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One day after having his morning Bat-coffee, Batman received the following message on his Bat-computer:

Hello there Dark Knight. I, the Riddler, along with Joker, Two-Face, and Penguin, are preparing to pull off the heist of the century. But before we go, we're going to play a little game. If you win, we'll surrender to the police. If we win, the heist goes off as planned. Here's how the game works, in 1 hour you'll call the number at the bottom of this message. One of us will pick up the phone, and you'll ask 1 yes/no question. Try to ask any other kind of question, and we hang up and you lose. After you ask and we answer, we pass the phone to another of our members. We know the order we'll pass in, but you won't. The following facts will hold:

  1. I, the Riddler, will always answer truthfully.

  2. Penguin will always lie.

  3. Two-face will flip his coin, if it comes up heads he will tell the truth, tails he lies.

  4. Joker will lie or tell the truth depending on what he feels will most disrupt you.

  5. Each of us will answer before anyone repeats, and the cycle will not change while we talk.(i.e. The order we answer will be the same each round)

  6. If we do not know the answer to a question, we will say "Pass". Note: Joker will never pass.

The task is simple, figure out our order. Good luck.

What is the smallest number of questions Batman will have to ask in order to figure out the villain's order, regardless of what it is?

Bonus: Does it matter if a villain decides to hang-up if they hear a repeated question?

Note: I just came up with this puzzle, so I do not know the answer at this point in time.

Clarification of Joker's behavior: If a) Joker knows the answer to a question AND b) There is a way to answer the question while remaining logically consistent, then the Joker answers the question (lying or truthing depending on how to most screw-up Batman)

If a) Joker doesn't know the answer OR b) There is no way to remain logically consistent while answering, then Joker will pick yes/no based on what he believes would most screw-up Batman.

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  • 5
    $\begingroup$ Nothing stops you from asking something like "Is the xth person after you Joker?" since that is a yes/no question. $\endgroup$ – Saladani Sep 18 '17 at 23:26
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    $\begingroup$ @Braydon The yes/no questions could be "are you the Riddler" or things of that nature, so if the Joker always answers truthfully, like the Riddler, then he would be less effective than "always answer truthfully, except for this question when it makes it too easy for Bats." $\endgroup$ – Mister B Sep 18 '17 at 23:27
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    $\begingroup$ @Braydon If you believe there is no solution, write it in an answer $\endgroup$ – Saladani Sep 18 '17 at 23:36
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    $\begingroup$ Let the heist go as planned. Batman will be there to save the day anyway. $\endgroup$ – Noldor130884 Sep 19 '17 at 6:56
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    $\begingroup$ Wait, you WROTE a puzzle and posted it without solving it yourself?? $\endgroup$ – feelinferrety Sep 19 '17 at 19:24

12 Answers 12

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At least half the time, Batman can get the order correct in only 4 questions. For certain sequences, it will take 5 questions.

Batman asks the first three villains in the sequence the same question:

If I were to ask you the question, "Are you The Riddler?", would your answer be "Yes"?
  • The Riddler's future answer would be "Yes", and since he answers truthfully, his answer is "Yes".
  • The Penguin's future answer would be "Yes", and since he answers untruthfully, his answer is "No".
  • Two-Face doesn't know what his future answer will be, so he answers "Pass".
  • The Joker doesn't know what his future answer will be, but since he never passes, he answers either "Yes" or "No".

At this point, Batman has asked 3 questions. Now for the fourth (and possibly the fifth).

What Batman does next depends on the answers that he's received so far.

  1. If the answers are either 2 "Yes" answers and 1 "No" answer, or 1 "Yes" answer and 2 "No" answers, the last villain in the sequence is Two-Face, since no one has passed yet. Batman will ask this question to Two-Face: (replacing [order number] with the order of one of the duplicate answerers)
    If I were to ask the villain in position [order number] if they are The Riddler, would they answer "Yes"?
    If Two-Face gives an answer, Batman knows that the character in [order number] is The Riddler or The Penguin, depending on whether the duplicate was "Yes" or "No". If Two-Face passes, that character is The Joker. With this information, Batman can figure out the order.
  2. If the answers are a "Pass" and either 2 "Yes" answers or 2 "No" answers, the "Pass" reveals Two-Face's location and Batman knows the last character is either Riddler (for 2 "No" answers) or Penguin (for 2 "Yes" answers). Batman can straight-out ask Riddler/Penguin about one of the duplicates (e.g. "Is The Joker the villain in position [order number]?"), and their Yes/No answer will reveal the entire order. This scenario is unlikely, as The Joker would have had to mess up by giving a duplicate answer.
  3. If the answers are "Yes", "No", and "Pass", the last character is either The Riddler, The Penguin, or The Joker. Batman will ask the last villain the following question:
    I am thinking of a number between 1 and 10. Is it the number 7?
    If the villain gives a Yes/No answer, the villain must be The Joker, since Riddler or Penguin would have to pass.

