Batman vs 4 villains

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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?").

Answer 6

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

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

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.