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You are a prisoner.
Posted: Posted June 19th by Xhin
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You are a prisoner in a room with 2 doors and 3 guards.

One of the doors will guide you to freedom and behind the other is a hangman - you don't know which is which.

  • One of the guards always tells the truth.
  • One always lies.
  • One will always give a random answer.

    You don't know which one is which.

    You have to choose and open one of these doors, and you can ask 2 two yes-or-no questions in total. Both of these can be addressed to the same guard or to two different ones.

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    I ask them:

    Am I a male?

    One will say yes, one will say no, and one will say a random answer.

    Posted June 19th by S.O.H.
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    S.O.H.
     

    Isn't this just an inverted Monty Hall problem?

    I ask them:


    Read carefully:

    Both of these can be addressed to the same guard or to two different ones.


    Edited June 19th by Cruinn-Annuin
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    I will just ask two of the 3.



    Posted June 19th by S.O.H.
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    S.O.H.
     

    If you ask the truth-teller and the random rando, they might both say yes. Now you've used up both of your questions without knowing anything other than who the liar is. If rando says no, then all you know is who the truth-teller is. You still know nothing about the doors and you have no questions left.

    Posted June 19th by Xhin
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    Xhin
    Sky's the limit

    If you ask a guard if both of the other guards would say that Door 1 is freedom, both the truth-teller and the liar will not be able to answer, as they don't know what the random answerer is going to say. Correct?

    Posted June 19th by Cruinn-Annuin
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    The questions have to have yes/no answers -- you can't ask recursive questions like "is your answer going to be yes" or ask the guards to predict the future ("Will my next answer be yes?") or inherent uncertainty ("Is randy the rando going to tell me the truth?")

    Posted June 19th by Xhin
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    Xhin
    Sky's the limit

    I'd select one of the guards and have him in my mind as I ask one of the other two something simple: "Am I a male?" Based on that guard's response, I'd either ask him or the third guard the following question: "Which door leads me to freedom?"

    To explain further, let's say I decided to choose Guard #1. I'd then ask Guard #3 the simple question: "Am I a male?" If he answers yes, I know then that I'm talking to the guy telling the truth. If he answers with anything else, I'd then ask Guard #2 which door leads me to freedom.

    Of course, this solution doesn't outright guarantee you freedom; even the original Monty Hall problem wasn't a 100% chance of success, as the MythBusters found when they tested it. It does, however, guarantee you the best chance of success.

    Edited June 19th by Black Yoshi
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    is fighting the guards an option'?

    Posted June 19th by S.O.H.
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    S.O.H.
     

    This reminds me of that "three lightbulbs" puzzle

    Posted June 19th by Fox Forever
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    I'd then ask Guard #3 the simple question: "Am I a male?" If he answers yes, I know then that I'm talking to the guy telling the truth.


    Alternately he could be Randy the rando answering yes. You would then ask guard #3 "is door #1 freedom", Randy would give the random answer "yes" and you'd lose your head.

    Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    behind the other is a hangman


    and you'd lose your head.


    Wait, wait, is it a hangman or an axeman? This is very important.

    Posted June 20th by Cruinn-Annuin
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    its simple ask the same question to the same guard twice.


    The Liar will lie

    and the Randomer will say something random

    Posted June 20th by S.O.H.
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    S.O.H.
     

    its simple ask the same question to the same guard twice.


    Um...the truth-teller and the liar will both say the same thing twice and the random answerer will say the same thing 50% of the time.

    Posted June 20th by Cruinn-Annuin
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    Can't beat that.

    Posted June 20th by S.o.h
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    S.o.h
     

    Alternately he could be Randy the rando answering yes. You would then ask guard #3 "is door #1 freedom", Randy would give the random answer "yes" and you'd lose your head.


    Oh damn, I didn't even think of that. I was more thinking that Mr. Random would answer something like "Maybe" or, in general, something besides yes or no.

