I found this math "problem" on the internet, and I'm wondering if it has an answer:

Question: If you choose an answer to this question at random, what is the probability that you will be correct?

a. $25\%$

b. $50\%$

c. $0\%$

d. $25\%$

Does this question have a correct answer?

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    $\begingroup$ You copied the question wrong. Option (c) should be 60%. See the original here: i.imgur.com/qvzU4.jpg. $\endgroup$
    – user18383
    Commented Oct 28, 2011 at 2:24
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    $\begingroup$ @Anna: Interesting -- so in that case the correct answer is 0%? :-) $\endgroup$
    – joriki
    Commented Oct 28, 2011 at 2:34
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    $\begingroup$ E. NONE OF THE ABOVE It's the answer I select for every multiple choice problem I encounter, regardless of whether it is offered as a choice or not. Usually I have to get very creative justifying my answer, typically based on some technicality which invalidates the allegedly correct response. But in this case, it is smooth sailing for me. $\endgroup$ Commented Oct 28, 2011 at 20:20
  • $\begingroup$ More discussion of this at metafilter: metafilter.com/108902/Try-not-to-think-too-hard $\endgroup$ Commented Oct 28, 2011 at 20:51
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    $\begingroup$ @sweetjazz An interesting adaptation - if we changed (c) to 25% and (d) to "None of the above", there still would be no answer! $\endgroup$
    – mboratko
    Commented Oct 29, 2011 at 13:11

6 Answers 6


No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth.

It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement

This statement is false.

is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.

Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%.

Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question

Is the correct answer to this question "yes"?

would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless.

One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on

True or false?

  • The Great Pumpkin exists.
  • Both of these statements are false.

where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed.

On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful.

In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.

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    $\begingroup$ Liar paradox is also known as the Epimenides paradox after Epimenides of Crete who claimed: "All Cretans are liars". $\endgroup$
    – GEL
    Commented Oct 28, 2011 at 1:39
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    $\begingroup$ Bertrand's paradox, the proof of the halting problem, and the proof of Godel's incompleteness theorem all rely on similar self-referencing statements. $\endgroup$ Commented Oct 28, 2011 at 7:29
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    $\begingroup$ @BlueRaja, are you on first names with Bertrand Russell? Usually, "Bertrand's paradox" is about failure to specify a distribution when saying "chose a random chord in a circle". An interesting distinction for Gödel's argument (and the halting problem, which is closely analogous) is that in Gödel's case the self-referential statement doesn't itself know it's being self-referential. It just happens to be. If we do some trivial rewriting on it (such as swapping two conjuncts somewhere) it still means the same but is not exactly self-referential. $\endgroup$ Commented Oct 28, 2011 at 10:13
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    $\begingroup$ Henning, I agree that @BlueRaja's formulation is overly ambiguous, see this other Bertrand paradox. $\endgroup$
    – Did
    Commented Oct 28, 2011 at 13:27
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    $\begingroup$ How dare you question the existence of the Great Pumpkin!? $\endgroup$
    – joriki
    Commented Nov 1, 2011 at 15:30

The question is underspecified since it doesn't say which distribution is used in choosing an answer at random. Any of the answers could be correct:

If I choose a. with probability 25% and b. with probability 75%, a and d are correct.

If I choose a. with probability 50% and b. with probability 50%, b is correct.

If I choose a. with probability 75% and b. with probability 25%, c is correct.

From the design of the question, it seems that whoever wrote it had in mind a uniform distribution over all four answers, but forgot to specify that. In that case Henning's answer applies.

