You are standing before 3 objects. You need to give the right one to the right monster or else you'll die.

There are 3 people in front of you for you to question.

One always lies, one always tells the truth, one varies between truth and lies, and you don't know which is which.

You can only ask 3 yes-or-no questions. How do you figure out which object goes to which monster?

EDIT: You are on a different planet, and you don't know the difference between "yes" and "no".

  • 1
    $\begingroup$ What do you mean by "varies between truth and lies"? Do they tell the truth or lie randomly, or do they always alternate true statements with false ones? $\endgroup$ Mar 4, 2017 at 13:52
  • $\begingroup$ I mean that he will never solidly tell truth, or lie, just randomly switches between them $\endgroup$
    – Joey Pandy
    Mar 4, 2017 at 14:13
  • $\begingroup$ @JoeyPandy Well that blows my strategy $\endgroup$
    – user41805
    Mar 4, 2017 at 14:13
  • $\begingroup$ youtube.com/watch?v=LKvjIsyYng8 kek $\endgroup$
    – user34009
    Mar 4, 2017 at 15:37
  • 1
    $\begingroup$ Do they answer something like "hufn" or "gawt" which I don't know which if is "yes" or which is "no" or I have no idea of what they might say until I hear two different sentences? $\endgroup$ Mar 4, 2017 at 15:55

2 Answers 2



This answer has a few assumptions:

  • Let's call the objects O1, O2 and O3, the guards G1, G2 and G3 and the monsters M1, M2 and M3, and let's suppose that the guards are able to understand that nomenclature. To make that happen, you might just announce what is what and who is who in the start when the guards are just silently listening, and since you aren't asking anything so far, you're ok.

  • Let's suppose that the guards know who of themselves is the truthful, who is the liar and who is the random.

  • Let's suppose that the guards won't answer anything other than just "yes" or "no" in some unknown language. They will never answer something like "I don't know", "maybe", "I refuse to give you an answer" or anything like that.

  • The guards knows which object should be given to which monster.

  • You can direct a question to only a single guard. I.e., you aren't allowed to give a single question to all the three and expect to hear three answers at the same time.

  • The guards are able to understand perfectly what you ask them.

  • The guards produce an answer promptly. I.e., they won't wait 10 minutes or so to produce an answer.

  • The random guard does not simply answer something randomly. First, he decides if he is going to lie or tell the truth, and then, he tries to find the answer to the question and spell it out. This is similar to him just mentally imitating either one of his peers when he answers the question.

  • However, the random guard decides to tell the truth or not by using a procedure that is in fact an ideal and perfect random data generator with an output that is absolutely unpredictable down to the quantum physics or dark energy level. There is no way to predict what would he choose to do even if you are able to map the position and the velocity of every subatomic particle in his brain and around him (Werner Heisenberg and Erwin Schrödinger say "hello guys from puzzling"). It is also impossible to influence, tamper or bias the random guard's choice.

  • Let's suppose that the guards are extremely intelligent and wise and will produce an answer to the best of their knowledge. I.e., they will not make any mistakes in their answers. Even if you ask them something really complicated and beyond the current best knowledge of the humanity like "is P = NP?", "is the Riemann hypothesis true?" or "are axions the main component of dark matter?" they will answer it promptly and without failure.

First question

Here it goes:

"If some person A asks you if blue and red are different colors, what would you give as an answer to A?"

This seems to be a pretty stupid crazy pointless question; what is going on?

The embedded question "blue and red are different colors?" is a very simple and trivial question that everybody knows is true. You may also replace it with any other simple and trivially true question.

1. If you ask this first question to the truthteller, he will answer "yes" because he would answer "yes" to A and he tells the truth.

2. If you ask this first question to the liar, he will answer "yes" because he would answer "no" to A, but he will lie about that.

3. If you ask this first question to the random and he decides to proceed like the truthteller, he will answer "yes". If he decides to proceed like the liar, he will also answer "yes".

WOW, this means that:

You are able to learn the word/sentence for "yes" with just one question regardless of who is being asked for! If you ever hear something different from now on, you can be sure that it is "no".

Is it really possible with 3 total questions?



There are six ways to assign an object to a monster:

1. O1 to M1, O2 to M2, O3 to M3
2. O1 to M1, O3 to M2, O2 to M3
3. O2 to M1, O1 to M2, O3 to M3
4. O2 to M1, O3 to M2, O1 to M3
5. O3 to M1, O1 to M2, O2 to M3
6. O3 to M1, O2 to M2, O1 to M3

So how many questions we need to separate those?

Exactly 3 yes-or-no questions. One of those questions separate the things in 2 categories, which is insufficient to completely distinguish 6 possibilities. With 2 questions, we have 4 categories, still insufficient. With 3 questions we have 8 categories, which is enough to completely separate that, but since we already used one question, we would be screwed. However...

