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Is it possible to solve this problem:

A prince wish to marry a princess. There are 3 princesses, one is young, one is a little older and one is old. The prince is able to tell the princesses apart. One of the princesses always tells the truth, one never tells the truth and one sometimes tells the truth and sometimes not.

The prince only wish to marry a princess whom he can trust. Therefore it must be the princess that always tells the truth or the princess that never tells the truth (he can just negate her answers for the rest of their marriage).

Before he chooses the princess he wish to marry, he can ask one and only one princess a single question. She must only answer the question by yes or no.

Which question must he ask to be sure he marries one of the right princesses?

Edit: I was not expecting the question "who is more truthful", so consider this change of rules. Suppose we remove the "random princess", and instead insert an "evil princess". The evil princess can choose her strategy for answering, after she has seen which princess we are asking. So asking "Who is more truthful", does not make sense anymore, since the evil princess could choose to answer correct to every question.

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    $\begingroup$ Hehe, "he can just negate her answers for the rest of their marriage". $\endgroup$
    – George Lowther
    Mar 27, 2011 at 21:44
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    $\begingroup$ Because in every logic puzzle I've ever heard, a "yes or no" question can only be answered with "yes" or "no" (tertium non datur), not with "I can't answer". $\endgroup$
    – user139000
    Feb 2, 2015 at 14:39
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    $\begingroup$ @GregMartin, I reversed the genders to address your concern. Given the lack of historical commoditization of male progeny, this should address it, and doesn't harm the problem in any way. $\endgroup$
    – Jaydles
    Feb 3, 2015 at 18:30
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    $\begingroup$ Users wishing to constructively discuss the original wording of this question and @Jaydles's edit are welcome to use this meta thread. Further discussion on this matter will be off-topic here. $\endgroup$
    – user642796
    Feb 4, 2015 at 9:01
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    $\begingroup$ @user 170039: please remember that this site is not intended for simply posting mathematical puzzles - it is not a place for "sharing" as the first sentence of the post says. $\endgroup$
    – Carl Mummert
    Feb 4, 2015 at 17:59

9 Answers 9

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Pick a random prince, then ask him:

Is he [pointing to one of his brothers] older than him [pointing to the second one]?

If you receive "Yes" as an answer, choose the second brother, otherwise choose the first. This ensures you will not marry the middle one.

Explanation:

If the prince you asked is the eldest, he will answer truthfully, and you will pick the youngest.

If the prince you asked is the youngest, he will lie, and you will pick the eldest.

If the prince you asked is the middle one, he will answer randomly, and you will pick one of the two others, as desired.

Edit:

More information about the many variations of this puzzle can be found by searching for the term "Knights and Knaves", which is the name logician and author Raymond Smullyan used to describe these problems.

Of particular interest is a problem known as The Hardest Logic Puzzle Ever, which has the same basic setup as the "princes(ses)" puzzle – lier, truther and random answerer – but with more information needing to be extracted and the additional complication that instead of "Yes" and "No", equivalent words in some esoteric language are used and it is unknown which word means which.

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    $\begingroup$ To optimise this solution, note that you never marry the princess to whom you ask the question. Therefore, ask the question of the princess you find the least attractive, rather than picking randomly. $\endgroup$
    – abligh
    Feb 2, 2015 at 23:02
  • $\begingroup$ @sidgate: You're actually quite correct with regards to at least you know where you stand with them stated in the question. $\endgroup$
    – barak manos
    Feb 4, 2015 at 8:19
  • $\begingroup$ oops.. where did my comment go? I had earlier commented that we will still not be able to guess whether wife is telling a lie or truth, as we cannot tell if married to eldest or youngest $\endgroup$
    – sidgate
    Feb 4, 2015 at 16:24
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    $\begingroup$ Surely this belongs on puzzling.stackexchange.com, recreational or not? $\endgroup$
    – AncientSwordRage
    Feb 4, 2015 at 16:40
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    $\begingroup$ This answer was already provided in 2011: math.stackexchange.com/a/29380/630 $\endgroup$
    – Carl Mummert
    Feb 5, 2015 at 0:30
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Ask the middle one if the youngest is more truthful than the eldest. If the answer is "no" then marry the youngest, otherwise marry the eldest.

That way, if the middle one is the liar, he is guaranteed to marry the most truthful of the remaining two, which is the 100% truthful princess. If the middle one is the truthful one, then he is guaranteed to marry the least truthful of the remaining two, who is the habitual liar. If the middle one sometimes tells the truth and sometimes lies then it doesn't matter which of the other two he marries (for the purposes of this riddle...).

This is just the answer I gave in the comment above, removing the mistake pointed out by Theo Buehler (and the implicit ageism in assuming the prince would rather marry one of the younger two).

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Label the brothers arbitrarily as $A$, $B$, and $C$ (as mathematical parents would). We can describe any strategy we might adopt as three pieces of information:

  • A question to be addressed to $A$. (Since the labeling were arbitrary, we assume $A$)
  • The brother we will marry if we receive "Yes" as an answer
  • The brother we will marry if we receive "No" as an answer.

