Ian Stewart tries to render
a service to mathematics
by proclaiming

paradox lost
lan Stewart

There is a hoary old "paradox" which most readers must, surely, have come across already; it is usually called the Paradox of the Surprise Examination. I make no apology for resuscitating it here; it lives after a fashion in that dank and dreary limbo to which all stock mathematical recreations are eventually consigned. My object is to perform a public service by disposing of it for all time. Hope springs eternal...

The death-blow has been attempted many times vide the bibliography. In the author's humble opinion these can only rate as attempts since they all obscure the true facts of the case; their common fault is taking the problem too seriously. But first we state the:

Paradox
A schoolteacher announces that there will.be a surprise examination during the next week (Monday to Friday). The pupils reason as follows: If it doesn't occur by Thursday we'll know it must occur on Friday. So Friday is out. But then, by the same reasoning, Thursday, Wednesday,Tuesday, and Monday are out. "Please, Sir, you can't give us a surprise examination."

O'Beirne, in (1), solves the paradox by showing that the pupils cannot logically assume that there will really be an examination at all; Woodall (2) shows that they cannot assume teacher tells the truth. Both work by a sort of reductio ad absurdum. But then, if the paradox is really a paradox, then logic contradicts itself and we can prove anything at all - in particular that the paradox is not a paradox ... for which reason I would reject their solutions on charges of sophistry.

Let us analyse the situation more carefully (at the end of the analysis the situation will be so transparently clear that no analysis will in fact be needed; but we'll do things the hard way first). Suppose we have a very obstinate teacher who doesn't agree with the pupils' analysis; how will they convince him? Clearly by predicting the day of the examination. And what happens? By exactly the reasoning used by the pupils (performing the induction backwards) they must predict Monday on Monday, Tuesday on Tuesday, Friday on Friday. At some point in the week they will be triumphantly correct ... but somehow their argument seems to have been devalued in the process. All it boils down to is that they predict all possible outcomes in such an order as to be always right in the end. There is no paradox for by similar reasoning they could prove:

Ultradox
Nothing surprising can ever happen.
Proof:
Expect anything to happen at all times.

The paradox has now vanished. The reason it has vanished is be- cause of a simple matter of definition, the meaning of the word 'surprise'. For the paradox to work we must adopt a definition of surprise something like this:

An event is a surprise if at no time before it occurs can it be predicted.

The paradox shows that this is a useless definition; surprises cannot occur, because we simply have to keep predicting them for each possible day.

Suppose now that the pupils ire. restricted to one guess; if they are claiming on Monday that there can be no surprise, then it is only fair that they say why on Moriday. But now everything breaks down; Friday can only be excluded by waiting until Thursday to make the announcement of the predicted day. And whatever day they predict on Monday, the teacher can simply choose another one. Indeed, if they are given four guesses, they can still be beaten.

To summarize: there is no paradox. The working of the paradox depends on a purely semantic confusion; the definition of surprise needed for it to work is valueless. With a more reasonable interpretation of 'surprise' the paradox breaks down.

I'll try to forestall some heated correspondence: I have not altered the conditions of the problem. I agree entirely that there can be no surprise examination - in the sense which the paradox assumes. That is what conflicts with experience; if teacher just decides to give an exam on Wednesday the pupils will not be able to predict with certainty that it will be on Wednesday; all they can say is that if it isn't they will be able to predict Thursday, and if that's wrong, it must happen on Friday. But this option is always open to them. The argument of the paradox in just this option reformulated.

I'm willing to justify my contention that this is the correct and canonical way of resolving the paradox to anyone who disagrees. But if anyone is thinking of writing an indignant letter to the editor with his own pet solution, I'd like to quote O'Beirne, referring to his own postbag on the same issue: "The paradox seems to bring out the worst in some people."

References
1. T.H. O'Beirne: Puzzles and Paradoxes, 0.U.P. p.189.
2. D.R.Woodall: The paradox of the surprise examination Eureka 30 p.31.
3. C.Hudson, A.Solomon, R.Walker:A further examination of the surprise examination paradox, Eureka 32 p.23.
4. M.Gardner: A new paradox, and variations on it, about a man condemned to be hanged, Scientific American March i963 p.144.
5. G.Gamow and M. Stern: Puzzle-math Viking Press 1958.

D. WOODALL REPLIES

The well known 'paradox of the surprise examination' has received attention from many authors, some of whom (1-4) have (in my opinion) sucessfully resolved it. However, tnere are still those who attempt to resolve it by resorting to arguments about the nature of time, surprise, prediction,etc. - see e.g. the preceding article, which comes to the surprising conclusion that nothing surprising ever happens! As a defence against this type of argument I thought it might be illuminating to formulate the paradox in mathematical language, as follows. Here n denotes an arbitrary but fixed positive integer, say ri=100.

Proposition T. (Teacher)
There exists an integer m (1 £ m £ n) such that, for each r (:i £ r £ m), there is no logical argument by which one can deduce, from the fact that n ³ m ³ r, the fact that m = r.

Theorem C. (Ohildren)
Proposition T is false.
Proof.
We prove the nonexistence of such an integer m by induction on n - m.
If n - m = O, choose r = n. Then from the fact that n ³ m ³ r = n we can deduce that m = r. So certainly it is impossible for n - m to be 0.
Suppose now that n - m = k > 0, and suppose we have proved by induction that it is impossible for n - m to be equal to any of 0, 1, 2, ..... k - 1. Then we have proved that m £ n - k. So if we choose r = n - k, we can deduce, from the fact that m ³ r = n - k, the fact that m = r. Thus it is impossible for n - m to be equal to k either. This holds for k = 0, 1, 2, .... n -1, and so we have proved that it is impossible for m to exist, i.e. that Proposition T is false. Q.E.D.

But if propisition T is false, then for each m (1 £ m £ n) there exists an r (1 £ r £ m) such that there is a logical argument by which one can deduce, from the fact that n ³ m ³ r, the fact that m = r. Suppose we take m = 23 and n = 100. We clearly cannot prove that m = r unless r = m = 23. Siould anyone care to produce for me a logical argument by which one can deduce, from the fact that 100 ³ m ³ 23, the fact that m = 23?