What are the chances of getting this correct if you pick at random?

  1. 1/4
  2. 1/2
  3. 1/3
  4. 1/4

You are not allowed to add more answers to this list

Note; This DOES have a correct, demonstrable answer!

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    $\begingroup$ Is it uniformly random among the choices? What does right mean? $\endgroup$ – xnor Dec 24 '14 at 3:25
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    $\begingroup$ This is meant to be a mathematically stated problem, right, not a lateral thinking problem? I'm not asking for a hint, I'm asking for clarification on the question. $\endgroup$ – xnor Dec 24 '14 at 3:30
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    $\begingroup$ Related to this question on math.SE $\endgroup$ – Julian Rosen Dec 24 '14 at 5:58
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    $\begingroup$ What is being picked and what is random needs to be more clearly defined. $\endgroup$ – McMagister Dec 24 '14 at 6:46
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    $\begingroup$ @warspyking That helps, the term "uniformly random" is used to mean random with equal chance. When you say "each answer" though, do you mean that each of the four labels is equally likely to be picked, or each of the three distinct values? $\endgroup$ – xnor Dec 24 '14 at 13:13

Probability of

picking `1/4` = `1/2`
picking `1/2` = `1/4`
picking `1/3` = `1/4`

Probability of

correct answer `1/4` = `1/3`
correct answer `1/2` = `1/3`
correct answer `1/3` = `1/3`

So the probability of me picking correct Answer is

( 1/2 * 1/3 ) + ( 1/4 * 1/3 ) + ( 1/4 * 1/3 ) = 1/3

Probability of me picking correct Label remains 1/4

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    $\begingroup$ This answer assumes that the prior probabilities of each of '1/4', '1/3' and '1/2' is 0.33333. $\endgroup$ – McMagister Dec 24 '14 at 11:34

By randomly picking one of the labels 1, 2, 3 or 4, my chance of being correct is $1/4$.

Correct here means the single label declared by tester (warspyking, in this case) as "correct". The content behind the labels is irrelevant to me as there is no question asked about those contents. I only see a question about my chance of picking the correct label.

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    $\begingroup$ My chance of picking the correct label when doing it randomly does not change (no matter what's behind the labels). Period. :) The other answers appear to go that vicious self-refering way. Good luck! :) $\endgroup$ – ir7 Dec 24 '14 at 5:44
  • $\begingroup$ Scratch my earlier comment. If it was 1/4 then it'd be 1/2 $\endgroup$ – warspyking Dec 24 '14 at 11:02
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    $\begingroup$ Cutting the Gordian knot. Not a bad answer at all. $\endgroup$ – jscs Dec 24 '14 at 20:58

I leave this question blank.

It can't be $1/4$ because there would be a $1/2$ chance of randomly selecting choices 1 or 4, and it can't be $1/2$ or $1/3$ either because the chances of randomly selecting either choice is $1/4$.

But if the correct answer is zero then there is zero chance of randomly selecting the correct choice. This matches, thus the correct answer is zero and I leave the question blank as I won't get it correct.

  • $\begingroup$ Answering the question is a very funny way of not anwering it! :-) Nice bit of paraleipsis there, McMag. $\endgroup$ – Rand al'Thor Dec 28 '14 at 19:37

The crucial point of this puzzle is what defines a correct answer.

Being a multiple choice question, one should possible assume the 4 answers should answer the question above.

With this in mind the correct answer

Is not available. Because there is a 1:2 chance for it being 1/4 (wrong) and a 1:4 chance for it being either of the other (wrong)

In short:

It is a paradox.


In the real world scenario, the probability of the answer being correct is 1/2. Two scenarios for this:

  • It is a multi choice question and thus can have only one right answer. As some one who do not know the answer(that's the reason we are choosing randomly or else we will pick the right one, right?), we will first discard the repetitive answer through logical reasoning. That is, if it is the right answer, such a blunder will not be seen in the question. Thus we will have two choices (2) and (3) and thus the probability will be 1/2(I am not noting the choice 1/2 but simply the probability of either choosing option 2 or 3)
  • Upon over simplification, the chance of choosing the right answer is going to be only 1/2. Either you choose the answer right or you choose the answer wrong. Which means your chance of getting a randomly picked answer right is 1/2.
  • $\begingroup$ How'd you get this? I don't see it. $\endgroup$ – warspyking Dec 24 '14 at 13:00
  • $\begingroup$ Which point you are asking for? first or second? :-) $\endgroup$ – thiruvenkadam Dec 24 '14 at 16:13
  • $\begingroup$ 1 ${}{}{}{}{}{}$ $\endgroup$ – warspyking Dec 24 '14 at 16:52
  • $\begingroup$ Let's say you are rapidly marking answers. Instead of solving the question, you might want to limit the choices. If same answer is given in option (1) and (2), most probably it means that those two options are not the answers. Possibly some one lazy to cross check the options just put on numbers that are closer to option (2) and (3). So it is a good reasoning to neglect those two options which leaves you with two options and probability of choosing the correct answer will be 1/2. $\endgroup$ – thiruvenkadam Dec 24 '14 at 17:01
  • $\begingroup$ There.... I added a long comment to average out your shorter comment :-) $\endgroup$ – thiruvenkadam Dec 24 '14 at 17:01
  • This is a multiple choice puzzle, that means a correct answer is a set of labels s such that "s are the labels of chances of getting this correct if you pick at random" is true ie "s = { x in {1, 2, 3, 4} | s={x} p(x)=1/4 }" is true ie "s = { x in {1, 4} | s={x} }" is true.

  • As you see only the empty set can make this true.


Random? Okay. I'll pick one of the three possible answers, 1/2, 1/3, or 1/4, uniformly at random. (There are two choices labeled "1/4", but that doesn't matter to me - they're the same answer.)

I think you see where this is going.


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