    Posted June 20th by Black Yoshi
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    its simple ask the same question to the same guard twice.


    The Liar will lie

    and the Randomer will say something random


    Even assuming that's how randomness works (it doesn't), you'd be out of questions and not know anything about the doors.

    I was more thinking that Mr. Random would answer something like "Maybe" or, in general, something besides yes or no.


    The random guy always answers yes/no like the other two guards, but whether he answers yes or no doesn't depend on anything -- sometimes he's lying, sometimes he's telling the truth, sometimes he just says yes or no for no reason.

    Wait, wait, is it a hangman or an axeman? This is very important.


    Maybe it's a hangman with piano wire instead of rope.

    Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    By the way, there is an actual solution here -- and it doesn't require a doctorate in math like the 200 people with blue/brown eyes riddle.

    Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    I’d choose 1 guard ask them them same question twice while pointing at the same door. “Am I going to die if I enter that door? Point at same door “am I going to live if I enter that door? if the answer is appropriate I would assume I have chosen the truthful guard and select the door I pointed to. If the answers differ, I’d assume I chose the lying guard or random guard and choose the door I didn’t point at.

    Edited June 20th by Brandy
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    Ask 2 out of 3 guards if the door to freedom is on the right or left side. I would determine which door to choose percentage wise

    Posted June 20th by Lead by Example
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    Wait can we address a single question to multiple guards? I thought it has to be individually

    Posted June 20th by Brandy
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    @Lead by Example: You're restricted to yes or no questions, unfortunately. Otherwise I'd just as soon ask the guards what gender I am rather than "Am I male?"

    The random guy always answers yes/no like the other two guards, but whether he answers yes or no doesn't depend on anything -- sometimes he's lying, sometimes he's telling the truth, sometimes he just says yes or no for no reason.


    Ah, that makes sense.

    Alternate solution: ask two of the guards in turn, "Would [other guard] tell me that [point at a door] leads to freedom?" So, for example, I'd ask Guard #1, "Would Guard #2 tell me that the door to my left leads to freedom?" To Guard #2 I'd ask the same question, only replacing Guard #2 with #3 in the question (making sure to mention the same door both times). If both guards give the same answer, one of them for sure is Mr. Random; if they give different answers, you're kind of back at square one, come to think of it, and you still don't know for sure which door will lead you to safety.

    A question: We're assuming that we and the guards are all in the room, or, at least, that we're all within earshot, so what's to stop us from asking all three guards the same question at once?

    Posted June 20th by Black Yoshi
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    I think if you are aware of the fact that two of the 3 guards are unreliable. You can just ask a single guard the same question about the same door and if the answers differ choose the opposite door

    Edited June 20th by Brandy
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    I think if you are aware of the fact that two of the 3 guards are unreliable.


    Only one of the three guards are unreliable: the one whose answers are not based on anything.

    The liar will always say the opposite of what is true and thus is reliable.

    Posted June 20th by Cruinn-Annuin
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    I’d choose 1 guard ask them them same question twice while pointing at the same door. “Am I going to die if I enter that door? Point at same door “am I going to live if I enter that door? if the answer is appropriate I would assume I have chosen the truthful guard and select the door I pointed to. If the answers differ, I’d assume I chose the lying guard or random guard and choose the door I didn’t point at.


    Let's say Door #1 is the hangman.

    You ask guard #1 "Am I going to die if I enter door #1?" and "Am I going to live if I enter door #1?". There are five possibilities:

  • Guard #1 is the truth-teller. They say yes twice.

  • Guard #1 is the liar. They say no twice.

  • Guard #1 is random. They say no and then yes or yes and then no.

    If you knew who was the truth-teller, the above would be enough to tell you what to do -- however you don't. And even if you did, you also have these two possibilities:

  • Guard #1 is random. They say yes twice.

  • Guard #1 is random. They say no twice.

    In either of these cases, the random guard is indistinguishable from the lying guard or the truth-telling guard.


  • Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    whats the correct way of not dying?

    Posted June 20th by S.O.H.
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    S.O.H.
     

    Alternate solution: ask two of the guards in turn, "Would [other guard] tell me that [point at a door] leads to freedom?" So, for example, I'd ask Guard #1, "Would Guard #2 tell me that the door to my left leads to freedom?" To Guard #2 I'd ask the same question, only replacing Guard #2 with #3 in the question (making sure to mention the same door both times). If both guards give the same answer, one of them for sure is Mr. Random; if they give different answers, you're kind of back at square one, come to think of it, and you still don't know for sure which door will lead you to safety.


    I like the way you're thinking. You're on the right track at least.

    However you can't ask these questions because if either guard #2 or guard #3 is random, then guard #1/#2 wouldn't be able to predict what they would say. Put another way, if you die if you ask a question without a yes/no answer, then with this strategy you die 2/3 of the time, which is worse than picking a random door. If you had only the liar and truth-teller to work with, though, this would be a pretty useful strategy:

    You ask guard #1 if guard #2 would say door 1 leads to freedom, and then ask guard #2 the same thing about guard #1:

    (( Assuming Door 1 is the hangman ))

  • Truth, Liar -- Yes, Yes

  • Liar, Truth -- Yes, Yes

    (( Assuming Door 1 is sweet sweet freedom ))

  • Truth, Liar -- No, No

  • Liar, Truth -- No, No

    From this, you can infer that, if you ask this complex question with door #1, then if you get yes twice then you enter door #2 and if you get no twice you enter door #1.

    Congratulations, you've solved the 2 doors, 2 guards problem! Definitely on the right track.

    However, you still have the pesky random guard to deal with. I suggest trying to eliminate him first.

  • Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    Wait can we address a single question to multiple guards? I thought it has to be individually


    Yeah it does. However you can ask the same question to two different guards individually.

    A question: We're assuming that we and the guards are all in the room, or, at least, that we're all within earshot, so what's to stop us from asking all three guards the same question at once?


    You have to ask them all separately. That's in the starting rules.

    You can just ask a single guard the same question about the same door and if the answers differ choose the opposite door


    Both the liar and truth-teller would give the same answer twice. The random guy has a 1/2 change of giving the same answer twice as well. You only have a 1/6 chance of getting different answers, and even if that happens, you still have no idea which door is correct.

    The liar will always say the opposite of what is true and thus is reliable.


    Very true. If you know he's the liar, he's basically a truth-teller, just in reverse.


    Posted June 20th by Xhin
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    Xhin
    Sky's the limit

    whats the correct way of not dying?

    Continuing to live, duh!

    This puzzle's really rattling my brain. I know about the 2 doors 1 question thing but this is really throwing me for a loop of how I'm to identify the random man and how to ask the right question to the right individuals (the other two). If it was 3 questions it could be easy but there's only two...

    Posted June 20th by Fox Forever
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    Trick question. The only freedom we truly have is death.

    Posted June 21st by Jet Presto
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    If it was 3 questions it could be easy but there's only two...


    How do you find out who the random man is with three questions?

    Posted June 21st by Xhin
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    Xhin
    Sky's the limit

    I give up there’s no way to determine the random guards identity your riddle is unsolvable and I consider myself good at riddles

    Edited June 21st by Brandy
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    your riddle is unsolvable


    It isn't though.

    Posted June 21st by Xhin
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    Xhin
    Sky's the limit

    If you guys want a hint, albeit not a very good one:



    Posted June 21st by Xhin
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    Xhin
    Sky's the limit

    there is no yes or no question you can ask that can eliminate the possibility of the random guard from the equation

    Posted June 21st by Brandy
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    Can you just pm me answer i want to know

    Posted June 21st by Brandy
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    Ask a “would you say this door” (indicate door) “leads to freedom?”
    Then ask another guard “would you say this door” (indicate same door) “leads to freedom?”
    If they both say “yes” that’s the right door.
    If they both say “no” the other door is the right door.