In any of the above scenarios, Batman solves the puzzle in only 4 questions.

In scenario 3 above, if the last villain does not give a Yes/No answer but passes instead, Batman knows that the last character is not The Joker, but doesn't know whether they are The Riddler or The Penguin. What Batman asks as his 5th question depends on who is first in the order.

  • If the first person is Two-Face (that is, the original answer was a Pass), Batman will ask him:
    If I ask the next villain the question "Are you The Riddler?", will they say "Yes"?
    Two-Face does not know what The Joker's answer would be, so if Two-Face passes, the next character is The Joker. If he answers, the asked-about character is The Riddler or Penguin (depending on a whether the first-round answer was Yes or No). Either way, Batman knows the order.
  • If the first person had a Yes/No answer, Batman will ask them the question from scenario 3. If they answer, the character is The Joker. If they pass, the character is Riddler or Penguin, depending on their original Yes/No answer.

In that scenario, Batman solves the puzzle in 5 questions.

Using this method, Batman is guaranteed to solve the puzzle in 4 questions if Two-Face or The Joker are the last villain in the order. Otherwise, Batman will solve it in at most 5 questions.

BONUS: With this technique, Batman will never repeat the same question to a single villain.

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  • $\begingroup$ Why is 8 the minimum? $\endgroup$ – boboquack Sep 19 '17 at 7:54
  • $\begingroup$ @boboquack You're right, I could actually refine it a lot. If I started with question #2 I could find Two-Face and The Joker faster. But I need to go to sleep. I'll take a look in the morning. $\endgroup$ – oobug Sep 19 '17 at 8:10
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    $\begingroup$ I don't think the qualifier 'on the next round of questions' is needed. I'd also hand-wave 2 away as Joker can't answer that way as it gives away too much information and lets Batman find him early. $\endgroup$ – Danikov Sep 20 '17 at 15:09
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    $\begingroup$ I think you're right, the more I think of it, Joker should be treated as a random selection of yes/no. He cannot predict Batman's strategy so, while sometimes he might answer later and there's an obvious answer that gives away information to avoid, sometimes he might go first and he cannot predict Batman will ask the same question again later on. $\endgroup$ – Danikov Sep 21 '17 at 10:34
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    $\begingroup$ Ah, maybe I'm wrong; the Joker's misinformation multiplies out the 24 possibilities into 48 possibilities, which is less than the 3^3 = 27 bits of information obtainable from the other three villains. $\endgroup$ – Danikov Sep 21 '17 at 13:42
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This answer is kind of cheap and relies on some technicalities, but here goes:

First round, ask everyone:

"Do I know how many more questions I'll ask?"
As per rule 6, all should say "PASS", except Joker. Now you know where he is.

Next, ask everyone:

"Does 1+1=2?"
Riddler says yes, Penguin says no, Two-Face is random, and we ignore Joker. If there are 2 "yes"es (ignoring Joker), then we know where Penguin is, and 2 "no"s mean we know where Riddler is.

Now, we have one person we know is either always lying or always telling the truth. Now, we ask everyone:

"Does Penguin answer before Two-Face?" (If we know Riddler's Position) OR
"Does Riddler answer before Two-Face?" (If we know Penguin's Position)
Riddler's answer to the first question, or Penguin's answer to the second, is enough to find everyone's order. (We already knew Joker's and one other's position, and their answer gives us the order of the other two)

So, at most, it would take:

12 questions, regardless of order.

I think I'll put a bit more work into this though, because the first question still seems cheap.

