|
MathOverflow has a nice take on the variant with 0%:
| quote: | No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth.
It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement
This statement is false.
is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.
Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%.
Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question:
Is the correct answer to this question "yes"?
...would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless.
One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on:
True or false?
The Great Pumpkin exists.
Both of these statements are false.
...where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed.
On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful.
In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics. |
http://math.stackexchange.com/quest...-have-an-answer
|