There is a little trick in our box:

What if you ask for something that has a contradiction and is paradoxical?

For example:

If I ask the liar guard if he is a liar, will he answer "yes" in your language?


There is no way that the truthteller or the liar guards can answer this, so they would just remain silent and perhaps show a scared and worried expression. The random guard would choose to either lie or tell the truth and try to produce an answer, but since he also can't answer that regardless of his choice, he would also remain silent.

How does this change our numbers?

This is a third possible answer. So in fact we can get a "yes", "no" or silence out of any guard. And with that, we need only two questions to fully separate all the 6 cases (in fact, we would be able to separate it in 9 cases). Fortunately, two questions is exactly what we still have left.

Of course, there is a special thing to note here:

The no-answer from a paradoxical question is also an assumption. If something other than something usefully identifiable as "can't be answered" sprouts out, you are doomed. For example, if the guard decides to answer paradoxical questions by shooting you, you are screwed. If they will just answer either "yes" or "no" regardless of the fact that it is paradoxical, you would be screwed too.

Second question

We have to separate our cases.

We have six categories. Since we have two remaining questions, we must produce at least a "yes" or "no" question and a "yes", "no" or "no answer" question that somehow manages us to solve the problem (or perhaps two "yes", "no" or "no answer" questions). So, let's find a "yes", "no", "no answer" question that is somehow helpful to us.

It is a very complicated and contrived question:

"Let's define H1 as the truthteller guard, H2 as the random guard and H3 as the liar guard, and let's also define $x$ as the index of the object O$x$ that should be received by the monster M1. If by tomorrow, somebody asks you what H$x$ would answer to the question "are red and blue different colors?", what would you answer?"

Ok, that was a really hard question. Let's analyze it:

The double negation trick from the first question that turns the liar in a truthteller while keeping the truthteller telling the truth is applied once again, so the answer, should you have one, is truthful. There is some object that M1 should receive. If it is O1, an imaginary question is directed to H1; if it O2, then to H2; if it is O3, then to H3. The imaginary question is "are red and blue different colors?", which is trivially true.

So, if $x = 2$, then the imaginary question goes to the random guard and the guard being asked won't produce an answer because he doesn't know the answer, even if he is himself the random guard – even the random guard doesn't know what he himself would answer by tomorrow! This way, if there is no answer, you should give O2 to M1.

If there is some answer, you can be sure that is a truthful one, so if it is "yes", then give O1 to M1. If the answer is "no", give O3 to M1.

Third question

Once you got this far, this should be easy.

Just a little change to the second question.


"Let's define H1 as the truthteller guard, H2 as the random guard and H3 as the liar guard, and let's also define $x$ as the index of the object O$x$ that should be received by the monster M2. If by tomorrow, somebody asks you what H$x$ would answer to the question "are red and blue different colors?", what would you answer?"

What changed?

It is now M2 instead of M1.

What to do with the answer?

If there is no answer, give O2 to M2. If the answer is "yes", then give O1 to M2. If the answer is "no", give O3 to M2.

And finally:

Give the last remaining object to the last remaining monster.

Final notes

This puzzle could be made a bit harder without changing much by the following modification:

You don't need three guards after all. You just need one! Even a single guard that you don't know if is telling the truth or not, and could even be choosing randomly to either tell the truth or lie (which is not the same thing as to answer randomly) would do. You would need to change the phrasing of the second and third question a bit by using imaginary guards as H1, H2 and H3 though.

The puzzle can be made a bit more surprising with this curious change:

Let's keep the three guards idea. If instead of being quiet, their heads explode and they die when presented to paradoxical questions (or any other thing that make them unable answer further questions without screwing you happens), you would still be safe, since you have 3 guards and you need 3 answers, regardless of the fact that any of guards is or isn't a truthteller, a liar or random.

What if the guards are allowed to answer "I don't know"?

If the guards are allowed to say "I don't know", "it is impossible to answer that right now" or something like that in their own language instead of just remaining silent, you would need 4 questions instead of 3. This happens because our assumption that they will remain silent (or do any other thing other than screwing you) that is clearly identifiable as "can't be answered" will break down and you would need to hear two different sentences with known meanings in order to separate the cases.

What if the random guard really just answers anything randomly instead of randomly choosing to either tell the truth or lie?

In this case, he will answer "yes" or "no" even if presented with a paradoxical question. His answers would be utterly useless gibberish, so the only solution would be to eliminate him as soon as possible. This can be done by using a paradoxical unanswerable question like "If I ask the liar guard if he is lying, what would he say?" and if he answers anything, you should direct further question to some other guard. Also, note that you would need to spend a question just to eliminate this guy, so the total question count should consider this.

What if the random guard decides if he will tell the truth or lie after he finds out the answer, instead of before?