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  • 2
    $\begingroup$ You could reduce the number of questions by at least one. If the Joker is the last, he is not required for the last round of questions. If he is not last, you can start the second round of questions at least one question earlier. $\endgroup$ – Apep Sep 19 '17 at 0:30
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    $\begingroup$ @Saladani How is the Joker answering to mess with you if he allows himself to be found by refusing to pass? If he wants to mess up Batman he should pass. If this is an acceptable answer the question was again unclear. $\endgroup$ – Braydon Sep 19 '17 at 2:30
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    $\begingroup$ @Braydon the puzzle clearly states that the Joker will never pass. $\endgroup$ – oobug Sep 19 '17 at 5:52
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    $\begingroup$ The problem is that if they "no" to your first question, it probably becomes true. I'm not certain about this but I think it makes sense. $\endgroup$ – Samthere Sep 19 '17 at 12:18
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    $\begingroup$ @Flater It works under the assumption that Batman does not have to ask the same question 4 times. The 'rounds' was those set up in the answer (asking the same question 4 times), not those in the question (the villains answering in order). Once the Joker is found, Batman can stop asking the first question and ask the second question 4 times. In the case that the Joker is not the 4th, this saves at least one question. In the case that the Joker is the 4th, Batman does not need to ask him the last question, so that also saves one question. $\endgroup$ – Apep Sep 20 '17 at 11:59
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The answer is

3

The reason is

Because he's Batman. The conversation is happening over the phone, which means he can hear their voices. He knows what each of them sound like so he just needs to ask three BS questions and he'll know who's answering simply by the sound of their voice. Once he knows the first three, the fourth is deduced from the process of elimination.

This will also work if

They attempt to disguise their voice. Batman is in his BatCave and has access to his highly-sophisticated audio analysis equipment. No doubt he would be recording the conversation to potentially use as evidence later. If the voices are masked, he'll be able to run them through his computer to remove that mask and determine who the speaker is.

And just for kicks,

Once he's identified the order, he won't actually reveal that he knows. Instead, he'll start asking questions designed to cause tension between the group. For example, he might ask the Riddler "Do you think TwoFace will get a bigger cut of this heist than the Riddler?" or maybe he'll ask Penguin "Did Joker tell you about that nuke he's planning on setting off in Gotham later today?". This is just to mess with them, possibly incite some in-fighting and more importantly, keep them occupied until he can show up and catch them. Because you see, they gave him their phone number and he'll have easily traced it to its location. He's asking these questions while en route and just needs to make sure they don't escape before he arrives.

However,

This won't work if they don't actually respond directly with their voices. Perhaps they have some machine that does the speaking for them and all they do is press the 'yes', 'no' or 'pass' buttons. But the question makes no mention of such a device so I think it's safe to say they won't be using one. I do realize this answer is probably going against the spirit of the question but this is Batman we're talking about. He's always one step ahead those bumbling buffoons.

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    $\begingroup$ HOLY out of box answer, bat-aleppke +1 $\endgroup$ – Mindwin Sep 19 '17 at 21:55
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    $\begingroup$ This was my first thought as an answer. Your "however" area can still be worked around, even if they are disguising their voice. Batman can probably still tell from their speech patterns. For example, Riddler will probably use lots of big words, Joker will use lots of terrible puns and manic laughter, Two-Face will probably make references to binary dichotomies, and Penguin will talk like Danny DeVito. Also make bird jokes. You know those 4 could never be content with a simple yes/no answer, they must taunt Batman whenever they can. $\endgroup$ – Cody Sep 19 '17 at 22:48
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    $\begingroup$ ... Because he's Batman. $\endgroup$ – vaxquis Sep 20 '17 at 15:59
9
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I can do (and this is minimal):

3 questions (for both variations, this is minimal because there are only 3 possible responses and 32<4!)

Ask the following questions to three people:

If I took the four letters J, P, R and T, put them in the order corresponding to the order you are in where J=Joker, P=Penguin, R=Riddler and T=Two-face, indexed it into the following dictionary (see dictionary below, Batman will have to recite this but I won't add it here), and took the first/second/third change this for each question - each corresponding to the question number asked letter of the identifier, will either the letter be Y and you are telling the truth, the letter be N and you are lying, or the letter be P and either you will lie and say yes or tell the truth and say no?

Then:

Yes means Y, no means N and pass means P since the option for P is a head-exploding statement. Then you can reconstruct the three letter identifier, back-index into dictionary and find the order!

