-3
$\begingroup$

1st Guard: Always lies all the time.

2nd Guard: Always tells the truth all the time.

3rd Guard: 75% probability of telling the truth.

4th Guard: 50% probability of telling the truth.

5th Guard: 25% probability of telling the truth.

By asking a single question to one guard of your choosing (although you don't know which is which), find which door has the highest probability of being safe.

The guards that lie, they lie about which door is safe, and they do not lie about other things.

For example, if a lying guard was asked "Is your door safe?", he would lie about this, but if a lying guard was asked, "Are you a liar?", he would admit that he's a liar.

| improve this question | |
$\endgroup$
  • $\begingroup$ Does the question have to have a Yes/No answer? $\endgroup$ – hexomino Sep 17 '16 at 22:21
  • 4
    $\begingroup$ @TheBitByte: hexomino is asking whether the questions we can ask to the guards are those questions which can be answered with a yes or a no (yes-no questions), or can we ask more open questions like "how many doors are safe?" $\endgroup$ – justhalf Sep 19 '16 at 4:04
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – GentlePurpleRain♦ Sep 20 '16 at 15:35
1
$\begingroup$

my try :

If I asked the 100%, 75%, and 50% liars, which doors is unsafe, which door would they point to ?

if the Truthful guard responds:

He will point to the safe door, because he knows the liars will say it is unsafe

if the 100% liar guard responds:

he will point the safe door too, because :
the 100%, 75% and 50% liars are the 3 most truthful guards.
the unsafe door means the safe door, in this case..

I Think this also work to the other guards so take the door.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 3 most liars means 100%, 75%, 50% liars. $\endgroup$ – Jamal Senjaya Sep 18 '16 at 0:06
  • $\begingroup$ I answer this using my handphone while cooking, so I can not give more answer immediately. $\endgroup$ – Jamal Senjaya Sep 18 '16 at 0:10
  • $\begingroup$ It's not possible to tell which door the partial liars would point to, unless they actually pointed to a door, because they have some chance of pointing to either a safe or unsafe door. Then, having an answer, it tells you nothing about which door is the safe one, since there are still four other doors, nor about which guard you asked. You've at best pushed the chance of a lucky guess from 20% to 23% or so. $\endgroup$ – Nij Sep 18 '16 at 0:40
1
$\begingroup$

Label the guards A, B, C, D, E.

Ask the following to A:

If I asked B, "If I asked C, "If I asked D, "If I asked E which door is safe, which door would he most likely point to?", which door would he most likely point to?", which door would he most likely point too?", which door would he most likely point to?

I think this has about,

50% of working.... The idea is that you want to cancel out lies. This only works if there are an even number of mostly liars... But then the one you ask has some chance of lying too... I can't think I'm going insane help

Sorry if I'm dumb.

| improve this answer | |
$\endgroup$
1
$\begingroup$

Summary:

$\frac{23}{40}$.
Question asked: "Which door would that person not say is the safe door?". Then choose that door.
Reason: The person you ask is taken out of the pool of other people, so if they are a liar, there are fewer other liars left.

My initial thoughts:

So you are going to randomly pick one guard to ask a question. As I see it there are 3 outcomes: A mostly liar (100%, 75% false), a neutral (50%) and a mostly truth (25%, 0%).
So what do we get out of this? You know that the person you ask is not going to be one of the other people!
Why is this important? Because if you have a mostly liar, you have better odds of getting a truthy person, and if you get a truthy person higher odds of getting a liar.
So we resort to the standard approach here: Asking one person, what another is most likely to say. (In the case of the 50% person, assume just a random answer about their likely answer). From my thinking the neutral case you are hosed, you have a 50% chance of then asking truthy people or Liar-y people. I believe this case (which happens 1/5th the time) is a 50% chance of victory.
Now for the other cases. If you get a truthy person, you know have 4 other people, a truthy, a neutral and two liary people. As such you best bet is going for a liary person. This could happen in two cases you ask you question to the 75% or the 100% liar. If you get the 100% liar, you can ask your question of 1 liary, 1 neutral and two truthy people. So we aim our question to be: "Which door would that person not say is the safe door?" and choose that door. We have 4 possibilities of who we ask about, and we will only be correct when they tell the truth. So our odds of winning are $\frac{1}{4}(0.25+0.5+0.75+1) = \frac{5}{8}$ and this will happen $\frac{1}{5}$ of the time. In the 75% liar case we use the same logic to get $\frac{1}{4}(0+0.5+0.75+1) = \frac{9}{16}$
Now we invoke symmetry on the Truthy cases and get: $\frac{1}{5}(\frac{5}{8}+\frac{9}{16}+\frac{1}{2}+\frac{9}{16}+\frac{5}{8})= \frac{23}{40}$

Thought I had embarassingly late:

You might just ask: "Which door would you say is the safe door?"
Liar fakes themself out, truthy truths, partials you get the times when they lie twice or truth twice. Seems like it would give better results...

| improve this answer | |
$\endgroup$
  • $\begingroup$ The one they answer with, woops, sorry if that wasn't clear. $\endgroup$ – Going hamateur Sep 18 '16 at 16:05
1
$\begingroup$

I can get 55%:

By including in the question a way of finding out which guard you're asking.

