  Now this seems like a real bargain. But think.  Obviously no rational person would buy the bottle for one cent for, by the condition of sale, he would be unable to sell it and so would go to hell.  Equally, no rational person would buy the bottle for two cents, because he would be aware that he would find nobody prepared to purchase the bottle from him for one cent, so he would be stuck with the bottle and would end up burning for ever. Similarly, there would be no rational buyer at three cents, nor at four.....nor at $1000.  Yet if you, a rational person, did buy the bottle for $1000 and obtained from the imp everything you ever wanted, surely your example would be sufficient to persuade someone to buy the bottle from you for one penny less than you paid.

(R.Sharvy, `The Bottle Imp', Philosophia 12 (1983), p.401).

 

Various attempted solutions to Sharvy's problem have been proposed in Philosophia 15 (1986), pp.421-444.

 

A Paradoxical Train of Thought

Two persons on a train, A and B, each think of a number and whisper it to a fellow rider C.  C gets up and announces `This is my stop.  You have each thought of a different positive integer.  Neither of you can deduce whose number is bigger.'  C then gets off the train.

A and B continue their travel in silence.  A, whose number was 157, thinks, `Obviously B didn't choose 1.  If he did, he'd know that my number was the bigger, just from C's statement that we chose different numbers.  Just as obviously, B knows I didn't choose 1.  So C did not whisper `1’ to either of us.  The smallest number that is even a possibility is 2.  But if B had 2, he'd know that I didn't have it either.  So 2 cannot be a number whispered by C…...'

If the train ride is long enough, both A and B can rule out every number.

 

Another Martin Hollis variant of the Surprise Examination

Sheikh Inbed says to his two oldest wives: `I am leaving my vellum copies of the Tractatus to be shared between you, each receiving at least one.  There are in total either 99 or 100.  Neither, on being told her own share, will be able to infer exactly how many the other has received'.  Whereat Fatima remarked `Then neither can receive 99.  So anyone receiving 1 could place the other with 98.  So anyone receiving 98 could place the other with 2.  So anyone receiving 2 could place the other with 97.  This reasoning continues until all possibilities have been eliminated.  Inbed, you are an old fool'

(M. Hollis, `More Paradoxical Epistemics', Analysis 46 (1986), pp.217-218.)

 

There are three claims that Sheikh Inbed makes regarding how his copies of the Tractatus are to be shared between his two oldest wives:

 

(i)                  Each wife shall receive at least one copy.

(ii)                There are in total either 99 or 100 copies

(iii)               Neither wife, on being told her own share, will be able to infer exactly how many the other has received.

 

Fatima, one of the wives, reasons that not all of these three claims can be true – specifically, that if the first two are true, the third must be false.  She reasons thus:

a.  Neither wife can receive 99.  (For if a wife received 99 then, because of (i) she could, contrary to (iii), infer the number of books the other wife has, namely 1.)

b.      So any wife receiving 1 copy would know that, since the other wife cannot have 99 (by a.), she must have 98.  This contradicts (iii).  Hence, if the Sheikh is consistent, no wife could have just 1 copy

c.       So anyone receiving 98, knowing that the other wife could not have just 1 copy, could know that the other wife has 2 copies.  But that would contradict (iii), hence no wife could have 98 copies.

d.      So any wife receiving 2 copies would know (by c.) that the other wife could not have 98 copies and so would know that the other has 97 copies. But that would contradict (iii), hence no wife could have 97 copies.

And so on.  So, if Fatima is right, then Sheik Inbed cannot divide up his books in the way that he proposes, i.e. in accordance with (i), (ii), (iii).  But HE CLEARLY CAN.

 

This paradox is obviously closely related to that of the Surprise Examination.  Despite Fatima's impeccable argument, it is quite clear that Inbed could have shared out the books in such a way that neither wife would know how many the other received.  For example, had Fatima received 57 books, she couldn't possibly tell whether the other wife had received 42 or 43.

 

[Note: In the present `politically correct’ times, it is doubtful whether an editor would allow the name `Sheik Inbed’ (`shake in bed’).]

 

 

How NOT to solve a paradox

Suppose that a theorist investigating a certain paradox hypothesizes that X is the hidden factor responsible for the paradoxicality.  Let S be the negation of this hypothesis.  The theorist then proceeds to argue that since S, taken as a premise together with the paradox generator yields a contradiction then, by reductio ad absurdum, we may conclude that X is the hidden factor responsible for the paradoxicality.

 

Our theorist is obviously employing a rotten argument and, for want of a better name, I'll call it The Rotten Argument.  One thing wrong with it is that since, from the paradox generator alone, we can derive a contradiction, it is of no significance that we can derive contradiction from the paradox generator plus X.  Further evidence of the rotteness of the argument is that it will certify any candidate as the hidden factor, and the reason for this is that it incorporates the paradox reasoning or a fragment thereof in a supposedly valid reductio argument - when, of course, it is precisely the validity of the paradox reasoning that is in doubt.  To try to show what is dubious about some dubious reasoning by the use of that very reasoning is a really iniquitous form of bootstrapping.

 

Versions of The Rotten Argument have been used to establish, as hidden factor in the Liar

 

(a)  that the principle of substitutivity of identicals is incorrect (Skyrms, 1970, p.68)

 

(b)  that the applicaton of a predicate to a well-formed sentence (or to a name of a proposition) does not always give a well-formed sentence as a  result (Fitch, 1970, p.77)

 

(c)  that certain token sentences are meaningless (Donnellan, 1957)

 

(d)  that certain token sentences do not express propositions (Parsons,  1974, pp.386-7)

 

(e)  that we cannot accept the dichotomy between true and not true (Stone, 1981)

 

(f)  that some sentences cannot say of themselves that they are not true (Cargile, 1979, p.253)

 

(g) that certain definitions (so-called `type-crossing definitions') are illegitimate (Pollock, 1970, p.84)

 

I'm not saying that none of these conclusions have any merit, but what merit they have will need to be demonstrated by means other than The Rotten Argument.  All that such instances of the argument deliver are hopeful suggestions.  What they do not offer is a diagnosis of what is responsible for the paradoxicality, and, as we (following Chihara) have already pointed out, without a diagnosis, no remedial measure can be satisfactory.  Without a proper understanding of the cause of a paradox, no treatment justified post hoc by its effectiveness can be relied on to be effective against related strains.  Russell, when he was still groping for a solution to the paradoxes, saw this clearly.  He remarks that, at that stage, he had found `no guiding principle except the avoidance of contradictions; and this, by itself, is a very insufficient principle since it leaves us always exposed to the risk that further deductions will elicit contradictions.' (Russell, 1906).