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    $\begingroup$ The writer of the question was definitely a programmer, then. We expect rand() to give numbers that are 'uniformly distributed'... Unlike some mathematicians. (*Rolls eyes.* :)) $\endgroup$ Commented Oct 28, 2011 at 5:08
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    $\begingroup$ nice but i think "at random" here exacty means uniform choice among the options given.. $\endgroup$
    – Nikos M.
    Commented Jan 5, 2016 at 23:36
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    $\begingroup$ @NikosM.: So choosing with a different distribution would not be "choosing at random"? $\endgroup$
    – joriki
    Commented Jan 6, 2016 at 4:49
  • $\begingroup$ it is the connotation of the phrase which is in most times synonymous to "unbiased coin" (or dice or whatever, aka uniform distr). Of course there is randomness in other distributions just lets say less ("biased distribution" as a phrase carries less randomness, but that is just the linguistics attached) $\endgroup$
    – Nikos M.
    Commented Jan 6, 2016 at 10:47

To offer up another perspective on Henning's answer, the question is essentially an elaboration of this (similar) multiple-choice question:

What is the correct answer to this question?

  1. Answer (2)
  2. Answer (3)
  3. Answer (4)
  4. Answer (1)

Note that there are some fine puzzles built around variants of the 'self-referential test'; for instance, this simple example:

Each of the following statements is either true or false. Which of them are true and which are false?

  1. All of these sentences are false.
  2. Exactly 1 of these sentences is true.
  3. Exactly 2 of these sentences are true.
  4. Exactly 3 of these sentences are true.
  5. Exactly 4 of these sentences are true.
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    $\begingroup$ I'm probably missing something subtle, but how can the answer be anything other than "Exactly 1 of these sentences is true" being the only true sentence? $\endgroup$
    – fluffy
    Commented Oct 28, 2011 at 0:53
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    $\begingroup$ @fluffy: that's exactly right (though it's often phrased in inverse - 'exactly 4 of these sentences are false', 'exactly 2 of these sentences are false', etc. - which makes it a little trickier). I said it was a simple example. :-) $\endgroup$ Commented Oct 28, 2011 at 1:36

If there is one right answer to the question, then you will answer this question statistically 25 percent of the time. If 25 percent is the "right" answer, then you actually have two options.

If you have 2 options, then 50 percent is the statistical answer. And if since 50 percent is the only option place to mark down, that means that you will only get this answer right 25 percent of the time because you have a 1 in four chance.

It is impossible without a miracle. Plus, if it is impossible then does that leave the option of 0 open because then there are no right answers? That is saying: "If there are no right answers, this is the right answer." What are you really saying there? Nothing.

I think maybe you can't find out how many answers there are in the first place. There can't be only one. There can't be only two. There can't be three. There can't be four, and therefore one is the right answer? No. Because then you start back at the beginning.

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    $\begingroup$ I haven't downvoted, but I believe it could be just because it doesn't add anything new to the previous answers. $\endgroup$
    – Mark Hurd
    Commented Jan 16, 2014 at 1:33

Well, I must be pretty insane if I start competing with these already heavily upvoted answers from high rep users. But although the following solution may sound a little creative or even frivolous, it could easily be the right one. You could say that this solution is reverse engineered, as follows:

The question instructs to only choose one single answer out of four. And assume a uniform distribution, since that is most likely intended, then each answer has a chance of 25% to become chosen.

So the correct answer should be: 25%.

This computes to answer A being correct, as well as answer D. Could that be? Yes, it can. The question does not reveal how many of the four given answers are correct, but since there is one to be picked, assume that at least one of the four answers is correct.

Let's call answer A + answer D the correct answer pair.

Now, there are two possible choices (A or D) that result in 50% of the correct answer (A and D). Secondly, there is 50% chance of picking one (A or D) of two (A and D) out of four (A to D). So whether answer A or answer D is chosen, in either case the probability of being correct (50% × 50%) is 25%, which evaluates true.

Thus, yes, the question has 2 correct answers.

And now I realize that this post is the long version of the by joriki ages ago given comment. ;)

Ok, to be clear about what I mean, I believe the question is a special variant of the following trick question:

What is the color of the car?

  1. Black
  2. Blue
  3. Gray
  4. Metallic

As owner of the car I know the correct answer is metallic black. But this would render the question unfair, because it is never possible to give this answer by only selecting one. The difference with the question in question is the equality of both answers to give, which makes it slightly more fair. But since you can select only half of the full solution, the probability is still 25%.