Dictionary:

JPRT: YNY
JPTR: YNN
JRPT: YNP
JRTP: YPY
JTPR: YPN
JTRP: YPP
PJRT: NYY
PJTR: NYN
PRJT: NNY
PRTJ: PPP
PTJR: PNY
PTRJ: PNP
RJPT: NYP
RJTP: PYY
RPJT: NPY
RPTJ: NPP
RTJP: PPY
RTPJ: NNP
TJPR: PYN
TJRP: PYP
TPJR: NNN
TPRJ: YYP
TRJP: PNN
TRPJ: YYN

Examples:

Riddler, letter=Y: "The letter is Y, and I am telling the truth. So true, and I say yes."
Riddler, letter=2: "The letter is not Y. The letter is N, but I am telling the truth, so false. The digit is not 0. So false, and I say no."
Riddler, letter=P: "The letter is not Y or N. The letter is P, and I am telling the truth. If I say no, then it's true, because I say no. If I say yes, then it's false, because I don't say no. So pass."
Penguin, letter=Y: "The letter is Y, but I am lying. So false, but I say yes."
Penguin, letter=N: "The letter is not Y. The letter is N, and I am lying. So true, but I say no."
Penguin, letter=P: "The letter is not Y or N. The letter is P, and I am lying. If I say yes, then it's true, and I should have said no. If I say no, then it's false, and I should have said yes. So pass."
Two-face/Joker: refer to Riddler if telling the truth, Penguin if lying. Note - the Joker will never be given a head-exploding question

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  • 2
    $\begingroup$ Could you please provide an example, I'm having trouble tracking the logic all the way through $\endgroup$ – Saladani Sep 19 '17 at 2:37
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    $\begingroup$ This raises the question of whether the "Joker never passes" rule is consistent with the "Joker lies or tells the truth depending on which messes you worse" rule (which isn't the same as "Joker answers whatever messes you worse", in which case it would indeed be impossible to force him to pass). $\endgroup$ – Oosaka Sep 19 '17 at 15:06
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    $\begingroup$ Also, I had to really struggle to understand this answer - maybe it's just me, but I feel it isn't completely clear that you aren't asking three times the same question, and "ones/threes/nines" represent a variable part where in the first question you plug in "ones", the second you plug in "threes" and the third you plug in "nines". $\endgroup$ – Oosaka Sep 19 '17 at 15:31
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    $\begingroup$ It doesn't say the villains are perfect logicians. What happens if all three say PASS because they can't figure out the question, and hence don't know the answer? Then all you have deduced is that the last person is the Joker. $\endgroup$ – Abigail Sep 19 '17 at 18:58
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    $\begingroup$ @oobug batman always uses oxford commas $\endgroup$ – Mindwin Sep 19 '17 at 21:54
5
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ETA: Just saw this is DqwertyC's answer. Sooo... I counted the questions better?

This answer relies on the fact that the Joker never passes in a way that makes things easier; so it really depends on whether it's true the Joker never passes (the question makes it look like the Joker makes things hardest for Batman and therefore never passes but that looks wrong to me).

[Looks like he does: "It is an additional rule, Joker likes the sound of his own voice too much to pass"]

Step 1

First ask a question that none of the villains know the answer to, forcing them to pass. Such as: "Did the coin I've just flipped fall heads?". The first person that doesn't pass is the Joker. Say it's the Nth person.

This takes 1 to 4 questions.

Step 2

Next, ask the next three people: "Is the Nth person the Joker?" (or any question both you and they know that the answer to is "yes"). The Riddler and the Penguin will answer differently, while Two-face answers like either of them. So you'll get two "yes"s and one "no", or one "yes" and two "no"s. Either way, you know that the answer you didn't get two of is the Penguin (if it was "no") or the Riddler (if it was "yes"). Actually if you're lucky and you get two "yes"s or two "no"s right away you know who the third person is without asking them. Say the newly-identified person is in position M, and one of the other two is in position L.

This takes 2 to 3 questions.

Step 3 (only necessary if Step 2 took 3 questions)

At this point you're at the Joker again; ask a dummy question and ignore the answer ("Are you that bad at makeup or is it on purpose?").

This takes 1 question.

Step 4

Now you just need to get to the Riddler or Penguin you identified, and ask them "Is the Lth person Two-face"? They'll tell the truth or lie, and since you know which they're doing that allows you to tell the last two people apart. If your first two answers in Step 2 were identical then you just need to ask this of the fourth guy; otherwise you need to do the round again after having passed the Joker, thus getting to the second or third.

