The Answer Man

The Answer Man claimed that he could answer over 96% of all questions correctly. He supported his claim as follows:

Ask me any question. If I know the correct answer, I will tell you. If I am unsure of the answer, I will answer the twenty six questions, 'Does (your question) start with the letter 'A'?', 'Does (your question) start with the letter 'B'?, and so on, until I answer, 'Does (your question) start with the letter 'Z'?'. In each case, '(your question)' is the text of the question I couldn't answer. Clearly, I can answer these 26 questions correctly (even if you start your question with non-alphabetic symbols, I will simply answer each question 'no'. Out of the 27 questions (the original question and the 26 others), I will answer at least 26 out of 27 correctly, which is over 96%.

Is the Answer Man's argument valid, or can you think of a way to prove him wrong?

Carl G.

Unless I'm missing something, it seems the Answer Man's argument is valid but silly. He doesn't need such a complicated formulation, either: to support the claim that he can answer at least (N-1)/N of all questions correctly, all he needs to do when confronted by 'Is chess a forced win for Black?' is to say

'Is 1 equal to 2? No.' 'Is 2 equal to 3? No.' ... 'Is N-1 equal to N? No.' 'Is chess a forced win for Black? Yes.'

Of course, the questioner might object that the first N-1 questions in this set were posed by the A.M. and not by the questioner himself. Such an objection would be, naturally, an aspersion on the A.M.'s honor, which can only be settled by a duel. Each duellist can shoot to kill, or can run away across an unlighted bridge which takes eleven minutes to cross, or can use a five-gallon and a three-gallon bucket to dump exactly three ounces of water in his opponent's powder horn, or ...

No need for all that. Just look him in the eye and tell him that he has green eyes, and he will kill himself.

Dennis Yelle

I think that you might be missing the point that the Answer Man can answer 96% of *all*possible* questions correctly, not just 96% of a finite or countable subset of questions.

Suppose that we wrote down every single question that makes grammatical sense in the English language. It's not clear to me that this is a countable set.

Using the original method we can imagine A.M. examining every single question one by one, and whenever he finds one question he can't answer he can also find 26 other questions that are already in the list which he can answer correctly. This reasoning works even if the set of all questions is uncountable.

Using your method we can imagine A.M. examining every single question one by one, and whenever he finds one he can't answer he uses the next 26 'Is N-1 equal to N?' questions. This method only works if the set of all possible questions is countable. It would be possible to extend this method if we allow the set of all possible questions to include things like 'Is R-1 equal to R?' where R can be any real number to give A.M. enough extra questions to match 'C', the cardinality of the continuum (most of these questions are infinitely long, since there's usually no shortcut for defining an arbitrary real number to an infinite number of decimal places) but it's not clear that the set of all questions would not then be larger than C.

*However*

Answer man's reasoning is somewhat flawed because you can't really get away with that sort of argument when considering infinite sets, and the argument doesn't work at all if we impose a limit which restricts us to a finite set of questions. If we limit ourselves to the finite set of questions that can be written down in N words or less, then A.M. can only find 26 answerable questions in the set under consideration corresponding to questions that can be written down in N-6 words or

I think it works only for countable sets, because there is no way (not even in principle) to examine all the members of an uncountable set 'one by one.' If there were, there'd be a way to count the set, right?

If I'm not mistaken, the Cartesian product of C with itself has the same cardinality as C. And (here I'm on shakier ground) are the 'quotients' 26* C / 27* C and C * C / C well-defined? They look to me like indeterminate forms

There are a finite number of words which have a currently acknowledged meaning.

Thus, for eanch strictly positive integer n, there are a finite number of questions consisting of n words.

The union of finitely many finite sets is at most countable.

Therefore, the number of questions is, at most, countable.

In practical terms, there must be some maximal length of a question in order that it be understandable, so, in practical terms, the number of such questions is finite.

If such a maximal length is imposed, then for questions that come close to that length then 'Does (that question) start with the letter 'A'?' becomes too long to be an acceptable question when we substitute the actual text of the long question into the brackets. So Answer Man's reasoning fails.

Godel considered questions (he called them theorems) of unlimited length, and showed that there must be considerably more undecidable theorems (unanswerable questions) than there are decidable theorems.

No; Mike Williams wrote the quoted paragraph.

Its easy to guarentee 96%. All he has to answer is one of the 27 questions correctly. Its a trick that anyone can do. Answer the first question wrong to get the other 26 questions. You will be able to answer one of those 26 questions. Just remember the first letter of the origional question. How,Who,When...

But the Answer Man can't answer 96% of those correctly, for to do so he has to extend the set to include 'Does <Q> start with the letter A?' for each element Q of the set.