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Multiple-choice question about the probability of a random answer to itself being correct
If you choose an answer to this question at random, what is the probability that you will be correct?

This question appeared on my G+ feed:

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%

My friend and I have differing views on how to answer it. While both of us agree that 0% is the correct answer if one was to connect the answer chosen with the actual question, he argues that 50% is still a correct answer if we understand the question in a broader sense: If a question is asked and four answers are given, while two of them are correct, what is the chance that the correct answer would be chosen at random? Could 50% ever be considered correct?

Here's how my friend thinks: There's a 1/4 chance that you've chosen the 'correct' answer. Out of given options, two of them are 'correct'. So, the chance that you're correct is 2/4.

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marked as duplicate by Henry, MJD, Did, user17762, Rahul Oct 21 '12 at 22:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I put in the question text into the post and replaced the link with the source of the problem. $\endgroup$ – Rahul Oct 21 '12 at 20:10
  • $\begingroup$ @RahulNarain Thanks, and sorry for my laziness. $\endgroup$ – programstinator Oct 21 '12 at 20:11
  • $\begingroup$ Short answer: yes. Longer answer: suppose any of the answers is correct (literally: try them all), and derive a contradiction for each of them. $\endgroup$ – akkkk Oct 21 '12 at 20:12
  • $\begingroup$ My reasoning would be along the lines that the answers given do not pertain to the question, and any mulitple choice question with 4 answers gives you a $25$% chance. So both A and D would be acceptable answers. Good question in general, though! (+1) $\endgroup$ – Envious Page Oct 21 '12 at 20:12
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    $\begingroup$ This may be useful. $\endgroup$ – Chris Oct 21 '12 at 20:35

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