You cannot marry the brother you ask the question to.

To show this, suppose that we marry $A$ if he responds "yes". It is possible that $A$ is the middle brother. Similarly if we marry $A$ for a response of "no" - it is, in fact, impossible for us to determine whether the person we ask the question to is the middle brother, since if it were, we could receive any answer - and in particular, we could receive the same answer as if he were not

Thus, we can assume, without loss of generality, that we will marry brother $B$ if we get a "yes" and brother $C$ if we get a "no" (since we should obviously not have chosen the brother we marry before hearing an answer!). Moreover, this means that we can assume that $A$ is not the middle brother, since if he is, the strategy works regardless of the question.

There are a number of questions that will let us determine, assuming $A$ is the youngest or the oldest, whether $C$ is the middle brother. For instance

Is $C$ older than $B$?

functions, since if we get the youngest and a "yes", we know $C$ is the middle brother and we should marry $B$ - and we can work out the other cases similarly. In particular, any question with the following truth table functions: $$\begin{array}[ccc]. &&\text{$B$ is the middle brother}&&\text{$C$ is the middle brother}\\\text{$A$ is the youngest}&&\text{true} &&\text{false}\\\text{$A$ is the oldest}&&\text{false}&&\text{true}\end{array}$$ To be very blunt we could ask $A$:

Are you the youngest exclusive or is $C$ the middle brother?

Another question, along a different line, would be:

If I asked you if $C$ was the middle brother, what would you say?

Since we essentially force the brother to tell the truth - but this is perhaps further from the spirit of the question.

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    $\begingroup$ This post is inspired by @pew's prior post, which gives the same answer. $\endgroup$
    – Milo Brandt
    Feb 2, 2015 at 21:53
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    $\begingroup$ @Jaydles Everyone knows that $A$, $B$ and $C$ are girl's names! If you ask me, it's downright silly to label brothers that way, as is the case in your edit (as opposed to, say, the more common $\alpha$, $\beta$, and $\gamma$ labeling of boys or $\mathscr a$, $\mathscr b$, and $\mathscr c$ for siblings of arbitrary gender) $\endgroup$
    – Milo Brandt
    Feb 4, 2015 at 0:11
  • $\begingroup$ I agree with Meelo. We should enforce a strict policy of choosing names from the set $\{'\emptyset','\emptyset \emptyset', ... \}$. $\endgroup$
    – copper.hat
    Feb 4, 2015 at 7:06
  • $\begingroup$ This is actually the better answer, because instead of pulling a rabbit out of a hat, it explains how to tackle these problems (there are many variations such as The Hardest Logic Puzzle Ever) in general. I love the question "Are you the youngest XOR is C the middle one?" $\endgroup$
    – user139000
    Feb 4, 2015 at 7:32
  • $\begingroup$ @pew: Nice link. $\endgroup$
    – user170039
    Feb 4, 2015 at 13:09
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Ask one of the siblings:

Are you someone who both has a younger sibling and is currently being as dishonest as you ever are?

The truth-teller has a younger sibling and is never dishonest, and so will honestly answer "yes."

The liar has no younger sibling, and so will lie and answer "yes."

The mischievous one either a) is telling the truth, and thus is more honest than sometimes, and so will truthfully answer "no," or b) is lying, therefore currently maximally dishonest, and so will falsely answer "no."

Note however that this answer relies on the mischievous one being logically mischievous (that is, arbitrarily choosing to be truthful or not, and then answering based on that choice), rather than being mischievous in the sense of answering with the intent to confuse you. In that sense the other answer (which works however the mischievous one decides what to answer) is superior.

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    $\begingroup$ This approach makes sure you marry the eligible princess you find more attractive, so +1 from me. $\endgroup$
    – Dennis
    Feb 3, 2015 at 1:48
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    $\begingroup$ Technically speaking, it is partially cheating by asking two conditions..sort of two questions. But still +1 for creativity :-) $\endgroup$
    – Michal
    Feb 3, 2015 at 14:28
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    $\begingroup$ @Michal I think it is valid to ask a conjunction in these sorts of questions, since you still only get one bit of information back. It would be cheating if you managed to get an answer and follow it up with something else, or somehow otherwise forced the answerer to give more than just "yes" or "no" - one's own question is allowed to be as complicated as desired. $\endgroup$
    – Mario Carneiro
    Feb 4, 2015 at 1:58
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    $\begingroup$ @Michal The answer can be summed up as a single yes or no, so it's one question. The answerer doesn't have to say "yes to the second but no to the first". $\endgroup$
    – Jack M
    Feb 4, 2015 at 21:00
  • $\begingroup$ @MarioCarneiro Of course you're both right about the fact, that the question is valid in the sense that only one output is produced. I only wanted to say that I see two separate questions creatively tied together - both of them could be asked separately, but no dispute about the fact that it is correct - it's the && that makes the question work. Really, it was just a comment, not critique :-) $\endgroup$
    – Michal
    Feb 5, 2015 at 9:29
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"Will I marry the middle brother?"