In this case, if you ask something that has no paradoxes, his answer will be gibberish. But if it has paradoxes, he will remain silent. With this, you can't rely on anything that he says other than the fact that he either said something or nothing at all. To get him out of the way, a question that is either paradoxical just for him or just not for him would be needed. One possible question would be "if I ask the other two guards if blue and red are different colors, would they both answer the same thing?" – the random guard knows that the answer is "no" and he will then answer something. The other guards have no way to know the answer and will remain silent.

  • $\begingroup$ −1 just for saying, “This is very easy.” …  … But seriously, $\endgroup$ Mar 13, 2017 at 8:20
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    $\begingroup$ Liars always lie, so a liar will have no trouble answering the question “Are you lying?” with the answer “No.” Similarly, if you address a truth-teller and ask, “If I ask the liar guard whether he is lying, what would he say?”, the truth-teller will have no trouble answering, “No,” and likewise the liar will easily answer that question in the affirmative.  (Although it’s debatable whether “what would he say?” is a legitimate yes-or-no question.) So either your logic is flawed or your explanation/presentation is unclear. $\endgroup$ Mar 13, 2017 at 8:39
  • $\begingroup$ @PeregrineRook Answer edited. Is it better now? $\endgroup$ Mar 13, 2017 at 16:03
  • $\begingroup$ (1) Thanks for trying, but, no, I believe that you haven’t really improved it. I believe that you’ve made only cosmetic changes, and we still have a fundamental misunderstanding about how the liar operates. You might want to read No honor amongst liars. (2) I appreciate the fact that you use G₁, G₂, and G₃ as actual names/labels for the three guards (you know which is which, although it’s arbitrary) versus H₁, H₂, and H₃, which are hypothetical formal names … (Cont’d) $\endgroup$ Mar 14, 2017 at 1:05
  • $\begingroup$ (Cont’d) …  — you know that all three exist, but you don’t know which is which. But I believe that your answer could be improved by adding that explanation. (3) I’m sorry; I should have caught this yesterday. Your second and third questions are three-level questions, and, upon further examination, I feel that the discussion of what level you’re talking about is lacking. $\endgroup$ Mar 14, 2017 at 1:05

I'm fairly certain this is not possible.

Even assuming we know what the two responses are (not what they mean), it takes two questions just to find one person that is definitely true or definitely false.
In any case, $3$ possible objects $\times 3$ possible monster targets amounts to determining which of 9 possible pairings is the correct one. That's not possible in 3 questions even if we had a single truthful person to ask who would answer with literal "yes" and "no", let alone in the scenario here where we must use up questions to find a non-random answerer.


I see from the TedEd video that this is just a butchered version of the question there, which is in turn a mildly broken version of the "Hardest Question" problem. The video includes the idea of supplying each overlord (here, "monster") with the right artifact (here, "object"), but glosses over the necessary prerequisite that you know which artifact belongs to which overlord (though not which overlord is which). Thus, identifying the overlords is the only requirement to completing the task. In the video's formulation of the problem, as in the original problem itself, the two possible answers are known in advance, but their meanings are not known.

Here, the problem neglects to state that either of these prerequisite pieces of information are known. If we can assume the prerequisites are provided, then the solution follows the well-known method of solving the problem in its original statement, which approaches a "trivia' exercise more than a logic/liars one.

Moreover, I read the requirement of "give the right one to the right monster" as having to determine which of the 3 objects had to be given, and determining to whom it had to be given, a very different thing than "give each monster the object corresponding to its identity". This puzzle needs some rewriting to make it clear what is being asked, and what is provided; as currently stated it is unsolvable without making assumptions unjustified by the puzzle's statement.

  • 2
    $\begingroup$ We only have 6 possible ways to assign 3 objects to 3 different monsters and $2^{2} < log_2{6} \leq 2^{3}$ - so if we had only a truthful person telling clear "yes" or "no" this would be doable with exactly 3 questions. Anyway, I partly agree with you. Having three persons and without knowing what is yes or no, I think that at least 2 more questions would be needed. $\endgroup$ Mar 4, 2017 at 17:19
  • $\begingroup$ Even if we allow "no answer", that still only gives us 5 possible cases, but we need 6 to solve this. $\endgroup$
    – edderiofer
    Mar 4, 2017 at 17:27
  • $\begingroup$ @VictorStafusa Free assignment of 3 objects to 3 monsters would have 6 possibilities, as you''ve said. I was considering a different challenge: with 3 objects A,B,C and three monsters 1,2,3, possible pairings of one object to one monster are: A1,A2,A3,B1,B2,B3,C1,C2,C3; if the challenge is to determine which of those pairings is the single correct one, then there are 9 possible ways, and $log_29 > 2^3$ needs more than 3 questions to find the right one. in any case I think the real issue here is just that we're looking at a clumsy restatement of a flawed version of the original problem. $\endgroup$
    – Rubio
    Mar 4, 2017 at 18:10

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