    If one says “yes” but the other says “no”, one of the guards you asked was the random guard; you don’t know which.
    You also don’t know which door is right.
    If you could ask the third guard, you’d find out.
    But as it is you just have to guess.

    Your odds are as follows.
    One chance in three the guards you ask are the truther and the liar. If so, you escape.
    Two chances in three one of the guards you asked is the randomer.

    If one of the guards you ask is the randomer, then half of those times both guards will give the same answer, and you’ll escape.

    So, in total so far, two-thirds of the time you’ll escape with knowledge.

    The other third of the time you’ll just be guessing. But you could as easily guess right as guess wrong.

    So this strategy lets you escape five times out of six.
    But you get executed one time out of six.

    —————

    So far I haven’t been able to do better.

    —————

    The “would you say...” question is recursive.
    If the answer is “yes” the truther would say “yes”, so his answer to “would you say ..l” is the truthful “yes”.
    If the answer is “yes” the liar would say “no”, so his answer to “would you say ... ‘ is the lie “yes”.
    If the answer is “no” the truther would say “no”, so his answer to “would you say ...” is the truth, “no”.
    If the answe is “no” the liar would say “yes”, so his answer to “would you say ...” is the lie, “no”.

    —————

    I don’t know anything about Monty Hall, and I haven’t read the 31 answers other people gave.
    So if I’m duplicating what everyone else already knows —— Sorry about that!

    ———

    Right now I don’t know of a way to solve the problem with just two questions; nor with only non-recursive questions.



    Posted June 21st by chiarizio
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    The Monty Hall paradox is based on the original version of "Let's Make a Deal!" hosted by Monty Hall. The game went pretty much as follows:

    The contestant would be presented three doors, and would be asked to choose one. Behind one of the doors was a prize of some sort; behind the other two was the "Zonk!", the game's way of saying you lost. Once the contestant had made his choice, Monty would reveal one of the "Zonks" and ask the contestant whether or not he wanted to switch doors. More often than not, contestants would choose to stick with their choice, and more often than not, they would lose.

    This is where the paradox lies: in fact, it was actually far, far better for the contestant to switch doors rather than stick with their choice. To the contestant, once one of the "Zonks" was removed from play, it looked like a 50/50 shot at the grand prize; however, that's not actually the case. When the contestant begins, there is a one in three chance that any individual door will house the grand prize, and a two in three chance that it will not. And so, there is a one in three chance that the door they select houses the prize, and a two in three chance that it doesn't. Even when one of the "Zonks" is removed, those odds don't actually change: the contestant still has only a one in three chance of winning by sticking to their original choice, and a two in three chance of winning by switching doors. Thus, by switching doors, the contestant doubles his chances of winning.

    More information here:
    https://en.wikipedia.org/wiki/Monty_Hall_problem

    Edited June 21st by Black Yoshi
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    That said, I wonder if the solution is literally as simple as asking one of the guards to open both doors...

    Posted June 21st by Black Yoshi
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    The Monty Hall paradox is based on the original version of "Let's Make a Deal!" hosted by Monty Hall. The game went pretty much as follows:




    The contestant would be presented three doors, and would be asked to choose one. Behind one of the doors was a prize of some sort; behind the other two was the "Zonk!", the game's way of saying you lost. Once the contestant had made his choice, Monty would reveal one of the "Zonks" and ask the contestant whether or not he wanted to switch doors. More often than not, contestants would choose to stick with their choice, and more often than not, they would lose.




    This is where the paradox lies: in fact, it was actually far, far better for the contestant to switch doors rather than stick with their choice. To the contestant, once one of the "Zonks" was removed from play, it looked like a 50/50 shot at the grand prize; however, that's not actually the case. When the contestant begins, there is a one in three chance that any individual door will house the grand prize, and a two in three chance that it will not. And so, there is a one in three chance that the door they select houses the prize, and a two in three chance that it doesn't. Even when one of the "Zonks" is removed, those odds don't actually change: the contestant still has only a one in three chance of winning by sticking to their original choice, and a two in three chance of winning by switching doors. Thus, by switching doors, the contestant doubles his chances of winning.