Assume the doors are numbered 1 to 5 and ask, "What number do you get if you add the position of the safe door plus the percentage chance you are telling the truth?". Since they can't lie about about their truthfulness you get their truth percentage plus a door position then you do the maths from there.

101 to 105 gives you the safe door 100%.
76 to 80 gives you the safe door with 75% chance.
51 to 55 gives you the safe door with 50% chance.
26 to 30 gives you the safe door with 25% chance.
1 to 5 means you're asking the liar so you choose one of the other 4 doors at random with 25% chance.
Average all that out and you have a 55% chance of choosing the safe door.

| improve this answer | |
$\endgroup$
  • $\begingroup$ @TheBitByte The latter. It was an initial best effort assuming there's one safe door out of 5, a naive attempt. $\endgroup$ – FreddyFlares Sep 19 '16 at 19:36
1
$\begingroup$

I'm assuming there is only one door that is safe since that is what I felt was implied from the original post.

Ask the first guard which doors you should not choose. If the guard points to a single door then he is lying and I select the door he pointed at. Otherwise he is telling the truth and I select the one door he didn't point at.

Example/Further Explanation:

After I ask my question the guard will either tell the truth or lie. If he tells the truth he will point to four out of the five doors. If he is lying then he will point to a single door out of the five doors. You must be asking yourselves what happens if he is lying and points to multiple doors? If the guard is pointing at multiple doors and is lying then some of the doors he pointed at must be unsafe since we assumed there is only one safe door. Meaning the guard didn't completely lie and we contradict the assumption. Since there isn't fuzzy logic involved he can only lie completely or tell the truth completely.

| improve this answer | |
$\endgroup$
  • $\begingroup$ @TheBitByte It addresses the outcome not the probability. I'll edit my answer with more information. $\endgroup$ – Jonathan Sep 19 '16 at 9:17
  • 1
    $\begingroup$ @TheBitByte The question is not clear about when the probability comes into play. In this answer the interpretation seems to be that each answer as a whole is either true or false, i.e., the 75% guard will either point a single door (25% chance) or at four doors (75% chance). Did you intend that the probabilistic liars consider each door separately when answering? If so, is it even permissible to ask this type of question? $\endgroup$ – Arkku Sep 19 '16 at 14:27
0
$\begingroup$

If you ask a random guard, the question

If you were the guard who always lies, then if I asked you which door was the safe door, which door would you never point at?

then you have

an equal chance that the door this guard points at, is the safe one.

because

the 100% truth guard will always point at the safe door. This gives 20% of a safe door; the 75% truth guard point at the safe door 75% of the time. This gives 15% of a safe door; similarly for the 50%, 25%, 0% truth guards, we obtain 10%, 5%, 0% of a safe door; in total, then, we have 50% of a safe door being pointed at.

I cannot be sure if this is optimal. There may be a more complex question along the same lines, combined with some logic, that allows the elimination of some possibilities and therefore a better chance of deciding the safe door.

| improve this answer | |
$\endgroup$
0
$\begingroup$

100% chance of success. One question. One guard.

Pick a guard, ask him, "Tell me, if I were to ask you to point to the safe door, where would you point?"

Results:
1) Always Truth: He would point to the safe door, so he'll tell you that door.
2) Always Lies: He would point to the wrong door, so he'll lie and tell you the right door.
3) I'll assume the mixes are either going to fully lie or fully tell the truth for that one question so they'll work out like either 1 or 2.

| improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ However, there are 5 doors, not 2. $\endgroup$ – Klyzx Sep 18 '16 at 3:15
  • $\begingroup$ @Caelan O: I meant "right" as in "correct" or "safe", not right/left. $\endgroup$ – Dark Matter Sep 18 '16 at 3:25
  • $\begingroup$ So if they point at a wrong door, how do you know which one of the remaining four is the safe one? This is not even near 100% to find the safe door. $\endgroup$ – Nij Sep 18 '16 at 7:32
  • $\begingroup$ @Nij They never point to any door, that's the point. It's a question about a question. In a world of perfect logic, if you're going to lie about a lie, you need to tell the truth. $\endgroup$ – Dark Matter Sep 18 '16 at 13:55
  • $\begingroup$ @DarkMatter How do you know a nondeterministic guard isn't imagining himself telling a lie, then being truthful about what he imagines himself saying, or vice versa? And two lies don't make a truth either, because the guard could say "Yeah I'd point you to WrongDoor2" when he would actually point to WrongDoor1. $\endgroup$ – Anon Sep 18 '16 at 15:27
0
$\begingroup$

This works:

Which door would the truthteller say is safe?

That isn't a question about which door is safe - it's a question about which door someone would say is safe, so any guard will give a truthful answer.

| improve this answer | |
$\endgroup$
0
$\begingroup$

This gets the right door almost 80% of the time, and does not depend on any guard knowing how any other guard would answer:

If I asked you a million times which was the correct door, which door would you say most often?

So long as you don't choose the liar, this will get you the right door with a probability of almost 80%. It's not a question about which door is correct, and so everyone will answer truthfully.

A variant that does require them to know what each other would say, and achieves almost 100%, is this:

If I asked you a million times each to say which is the correct door, which door would be identified as correct most often?

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.