A twist on the old one liar on truth teller riddle

There is a king. Not a crazy, murderous king like the other riddles, this one is actually kind and wealthy, and as such you would like to marry one of his three daughters. These daughters, while lovely creatures, can only answer yes or no questions. The eldest daughter always tells the truth, the middle daughter always lies. The third daughter is crazy and will answer any question randomly, either yes or no. You crave stability in life. You would like to marry the one who always tells the truth, but you are willing to settle for the one who lies all the time just because at least you will know where you stand. You do NOT want to marry the crazy one under any circumstances.

The big day comes when you make your choice. The king brings out the 3 ladies but you have no idea which one is which. You are allowed to ask ONE of the ladies ONE question, and then make your decision based on the answer. Note that your question must be answerable (ie you can’t ask unknowable questions like “Is there a God?” or logically unanswerable questions like “would the crazy person answer yes to this question?”). What question will you ask to guarantee marrying either the liar or the truth-teller?

No spoiling if you already know this one, please!

We did essentially the same problem very recently.

[spoiler]Any question you want to ask, you can easily reword into such a fashion as that you’ll get a straight answer from anyone except the crazy one; instead of asking “Is P true?”, ask “Is it the case that either you’re the truth-teller and P is true, or you’re the liar and P is false?”. You’ll get the right answer to “Is P true?” from either the truth-teller or the liar, and you’ll get a random answer from the crazy one.

Using that technique, go up to Person A and ask “Is Person B the crazy one?” (i.e., actually ask “Is it the case that either you’re the truth-teller and Person B is the crazy one, or you’re the liar and Person B is not the crazy one?”). If the answer you get is “Yes”, then you know either A or B is crazy, and you can marry C. Otherwise, either A or C is crazy and you can marry B.[/spoiler]

I don’t feel like going through the gymnastics of actually constructing the questions, but first, we need to use one of the standard tricks to get good information out of a liar, common in puzzles of this sort. Then, we need to ask one of the girls which of her sisters I should marry. If the one I ask is the truth-teller, then she’ll answer honestly, and I’ll get the liar (whom I can settle for). If the one I ask is the liar, then I’ll still get good information out of her, and end up marrying the truth-teller. And if the one I ask is the crazy daughter, then I don’t know if I’ll get the liar or the truth teller, but since I’m taking one of the questioned girl’s sisters, I still won’t end up with the crazy one.

I presume that there will be time after the wedding to figure out whether the lass I got is the liar or truth-teller, but that can’t be determined from the single question.

Chronos has it. You don’t need the compound questions to solve this one (and had I thought about it I would have explicitly forbid them anyways since they are cheating :D)

Took me an hour to figure this one out. The correctly phrased question is Assuming you ask person A, “Does person B lie more than person C?” If they answer yes, marry that one else the other one.

XKCD’s take on this riddle.

Oh, that one’s easy. Anyone who follows XKCD knows that you can’t trust the guy in the hat.

You just marry the one with the biggest tits.

This doesn’t seem right for Cafe Society, so I’ll move it over to the Game Room.

twickster, Cafe Society moderator

And the Order of the Stick take the Gordian Knot approach.

Does that work? The answers I’d expect from that question (“which of your sisters should I marry?”):
Truth: Liar
Liar: Crazy
Crazy: Truth or Liar

Crazy’s still an option. Hobo’s question eliminates that.

In this case, you want to reach the native village.
You arrive at a fork in the road, where there are three natives gathered : one truthteller, one liar and one answers at random.

You ask the single question:

Did you know they are serving free beer in the village?

Then you simply follow everyone to the village!

Keep in mind you can only answer yes or no questions, Chronos was using his question as shorthand for mine, but they are essentially the same question.


Well, there’s no use outlawing them, if you’re allowed to ask any questions about the role-assignments at all, since they amount to the same thing. After all, any such question amounts in the end to “Is it the case that the role assignments are like this or this or …?”, where the various "this"es are the role assignments on which the answer is “Yes”. Which is to say, “Is it the case that (you’re a truth-teller and everyone else’s roles are like this or this or …) or (you’re a liar and everyone else’s roles are like this or …) or (you’re crazy and everyone else’s roles are like this or …)?”, just organizing the assignments like so. Phrasing a question as a “compound question” is just making this explicit.

For example, with yours:

[spoiler]“Does B lie more than C?” amounts to “Is it the case that either B is the liar or C is the truth-teller?” which amounts to “Is it the case that either (I’m not the liar and B is) or (I’m not the truth-teller and C is)?” which amounts to “Is it the case that either ((I’m the truth-teller or crazy) and B is the liar) or ((I’m the liar or crazy) and C is the truth-teller)?” which amounts to “Is it the case that either (I’m the truth-teller and B is the liar) or (I’m the liar and C is the truth-teller) or (I’m crazy and either B is the liar or C is the truth-teller)?”.

And, of course, the final clause about what the right answer is when A is crazy can be modified willy-nilly without changing what kind of responses you might get, since, when A is crazy, she’ll just ignore your question and give a random response anyway. So getting rid of it for simplification, you end up with “Is it the case that either (I’m the truth-teller and B is the liar (and thus C is the crazy one)) or (I’m the liar and C is the truth-teller (and thus B is the crazy one))?”. Which [swapping B for C] was my question. [In fact, every question that works is isomorphic to my question in this sense][/spoiler]

Right, the actual question would have been something like “If I were to ask you if sister B was consistent, what would you say?”.

HoboStew’s question is a bit more elegant, but it runs the risk that they’re not all equally talkative: The crazy lady might lie more than the actual liar, if she talks twice as much.

What if the liar is wearing falsies?

I do not have any new contribution to offer to the logical puzzle, which seems to me to have been handled admirably. I would like to commend to your attention the quoted sentence of the puzzle, which seem to me to be the best advice of any kind that I have ever seen.

But often hard to put into practice. Those who always tell the truth or always lie typically are thin on the ground, whereas crazy seems to be eternally popular.

Hobostew’s answer does not work.

The crazy daughter answers randomly, therefore you cannot be sure that she has not lied more than the liar. For that matter, you cannot be sure that she hasn’t been consistent so far in her life by pure coincidence. Maybe she rarely speaks, which makes sense given her limited vocabulary.

Further, it is questionable to assume the daughters know that they fall into the three categories, or who falls into which. The king knows, but maybe they don’t.

The XKCD forum has a working answer, which instead relies on everyone being aware of the relative ages of the daughters.

I would prefer an answer that does not rely on that though. What if absolutely all you know is that one daughter always lies, one always tells the truth, and one is crazy? Nothing about ages, nothing about the daughters knowing who is what.

Just rephrase Hobostew’s answer then; reword the binary relation “x lies more than y” as “either x is the liar or y is the truth-teller”, for example. As I said, this is the unique solution, up to isomorphism.

I can’t see how this would be possible (see below). But perhaps I misunderstand what you are stating. What’s the answer the xkcd forum gives?

Then you can’t do it. You put your one question to one of the daughters (let’s call her A) and she gives you information which is supposed to guarantee you the ability to figure out a non-crazy one. No matter what she says, it would be compatible with the possibility that she herself is crazy and just answering randomly; thus, since you cannot hope to rule her craziness out, one of her answers has to indicate that B is non-crazy and the other has to indicate that C is non-crazy. But if she lacks any information about which is which, she cannot modify her behavior accordingly; e.g., she would have to give the same answer whether she was True and B was crazy or she was True and C was crazy.