    More information here:




    https://en.wikipedia.org/wiki/Monty_Hall_problem




    Oh! I knew that; but I didn’t know it had anything to do with Let’s Make a Deal or Monty Hall!

    Posted June 21st by chiarizio
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    How do you find out who the random man is with three questions?


    You can easily find out which door is the right door with only three questions.
    But you have only a 50%-50% shot at finding out which guard is random with only three questions.
    Ask each of the three the “would you say “ question.
    If two of them give the same answer that tells you which door is right.
    If all three give the same answer you don’t know which guard is random.
    If one guard answers differently he is the random one.
    There’s no way to tell which of the other two is the truther and which is the liar, though.

    Posted June 21st by chiarizio
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    @Xhin:

    I can’t find the information to which you refer in the spoiler. It’s not in this thread, is it?



    Posted June 21st by chiarizio
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    Ask a “would you say this door” (indicate door) “leads to freedom?”
    Then ask another guard “would you say this door” (indicate same door) “leads to freedom?”
    If they both say “yes” that’s the right door.


    Let's say door #1 is the hangman. You ask Guard A whether door #1 leads to freedom. Guard A, being the liar, says yes. You then ask Guard B the same question. Guard B, being the random guy, also says yes. Happily, you pull open door 1 and lose your head.

    So, in total so far, two-thirds of the time you’ll escape with knowledge.


    There's a solution where you escape 100% of the time.

    The “would you say...” question is recursive.


    That's another good way to find out the truth when you just have the liar/truth-teller left. However you still haven't found the random guy.

    That said, I wonder if the solution is literally as simple as asking one of the guards to open both doors...


    No, you have to ask two yes/no questions to figure out which of the doors to open yourself.



    Posted June 21st by Xhin
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    Xhin
    Sky's the limit

    @xhin:

    Let's say door #1 is the hangman. You ask Guard A whether door #1 leads to freedom. Guard A, being the liar, says yes. You then ask Guard B the same question. Guard B, being the random guy, also says yes. Happily, you pull open door 1 and lose your head.


    You don’t ask him whether door #1 leads to freedom. You ask him whether he would say door #1 leads to freedom. Since it doesn’t, he would lie and say “yes”. So he has to lie in answering your “would you say” question, and say “no”.

    If guard B is the random guy, and randomly says “yes”, you have one “no” and one “yes”. You have to guess or flip a coin or something. You have a 50%-50% chance of guessing door #2. Whichever guess you make it will probably be with trepidation, rather than happily.

    The strategy violates your “no recursive question” restriction which you imposed when you moved the goalpost some time ago. But even without that restriction it works only five times out of six. About the same odds as Russian roulette.



    Edited June 21st by chiarizio
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    @xhin:

    What if you ask the Truth teller or the False teller an unknown yes or no question?

    Edited June 21st by Fox Forever
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    Do you know what Xhin was referring to in
    this

    spoiler?


    Posted June 21st by chiarizio
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    I believe he meant Black Yoshi and not me. But I think I found out the answer but it all depends on Xhin's answer to my question.

    Posted June 21st by Fox Forever
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    Regardless I'll type it out here and now I guess.

    Go up to a guard and ask "Will the other two guards have the same answer to a question I ask?"

    If the truth teller and false teller give an "I don't know" kind of answer when they answer an unknown question or just stand there because the question is unknown you know they are not the random guy. Ask the random guy the same question and if he definitively says yes or no then he is the random guy. Then use the answer Black Yoshi gave once you can write off the random guy.

    Remember: "The random guy always answers yes/no"

    Posted June 21st by Fox Forever
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    Go up to a guard and ask "Will the other two guards have the same answer to a question I ask?"