Because the liar and the truther don't know the answer, they can't answer without having a chance of betraying their nature.

The middle brother will be the only one able to answer.

Marry either brother that doesn't answer the question.

Edit: Due to the recent edit, this answer is no longer valid to the question. It depends on the fact that you can force the princesses to be unable to answer.

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  • $\begingroup$ Why the -1? It works. $\endgroup$
    – Axoren
    Feb 2, 2015 at 14:33
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    $\begingroup$ @Joffan, If the sister I asked can't answer: marry her. If she can: marry any other sister. $\endgroup$
    – Axoren
    Feb 2, 2015 at 14:48
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    $\begingroup$ Well I like your answer then :-) $\endgroup$
    – Joffan
    Feb 2, 2015 at 14:49
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    $\begingroup$ Do you mean "If I married you, will I be marrying the middle aged sister?" Currently your question doesn't appear to be directed at someone, and is a question of an outcome based on a decision that is not within their ability, rather than on an outcome for a known choice. $\endgroup$
    – user142198
    Feb 2, 2015 at 14:58
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    $\begingroup$ Note that the question doesn't say the middle sister will choose between saying "yes" or "no". It says she will choose between telling the truth and lying. So if she cannot tell the truth (for the same reason the eldest sister cannot tell the truth -- she doesn't know the answer) and she cannot lie (for the same reason the youngest sister cannot lie -- she doesn't know the answer) then I don't see how the problem as stated permits her to answer. $\endgroup$
    – Steve Jessop
    Feb 2, 2015 at 18:46
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There are often brute force solutions to these kinds of questions that can be constructed by simply throwing enough logical disjunctions and conjunctions at the problem.

Let $M$ denote the proposition "You are the middle brother", and $L$ denote the proposition "You are lying". Then ask any brother:

$(M\wedge\neg L)\vee(\neg M\wedge L)$ ?

Assume $M$ is true. Then the above simplifies to $\neg L$ - "Are you telling the truth?" - to which the answer is always Yes.

Assume $M$ is false. Then the above simplifies to $L$ - "Are you lying?" - to which the answer is always No.

Thus the answer will be Yes iff you asked the middle brother (Note that the middle brother doesn't randomly say yes or no, he randomly answers truthfully or falsely, otherwise this wouldn't work).

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Ask the youngest princess, "If I were to ask you if the middle princess is the evil one would you say yes?" This formulation causes the false one into a double negative thus if the middle princess is evil both true and false will answer yes, otherwise both true and false will answer no. Thus if the youngest princess answers yes to your question either she is evil and trying to trick you or the middle princess is evil, in any case you can be sure that the oldest princess is not evil and thus marry her. If the youngest princess answers no to your question than whether or not she is evil the middle princess is definitely not evil thus you marry the middle one.

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You can't pick the one you ask the question to, as there is no yes/no question you can ask that will let you determine that the asker is not random.

From this, there are 3 cases to consider:

  1. If the one you ask is random, it doesn't matter who you pick
  2. If the one you ask is true, you want to pick the false one
  3. If the one you ask is false, you want to pick the true one

We only need to worry about cases 2 and 3, since case 1 solves itself. Thus, the question boils down to these 4 scenarios:

  1. You get a true answer about false
  2. You get a true answer about random
  3. You get a false answer about true
  4. You get a false answer about random

From these cases, let's constrain it so that you'll get a "yes" for 1 and 3, and a "no" for 2 and 4. That way, you can pick the one they say "yes" about, or pick the other if they say "no". There are many possibilities, but here's one such question:

If you are true is that brother false, otherwise is he random?
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Let's use the term "consistent" for a person who either always speaks the truth or always lies, and let's assume that the princesses themselves use and recognize this term with the same meaning.

Given that exactly two of the three princesses are consistent, we are to find one of them, using one question.

The simplest question that I've been able to come up with is to ask one princess (call her $\;A\;$)

"Can you say that she [point to another princess, $\;B\;$, here] answers consistently?"

If the answer is "yes" then $\;B\;$ is consistent, otherwise the third princess, $\;C\;$, is.

Why does this work? Well, for any consistent person, if you ask "Can you say that $\;P\;$?", the answer "yes" means that $\;P\;$ is true, and "no" means that $\;P\;$ is false.

So if $\;A\;$ is consistent, then we can choose $\;P\;$ as "$\;B\;$ is consistent", and know whether or not $\;B\;$ is consistent, and if not then $\;C\;$ is (because there are two consistent princesses).

And if $\;A\;$ is not consistent, then both $\;B\;$ and $\;C\;$ are consistent (again, because there are two consistent princesses).

For a more formal argument, see my answer to another math.se question, up until the section "The first question: determinism".

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