This takes 1 to 2 questions.

Total questions : 10 at most, 4 at least.

BONUS QUESTION:

This already doesn't repeat any question to a single villain. It can also be made so that no question is repeated at all:

make three questions they don't know the answer to ("Did the coin I just flipped fall tails?", "Did I just flip a coin?", "Do I have a coin on me?") and two answers that you all know the answer to is "yes" ("Is 1+1=2?", "Am I Batman?"), and different dummy questions ("Did you really think this stupid plan would work?").

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  • $\begingroup$ Aw, why you gotta be mean to the Joker like that? :( $\endgroup$ – Nic Hartley Sep 21 '17 at 15:22
2
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These types of questions assume that the villains are perfect logicians and either tell the truth completely, or lie completely. No partial truths allowed.

Therefore, a simple way to get them to answer consistently is:

To the question "Are you X?", is your answer "yes"?

If the villain is not speaking the truth, then if you were to plainly ask them if they are X, they would lie. But in this question, they would lie about lying, and therefore end up giving the same answer as someone who speaks the truth.

Given that, it's trivial to simply

find out from everyone if they are the Joker, then if they are the Riddler, etc.

The worst case number of questions for this is

12, since if you know three villains, you can deduce the fourth

The best case number of questions for this is

3, if you get lucky and guess right the first three times

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    $\begingroup$ I see how this allows you to get the identity of The Riddler and The Penguin, but both Two-Face and The Joker answer randomly. How would you extrapolate this logic to identify them? $\endgroup$ – oobug Sep 19 '17 at 6:01
  • $\begingroup$ @oobug They do not randomly decide to give a yes or no answer, they randomly decide to tell the truth or lie. Therefore, this should work just as well for them, if I'm understanding correctly. $\endgroup$ – hvd Sep 19 '17 at 6:11
  • $\begingroup$ My understanding is that they can randomly choose each time they are asked a question whether to tell the truth or lie. Two-Face and The Joker's answers will not be consistent. $\endgroup$ – oobug Sep 19 '17 at 7:38
  • $\begingroup$ @oobug I can sort of see that the wording I picked allowed it to be treated as two separate questions, so they could tell the truth for one and lie for the other. Re-worded so that it's no longer a hypothetical separate question. $\endgroup$ – hvd Sep 19 '17 at 7:55
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    $\begingroup$ The Joker could still answer "Yes" or "No" because he can still choose each time the question is asked to tell the truth or lie. But the specific wording of your question could be interpreted as always forcing Two-Face to pass, which is interesting. $\endgroup$ – oobug Sep 19 '17 at 8:03
2
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It can be done in 5 questions, depending on if Two-Face is first or not.

Round 1:

"Next round, will you say yes if I ask if you are the Joker or the Penguin?" Two-face immediately identifies himself by passing, since he doesn't know. The Riddler will truthfully answer no. The penguin will say yes. At this point, the Joker needs to decide whether he will tell the truth or lie next turn ahead of time. If his truthiness is the same next turn as the current turn, he will say yes, and the Riddler will be identified. Otherwise he will say no, and the Penguin will be identified.

Round 2:

This depends on the order now. If Two-Face is not first, and, if we have someone we've identified we just ask them if the next non-Two-Face is the Joker, and it's solved. Otherwise, ask "Are you the joker and did you tell the truth last time I asked you a question?" If you didn't identify the Riddler yet, the Riddler will always say no, and the Joker will always say yes. If you didn't identify the Penguin yet, the Penguin will say yes and the Joker will say no. If Two-Face is the first answerer, ask "If the coin flip were reversed, would you say [the next person I haven't identified] is the Joker?" Two-Face is guaranteed to lie in response, so you invert his answer and the problem is solved.