    I already tried that:

    If you ask a guard if both of the other guards would say that Door 1 is freedom, both the truth-teller and the liar will not be able to answer, as they don't know what the random answerer is going to say. Correct?


    The questions have to have yes/no answers -- you can't ask recursive questions like "is your answer going to be yes" or ask the guards to predict the future ("Will my next answer be yes?") or inherent uncertainty ("Is randy the rando going to tell me the truth?")


    Posted June 21st by Cruinn-Annuin
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    Ok so if they don't answer then you know they are reliable (i.e. always truth or always a lie). Unless you specifically are not allowed to ask this question.

    Posted June 21st by Fox Forever
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    Ok so if they don't answer then you know they are reliable (i.e. always truth or always a lie). Unless you specifically are not allowed to ask this question.


    you can't ask


    Posted June 21st by Cruinn-Annuin
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    What about that question breaks the conditions?

    Posted June 21st by Fox Forever
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    You can't ask that question because no-one knows what Randy is going to say. With your question, there is the potential that Randy is one of the other two guards and his answer cannot be predicted. You cannot ask a question that has an inherent uncertainty to it.

    Posted June 21st by Cruinn-Annuin
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    Even if I ask it to Randy himself? He'll never have an issue answering an uncertain question because he randomly answers.

    Posted June 21st by Fox Forever
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    Regardless of what the answer is here we have 1 question to determine if someone is the random guy or not. It will likely be a question asking guards about other guards.

    Posted June 21st by Fox Forever
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    I will eagerly await to hear what Xhin's intended solution is because no matter how I look at this I keep coming back to the conclusion that it's impossible.

    The only chance you have to identify which guard is which would be to ask all three a question... Which uses up your two questions on getting them to answer a question you know the answer to (eg, am I human). Even then there is a chance that if you were to ask the random guy twice he could answer yes both times so you'd never know which was the truthful guardian and which was the random.

    Alternatively if you try to use your two questions to identify which door is which you have no idea which guard is which so you're kinda right back at the start. Again you can also end up in a situation where the random guy misleads you by providing the same random answer both times, not that it makes much difference because you still don't know who truth and liar are in order to determine which to believe at all.

    If you try to split your questions between both. Eg, first ask am I human to two guards and then ask about the doors to try and work them out. You still are none the wiser because you don't know if the random person replied to your first question and randomly hit on the right answer so you then don't know if you can trust the answer to your question about the door.

    There's just no way to identify who is who to then work out which one to trust in regards to the doors. The original problem; two guards, two doors (or paths), relies on you being able to ask a question that has a guaranteed answer every time. Adding a random element makes it impossible to complete 100% of the time as far as I can figure out.

    --------------------------------------

    There are several questions you could ask that in a lucky scenario random won't break the results for, but because random exists you can never know with 100% certainty that your question's answers are as they should be.

    Happy to be proven wrong though.

    Posted June 21st by Moonray
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    Can I ask "Are any of these statements true?" and then proceed to give a list of statements?

    Posted June 21st by Fox Forever
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    He'll never have an issue answering an uncertain question


    But you can't ask an uncertain question.

    Posted June 21st by Cruinn-Annuin
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    Inspired by Fox's question it I think I have an answer...



    I apologise for saying it was impossible. I was wrong :(

    Also if this isn't the solution then I'm out as it took me a while to think this out (I even made a flowchart and everything).

    Posted June 21st by Moonray
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    According to the usual rules in this type of problem, the questionees will not answer (ie will ignore) a question unless:
  • it’s a yes/no question
  • they know the answer
  • they don’t think you know the answer.

    In Martin Gardner’s “colorful isles” problems, these rules applied to only the first day after they’d met the foreigner.
    White tribe always told truth (on first day)
    Red tribe always lied (on first day)

    Pink tribe always alternated truth and lie
    Pink tribe had three branches, one of which was Coral, and I don’t remember the other.
    One branch always started with the truth. Another branch always started with a lie. The third branch answered the first question randomly.

