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

A) 25%

B) 50%

C) 60%

D) 25%

On internet you can find the problem here.

  • $\begingroup$ Ben Millwood and AndreasT have suggested one could argue that the correct answer is $0\%$. I would agree, though that possibility would be removed if option C became $0\%$ instead of $60\%$. $\endgroup$ – Henry Oct 3 '13 at 10:33
  • $\begingroup$ @Henry I thinkt that 0% is 'probably' the right answer. You can indeed make variants. If C) would be 50% for instance then 50% becomes 'probable' too. That gives extra puzzle-fun. $\endgroup$ – drhab Oct 3 '13 at 10:45


Assume this question has a correct answer.

If the correct answer appears 1 times on the possible answers then its probability should be 25%. Cannot be since 25% appears twice

If the correct answer appears 2 times on the possible answers then its probability should be 50%. Cannot be since 50% only appears once.

No answer is repeated 3 or 4 times.

So if the correct answer appears 0 times on the possible answer then its probability should be 0%. 0% does not appear on the possibilities so it is the correct answer.

  • $\begingroup$ sounds good. Can you also answer it if behind C) you would find 50%? In the Original every assumption of the sort 'x) is the correct answer' leads to a contradiction. But not anymore in the adapted case. $\endgroup$ – drhab Oct 3 '13 at 13:02
  • $\begingroup$ My thinking was: Let X be a random variable with P{ X=A} =P{ X=B} =P{ X=C} =P{ X=D} =1/4. Let Y have the same distribution and let X and Y be independent. For x in{A,B,C,D} let p_{x} denote the probability mentioned behind letter x. To determine is then: P{ P{ p_{Y}=p_{X}} =p_{X}} In words:What is the chance that the probability that you find at your chosen letter agrees is correct, i.e. is exactly the probability that this would be the outcome? $\endgroup$ – drhab Oct 3 '13 at 13:08

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