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    $\begingroup$ The Joker cannot be forced to give a single answer. For one thing, nothing's keeping the Joker from changing his mind about lying or telling the truth from one round to the next. But it's also been updated in the rules that the Joker can answer yes/no regardless of truth value if he doesn't know an answer or can neither lie or tell the truth (which in this case can be the case with the first question) $\endgroup$ – Oosaka Sep 20 '17 at 11:24
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    $\begingroup$ "Next round, what will be you answer if I ask if you are the Joker or the Penguin." might be considered not to be a "yes/no" question... You are asking for their answer after all! Better ask "... would your answer to the question ... be 'yes'?" $\endgroup$ – M.Herzkamp Sep 20 '17 at 15:39
  • $\begingroup$ @M.Herzkamp Thanks, I cleaned that up. $\endgroup$ – Ethan Sep 20 '17 at 17:18
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    $\begingroup$ @Ethan: Nothing requires the Joker to decide in advance what he will answer on a later questions, meaning you can't guarantee that he will know the answer to the question "How will you answer next round", and if he doesn't know the answer he can say whatever. Also I don't see how changing one's mind is logically inconsistent. Truth or lies depend on your state of knowledge; if you make a statement you think is true, and later new information shows it to be false, it doesn't retroactively turn it into a lie. And changing your mind is new information that wasn't available before the change. $\endgroup$ – Oosaka Sep 21 '17 at 8:42
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    $\begingroup$ I didn't talk about a strategy of changing his mind; that would come down to him not deciding in advance what to answer. I talked about just... changing his mind from one round to the next. The kind you don't predict in advance because that's what "changing one's mind" means. Also, the rules don't say "if he CAN know the answer", they say "if he knows the answer", on the same priority level as "if he can be logically consistent". No suggestion that being logically consistent is a higher constraint that forces him to decide things just so he can know them when he wouldn't know them otherwise. $\endgroup$ – Oosaka Sep 21 '17 at 22:53
1
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The minimum questions is 8, but this is only possible with a specific order and behavior from Two-face.

Explanation:

The Joker will likely imitate the behavior of another villain in order to make it hard to distinguish between the two of them. In order for him to be the most problematic he will either mimic the behavior of the Riddler or Penguin. In order to find the joker we must first identify either Riddler or Penguin, so we begin by asking all 4 questions with known true of false answers.

After one round we will not have verified anyone's identity, so we ask again questions with known answers. In this second round we hope to get either two repeated lies and one truth, or two repeated truths and a lie, plus a behavior change from Two-face. (However if Two-face repeats his answer from chance, you must repeat this round of questioning until Two-face repeats.)

Assuming however that Two-Face, Joker, and the villain he is mimicking are the first three, we will then be able to identify the villain Joker is not mimicking after the 7th question. Facing two repeating answers and one non-repeating you know the non-repeating is Two-Face and the other two are the Joker and the villain he mimics, this gives us the identity of the fourth based on if he lies or tells the truth. You must then ask the predictable villain with the known identity if one of the two identical villains is the Joker, and then interpret his answer to deduce which is the Joker, and then which must be the last remaining unknown villain.

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  • $\begingroup$ Why is 8 the minimum? $\endgroup$ – boboquack Sep 19 '17 at 7:54
  • $\begingroup$ Umm. . . See the explanation? If you have a more specific question about it then maybe I can explain. $\endgroup$ – Braydon Sep 19 '17 at 13:38
  • $\begingroup$ I see why it is the minimum in this answer, but why would a different set of questions not be able to reveal the identities sooner? $\endgroup$ – boboquack Sep 19 '17 at 23:02
  • $\begingroup$ @boboquack Well at the time of this answer it was not yet clarified that the Joker always answers, since it simply said he would answer in a way intended to mess you up. (This would seem to indicate he would not reveal himself by refusing to pass.) Without the "Joker never passes" rule I had to find him last, which makes it longer. $\endgroup$ – Braydon Sep 19 '17 at 23:20
1
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Best Case Scenario: 3 questions
Average: 4.5 questions
Worst Case Scenario: 6 questions
The Questions:

"How many fingers am I holding up" until you find the Joker "Would you say yes if I asked you next round if the penguin was ahead of the riddler?"
"Would you say yes if I asked you next round if the riddler was ahead of the penguin"

The Logic:

First you have to figure out who the Joker is, so ask some arbitrary question that the villains can't possibly know the answer to. The Joker is the only one who says he doesn't know. Worst case scenario is that the Joker is last so that's 4 questions.

The second part relies on the fact that a double positive and a double negative is a positive. By asking if they would say yes to an embedded question, it doesn't matter if they're lying or not. Consider the example "If I asked you if 1 plus 1 is two, would you say yes?" The Riddler would say yes to 1 plus 1 being two, so he says yes. The Penguin would say yes because he would say no to 1 plus being 2, so he lies and says yes. Now consider if I asked "If I asked you if 1 plus 1 is three, would you say yes?" The Riddler would say no to 1 plus 1 being 3, so he would honestly say no. The Penguin would say yes to 1 plus 1 being 3, so he lies and says no. Since we've proven we can ask questions without worries if they're lying, we just need to find the order of the three. Since Two-Face doesn't know what he's gonna say next round, he has to pass. So with 2 questions, you can find out if the penguin is ahead of the riddler, and where two-face is.