    There were also Blue(?) tribes who answered randomly.
    The main Blues answered truth or lies about in a 50%-50% split.
    Sky-blues answered true about 75% of time, false about 25% of time.
    Navy-blues answered false about 75% of time, true about 25% of time.



  • Edited June 23rd by chiarizio
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    Inspired by Fox's question it I think I have an answer...

    Whooh, that's where I might have gone. Reading that made my head hurt a little (maybe if I saw that flowchart, lol). I didn't want to go into all of those different answers just to have Xhin tell me I couldn't ask those questions but they don't seem to break any of the conditions.

    Posted June 21st by Fox Forever
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    Moonray's definitely on the right track. I don't have time atm to review the rest of the post or his specific answer and if it works, will do later.

    Posted June 21st by Xhin
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    Xhin
    Sky's the limit

    Okay, I worked it out because fuck the other project I was working on

    MOONRAY HAS FOUND THE SOLUTION


    Question 1: Is only one of these true:
    1. You (Guard #1) tell the truth.
    2. Guard #2 gives random answers.


  • Truth, Liar, Random -- Yes
  • Truth, Random, Liar -- No
  • Random, Truth, Liar -- Yes/No
  • Random, Liar, Truth -- Yes/No
  • Liar, Random, Truth -- No
  • Liar, Truth, Random -- Yes

    Guard #1
    Yes -- Truth, Random, Liar
    No -- Truth, Random, Liar

    Guard #2
    Yes -- Liar, Truth
    No -- Random, Truth, Liar

    Guard #3
    Yes -- Random, Liar, Truth
    No -- Liar, Truth

    From this you can infer, if you get the answer "Yes", then Guard #2 isn't random, and if you get the answer "No", then guard #3 isn't random. So you just ask the non-random guard the question Black Yoshi came up with.

  • Edited June 21st by Xhin
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    Xhin
    Sky's the limit

    Yea I realised from the start that the goal was to eliminate the random guy because he basically breaks any chance of figuring out which door is which.

    Once Fox gave me the idea of checking if statements were true I spent a good while figuring out what combination of statements would work but every time the liar would break it because I was just asking if "any" were true. That's when I realised if you make it so that only one is allowed to be true you can trick the liar into playing ball.

    Posted June 21st by Moonray
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    From this you can deduce that if you got "No" then the guard you pointed at has to be random if you asked either Truth or Liar.

    Wrong the random could answer correctly on both statements


    Posted June 21st by Brandy
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    There is no solution to this problem that is a 100% guarantee . Or if there is put it in layman’s terms because moonrays solution makes no sense

    Posted June 21st by Brandy
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    Wrong the random could answer correctly on both statements

    You're not asking if BOTH of the statements are correct, you're asking if ONLY ONE of the two is correct. If they're both correct the truth teller says "no" because two are correct and not "only one".

    If you end up asking the random guy first then it literally doesn't matter what he says because you'll ask the other two (who are both completely reliable, i.e. reliable liar or reliable truther).

    Posted June 21st by Fox Forever
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    Wrong the random could answer correctly on both statements


    Right. But for the first question it doesn't matter what Random says. As long as you follow the rules under the assumption you're talking to Truth or Liar then you'll move on to the next step and have eliminated Random from the puzzle.

    I think my solution doesn't make sense because I didn't explain it properly. It's easier to work out if you assign the guards as 1, 2 and 3 and ask the questions based on the numbers.

    On reflection I think got my final question wrong. What you should do is ask the one guard you know isn't random "If I asked your opposite if this is the safe door, would they say yes or no". At which point No is still always the safe door. You can't actually risk asking two because technically the steps before didn't totally eliminate the random guard, they just designated one guard that wasn't a random.