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    $\begingroup$ If I'm reading your response correctly, you're both giving meaning to the answers (answers A and D are possibly correct because they are both 25%), and ignoring this meaning (by assuming that one of A or D is the "correct" answer, as in on an answer key or in a computer). This doesn't make much sense... $\endgroup$
    – mboratko
    Commented Oct 28, 2011 at 2:28
  • $\begingroup$ @process91 Imagine the answers not being radio buttons as in a conventional multiple choice question, but instead being check boxes: A and D must be checked for the full solution, but you can choose only one. $\endgroup$
    – NGLN
    Commented Oct 28, 2011 at 2:32
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    $\begingroup$ That doesn't seem to be a fair analogy, since "Metallic Black" is not an option, and "Metallic"$\ne$"Black". Based on Joriki's answer above, this doesn't seem to be what he was thinking. $\endgroup$
    – mboratko
    Commented Oct 28, 2011 at 19:13
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    $\begingroup$ So, fuzzy logic then? $\endgroup$
    – kinokijuf
    Commented Jan 23, 2012 at 20:30

"choose an answer to this problem at random" does not specify a distribution. Let $c$ be a random variable on the choices, "$A$", "$B$", "$C$", and "$D$", representing the probability that we choose that answer to the problem. We have $0 \leq c(x) \leq 1$ for $x \in \{A,B,C,D\}$. Since the problem is conditioned on actually making a choice, $c(A)+c(B)+c(C)+c(D) = 1$. Let $m$ be a random variable on the choices, representing the probability that the choice is marked "correct". (For instance, when more than one grader is grading the pile of work, it is easy to imagine two graders, one who decides "$A$ is correct" and one who decides "$D$ is correct" and grades the papers they receive that way. Alternatively, we may imagine an automated grading machine that has various defects leading to the specified distribution of markings.) Here, each $m(X) \in [0,1]$, but also $m(A)+m(B)+m(C)+m(D) \in [0,1]$ since any given instance of the problem could have no answer marked "correct".

Now add the constraints \begin{align*} c(A) m(A) &= 0.25 \text{, } \\ c(B) m(B) &= 0.5 \text{, } \\ c(C) m(C) &= 0.0 \text{, and } \\ c(D) m(D) &= 0.25 \text{. } \end{align*} Either spend a while churning up the algebra, or ask your favorite CAS to do the work, and discover there is no solution.

A variant of the problem has choices "A. 25%, B. 50%, C. 25%, D. 30%". This variant also has no solution.

A way to find if there is a version of this problem that has a solution is to write $c$ and $m$ as vectors: $c = (c(A), c(B), c(C), c(D))$ and similarly for $m$. Then the four constraints supplied by the problem choices are the components of $c \cdot m = (c(A) m(A), \dots)$. Given the constraint on the sum of the $m$s, the sum of these components is less than or equal to the sum of the $c$s, which is $1$. So if any problem has four choices whose sum exceeds $1$, there are no distributions of choosing answers or marking answers that is compatible with the given choices. This criterion immediately rejects the solvability of the variant described above.

It should come as no surprise that the dot product is maximized when the vectors $c$ and $m$ are parallel. One randomly generated (by CAS) such set of problem choices is "A. $\frac{45}{64}+\frac{1}{128} \left(\sqrt{1541}-45\right)$, B. 1/4, C. 3/64, D. $\frac{1}{128} \left(45-\sqrt{1541}\right)$". (Here, the expectation of credit is $1/2$ with the implied pair of identical PMFs.) Not all such consistent sets of choices have $c = m$ or even $kc = m$ for $k \in [0,1]$, but it is "easy" to generate such examples. A consequence is that there are variants of this problem that are not internally inconsistent.

  • $\begingroup$ And what is a CAS? $\endgroup$ Commented Jan 4 at 14:01
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    $\begingroup$ @bballdave025 : "Computer Algebra System". Examples: Derive, Macsyma, Maxima, and many others. $\endgroup$ Commented Jan 5 at 18:37

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