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  • $\begingroup$ What if in the hypothetical scenario two-face flips heads, but in reality flips tails? $\endgroup$ – boboquack Sep 21 '17 at 0:28
  • $\begingroup$ He would only flip once since you're only asking one question $\endgroup$ – Qwepoiwo Sep 21 '17 at 0:49
  • $\begingroup$ Actually after some additional thought, Two-Face has to answer pass for the question, Batman can find their order reliably with 6 questions. $\endgroup$ – Qwepoiwo Sep 21 '17 at 12:38
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I can do it reliably in 6 Questions:

Ask the following question four times:

If I ask the next person if they are the Riddler, will they say 'Yes'?

If the Riddler answers the question and the next person is the Penquin, he will say 'Yes' otherwise he will 'Pass'.

If the Penquin answers the question and the next person is the Riddler, he will say 'No', otherwise he will 'Pass'

If TwoFace answers the question, he will 'Pass' if the next person is the Joker, otherwise he will answer 'Yes' or 'No'

If Joker answers the question, he will just answer 'Yes' or 'No'

This will tell us some thing about the order and might actually tell us what the order is:

If the Joker follows TwoFace, then we will have two Passes together, the first from either the Riddler or the Penguin, and the second from TwoFace.

The previous answer, before the first 'Pass' will tell us whether the Penguin is before or after the Riddler, e.g.

Yes,Pass,Pass,No - would be Riddler, Penguin, TwoFace, Joker
and ...
No,Pass,Pass,No - would be Penguin, Riddler, TwoFace, Joker

alternatively:

If TwoFace follows the Joker, then there will be only 1 'Pass' (the Penguin or the Riddler) which will identify the Joker, but TwoFace will be the next person, so:

Yes,Pass,No,Yes - would be Riddler, Penguin, Joker, TwoFace
Likewise,
No,Yes,Pass,No - would be TwoFace, Riddler, Penguin, Joker

So anything ordered similarly to the above will only need 4 questions.

Other orders are dealt with as below:

Otherwise, the Riddler and the Penguin will be alternating with TwoFace and Joker, and there will be two Passes, one from the Riddler, and one from the Penguin. So in this case it is simply a matter of telling the Joker from TwoFace, after the second 'Pass' answer is given - and then telling the Riddler from the Penguin

If you ask TwoFace/Joker, 'Will the Joker tell me the truth?', TwoFace must always 'Pass', the Joker must answer 'Yes' or 'No'

If you ask the Riddler/Penguin, 'Will the Penguin tell me the truth?', the Riddler will always answer 'No', the Penguin will always answer 'Yes'.

Bonus: No one will be asked the same question twice.

So, Batman can work the order out in a maximum of 6 questions.

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0 Because it is possible to nullify some character's abilities to answer correctly, making logical deduction uncertain.

Ask each person:

"Are you going to pass? "There is no way for the Riddler to tell the truth, so there is no one correct answer that is always true. The puzzlemaker has to eliminate some question(s) beforehand. In other words, if you ask the Riddler, and he says yes, then he cannot pass because he already said yes. In fact, none of them can be relied on, since they can only answer yes or no.

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  • $\begingroup$ Riddler would answer "No", truthfully since he didn't pass $\endgroup$ – Saladani Sep 20 '17 at 21:01
  • $\begingroup$ So if anyone says yes, they are lying, right? $\endgroup$ – johnny Sep 21 '17 at 4:29
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Well, I would say 0. Bats got a phone number to dial in 1 hour. In that time, he finishes his Bat-coffee, suits up, enters the Bat wing and flies over Gotham, searches the number in the Gotham Communication Services Database and knows the exact address of the holder of said phone number. He will fly to that address, jump out of the wing and tadaaa... after some punches, there is a Batman in the middle of 4 knocked out villains. Should it be a mobile number, Bats will use his technology to trace back the belonging cell phone by the providers mobile network. He is Batman, he has access to everything.

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protected by Community Sep 21 '17 at 10:13

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