    Edited June 21st by Moonray
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    T = Truth, F = False, R = Random

    Question: Is only one of these statements true:
    1. You tell the truth.
    2. That guard -points at one of the other two- gives random answers.

    Scenario 1:

    You are talking to T and pointing at F

    T would say yes because statement 1 is true but statement 2 is false. The one you are pointing at is a safe option.

    Scenario 2:

    You are talking to T and pointing to R

    T would say no because both statements are true. You are pointing at an unsafe option.

    Scenario 3:

    You are talking to F and pointing to R

    F would say no because statement 1 is false but statement 2 is true, so he lies about it. You are pointing at an unsafe option.

    Scenario 4:

    You are talking to F and pointing to T

    F would say yes because no statements are true but he lies and says "yes". You are pointing at a safe option.

    Scenario 5

    You are talking to R which it doesn't matter what he says because you're going to talk to one of the other ones after who are both reliable.


    Edited June 21st by Fox Forever
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    i would ask the guards for a blowjob

    Posted June 22nd by poptart!
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    If you ask the question to the random guy first you’ve wasted a question. The 2nd question is irrelevant because you cannot deduce the safety of the door in one question unless you get lucky and pick the truth teller. There is no 100% guarantee to this riddle

    Posted June 22nd by Brandy
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    Chin said that after moonrays 1st question you ask any other the guards black Yoshima question which he even said at the end of the sentence isn’t 100 percent reliable

    Posted June 22nd by Brandy
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    The point of moonray's first question is to figure out whether to ask the second or third guard your next question. You want to ideally ask someone who isn't random -- and if your first guard is the liar or the truth-teller you can narrow down which one is safe from that.

    If the first guard is random, then it doesn't matter -- you know both the second and third guard are safe.

    So, if you get a yes, you know to ask guard #2, because the only possibilities are that they're the liar or the truth-teller. If you get a no, you know to ask guard #3 because the only possibilities are that they're the liar or truth-teller. If you get your yes or no from the random guy, then it still doesn't matter because in that case, both guard #2 and #3 are safe.

    Posted June 22nd by Xhin
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    Xhin
    Sky's the limit

    Chin said that after moonrays 1st question you ask any other the guards black Yoshima question which he even said at the end of the sentence isn’t 100 percent reliable


    The first question eliminates the random guard from the equation.

    The second question you get the same answer from both truth and lie so it doesn't matter which you are talking to. They will both say the Safe door is unsafe.

    Posted June 22nd by Moonray
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    Every one here is wrong.

    This is the real solution.

    Pick two of the three Guards, smile and ask them to their face: "Am I locked in here with you?"

    Depending on who you choose:

    One Guard will Lie

    One will give a random answer

    The other will tell the truth

    You then turn to all three of the guards and say: "Wrong, you are all locked in here with me!"

    Before killing the three of them.

    It is science.


    Posted June 22nd by S.O.H.
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    S.O.H.
     

    Verily. S.o.h.'s method now seems to be the only true way to solve this puzzle!



    Posted June 24th by Fox Forever
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    The only problem with SOH's solution is he just killed the people who know which door is safe. Now he only has a 50/50 chance of surviving himself!

    Posted June 24th by Moonray
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    or you can wait until they have to change guards D:


    tbh I am not sure what the real answer is. What if both doors lead to death?

    Posted June 24th by S.O.H.
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    S.O.H.
     

    No one said you had to walk through the first door you opened; you could just open one to see what's behind it and then if it's the gallows, open the other and walk out.

    Posted June 24th by Black Yoshi
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    No one said you had to walk through the first door you opened; you could just open one to see what's behind it and then if it's the gallows, open the other and walk out.


    That is not how doors work!



    Posted June 24th by Moonray
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    A STRANGE GAME.
    THE ONLY WINNING MOVE IS NOT TO PLAY.

    Posted June 24th by GH0ST IN THE MACH1NE
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    GH0ST IN THE MACH1NE
    GH0ST
    IN THE MACH1NE
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