If you choose an answer to this question at random, what is the chance you will be correct?

A) 25%

B) 50%

C) 60%

D) 25%

  • $\begingroup$ Related: puzzling.stackexchange.com/questions/6372/… $\endgroup$ – GOTO 0 Jan 4 '15 at 0:38
  • $\begingroup$ Looking at the other question, there's not going to be a demonstrably correct answer because the question is not well-specified. So, voted to close. I'd also vote to close as a duplicate except I think the original is also close-worthy. $\endgroup$ – xnor Jan 4 '15 at 0:43
  • $\begingroup$ As far as I can tell the correct answer must be 0%. It isn't one of the listed options, so if you chose randomly form the listed options, there is indeed 0% chance of choosing the correct answer. $\endgroup$ – kasperd Jan 4 '15 at 15:58
  • $\begingroup$ Define 'correct'. $\endgroup$ – chasly from UK Aug 25 '15 at 11:33

When limited to the 4 provided answers, there is no correct answer, because this question is a paradox.

When there is one correct answer, the chance to pick that answer would be 25%. However, the answer 25% exists twice, so there is a 50% chance of picking it. But there is only one answer which says 50% and the chance to pick that answer is 25%.

However, there is no instruction in the question that you need to pick one of the provided answers. In that case "picking an answer at random" would mean that I would respond with a completely random statement like "A", "Y", "3.14159", "Warshaw", "jksdaskfa" or "Bob is the one who stole the strawberry cake". My answer could be literally anything, so there is an unlimited number of possible responses. Assuming that there is an unlimited set of incorrect answers but only a limited set of correct answers, the chance of picking a correct answer is 0% which would be my answer.

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You did not state the answer had to be uniformely random so therefore I can look at the answers, recognize the paradox, and limit the answers to the 2 possible, 25% and 50%. Then the answer becomes 50% since there are 2 choices.



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  • $\begingroup$ Nice solution. For that matter, you could make any of the answers work out. $\endgroup$ – Arel Jun 14 '17 at 3:54

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