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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
    Commented Dec 24, 2014 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
    Commented Dec 24, 2014 at 3:30
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    $\begingroup$ Related to this question on math.SE $\endgroup$
    – Julian Rosen
    Commented Dec 24, 2014 at 5:58
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    $\begingroup$ What is being picked and what is random needs to be more clearly defined. $\endgroup$
    – McMagister
    Commented Dec 24, 2014 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
    Commented Dec 24, 2014 at 13:13

10 Answers 10

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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
    Commented Dec 24, 2014 at 11:34
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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
    Commented Dec 24, 2014 at 5:44
  • $\begingroup$ Scratch my earlier comment. If it was 1/4 then it'd be 1/2 $\endgroup$
    – warspyking
    Commented Dec 24, 2014 at 11:02
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    $\begingroup$ Cutting the Gordian knot. Not a bad answer at all. $\endgroup$
    – jscs
    Commented Dec 24, 2014 at 20:58
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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.

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  • $\begingroup$ Answering the question is a very funny way of not anwering it! :-) Nice bit of paraleipsis there, McMag. $\endgroup$
    – Rand al'Thor
    Commented Dec 28, 2014 at 19:37
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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.

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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.
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  • $\begingroup$ How'd you get this? I don't see it. $\endgroup$
    – warspyking
    Commented Dec 24, 2014 at 13:00
  • $\begingroup$ Which point you are asking for? first or second? :-) $\endgroup$
    – thiruvenkadam
    Commented Dec 24, 2014 at 16:13
  • $\begingroup$ 1 ${}{}{}{}{}{}$ $\endgroup$
    – warspyking
    Commented Dec 24, 2014 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
    Commented Dec 24, 2014 at 17:01
  • $\begingroup$ There.... I added a long comment to average out your shorter comment :-) $\endgroup$
    – thiruvenkadam
    Commented Dec 24, 2014 at 17:01
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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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  • 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.

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There is no paradox, just an illusion in the concept. That is, you are all deluded with the right answer because you are seeing the answers. If you assume that you don't know what the answer is, that is, it is hidden, even if all 4 answers have the same result, the probability of getting the correct answer it will be always 25%, why? Because it's the number of options you have to choose from.

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The "question" is meaningless.

Consider the following Python function:

def f():
 result = f()
 if result == 1/4: return 1/3
 elif result == 1/3: return 1/3
 elif result == 1/2: return 1/3

What does f() evaluate to? One might argue that the only consistent answer is that it evaluates to 1/3. But that would be wrong: the function just calls itself endlessly and never evaluates to anything at all.

Similarly, the way that human language works is that you read a string of text and attempt to parse it into something meaningful. But when one tries to parse the given text, the process of parsing never terminates:

If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to this question?

If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to the question, "If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to this question?"

If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to the question, "If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to the question, "If you pick one of 1/4, 1/3, or 1/2 at random, what is the probability that it will be the correct answer to this question?""

... and so on. So no meaningful question ever emerges, any more than it would from a string of random words with a question mark at the end.

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Seems there is a 2 different scenario.

If we have to answer this question only once then the chance of being correct is 0. Because chance to pick the answer 25%(a or d) is 50% Chance to pick 50%(c) is 25% Chance to pick 60%(b) is 25%

But if we have to answer this question N times, sitation changes. If we do random picks N times then: Chance to pick the answer 25%(a or d) in 25% cases is 1/4(25%) * 1/2(picking a or d randomly) = 1/8 Chace to pick an answer 50%(c) in 50% cases is 1/2(50%) * 1/4(chace of picking c) randomly) = 1/8 and vor b) it's 60/100 * 1/4 = 3/20

So the chance of picking the answer 25% in 25% cases or chance of picking the answer 50% in 50% cases or chance of picking the answer 60% in 60% cases is 1/8 + 1/8 + 3/20 = 2/5 = 40%

in case of N is divisable by 40, if covers all 3 cases. The answer is 2/5=40% in case of N is divisable by 2, but not by 4 and 40, then it covers only case of c)50%. The answer is 1/8=12.5% in case of N is divisable by 4, but not by 40, then it covers a,d)25% and c)50%. The answer is 1/8+1/8 = 1/4=25% In other cases of N, the answer is 0.(This covers first scenario too, in case of N=1)

So, the answer depends on how many times we pick the answer.

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