If anyone thinks it might be possible to solve the riddle in 5 questions or fewer, please express your interest in arguing as much. Otherwise, I might resolve this in the near future.

It doesn't work in the main riddle, but if the coin flip were to decide between acting like True and False you could get it in five. You could ask anybody a question and get them to answer ja for yes and da for no by including a clause for "[convoluted question other comments have stated] xor you just flipped a coin secretly and it came up tails". Then you could go up to Amanda, ask her what her favorite color is (uses two questions). Then you could go up to Bill and ask him "Is True's favorite color Blue", "is True's favorite color Red", "is Random's favorite color Green (or Red if True's favorite color is Green)", and you will know the favorite color of every truth-teller in five questions. Note that Bill now can answer this question because you can tell him Amanda's favorite color, and he knows his own favorite color, so he can figure out Cindy's favorite color by process of elimination. If you make it so he can't know other people's favorite colors by fiat you can create questions that produce the same result yourself but it's annoying and I think one of those questions would be about as long as this comment given all of the extra conditionals you have to add for different scenarios.

Again, this doesn't actually apply to the question since it assumes Random gives us useful information, but I think it has some of the same general qualities a real five solution would need, as well as being something I find interesting

@CollinMatthews If it turns out we can't get lower than 6 I'd be interested what happens if, instead of answering randomly, the "Random" person just answers Da to everything. I think it adds a lot of extra complexity, but might end up with a lower number of necessary questions

@CollinMatthews If Random always answered "Da" to everything, I think the minimum required number of questions would still be 6. In the solution to the riddle I proposed, the first question is spent identifying a non-Random person, and then never interacting with Random afterward. In a Random-always-says-Da version, the only way you could obviate this first step would be if you expect True and False to answer all of your questions with "Ja", in which case you wouldn't be gaining any information from their answers.

I could be wrong, of course, and I haven't thought about it super hard. Would be curious to know what you have in mind.

Add the topic Riddles for more traders

@Chumchulum Thanks!

I can do it in 5, and I am pretty sure that is minimal.

Firstly, we can get an honest answer to a question Q by asking "Is it true that (you would say ja to 'is 1+1=2?') xor [Q]?". Both True and False will answer ja if Q is true regardless of their language and ja if Q is false. Call this question Honest(Q).

We can start by asking Amanda Honest("Does Bill answer Randomly"). If the answer is ja, we know that Cindy does not answer randomly. If the answer is da, we know that Bill does not answer randomly.

We start by asking the person we know does not answer randomly 2 questions to determine their favorite color, then 1 question to determine the other person who does not answer randomly. We ask that other person 1 question to determine their favorite color. We know Random's favorite color by process of elimination.

This is (probably) minimal because we have to know who random is as we must ask both deterministic people individuals what their favorite color is. Therefore, there are (3 locations of random) x (6 sequence of favorite colors) = 18 things we must distinguish between, which cannot be done with 4 or fewer questions.

@robert I believe this is directionally correct, though I think you need another question to determine whether the first non-Random person is True or False.

I would have to sit down and make a truth table to determine if your Honest(Q) is actually a working formula, but if you'll allow me to be lazy, I have another formula that I know works:

Decipher(Q) = "If I asked you, '[Q]', would you say, 'Ja'?"

When posed a question of this form, True and False both answer with "Ja" precisely when the answer to Q is "Yes", and they both answer with "Da" precisely when the answer to Q is "No", regardless of what language they speak.

From that point on, the process is as you say (with one additional question to determine if the first non-Random person is True or False):

Q1: Ask A Decipher("Is B Random?").

Based on their answer, you identify someone who is definitely not Random (because either A is Random or they answered your question truthfully). For the sake of example, let's say you determine that B is not Random. You would then go to B for Q2.
Q2: Ask non-Random B Decipher("Are you True?")
This tells you if they're True or False.
Q3: Ask non-Random B Decipher("Is A Random?")
You have now identified everyone.
Q4: Ask non-Random B Decipher("Is your favourite colour red?")

Suppose it's not red.

Q5: Ask non-Random B Decipher("Is your favourite colour blue?")

You now know their favourite colour.

Q6: Ask the other of True/False Decipher("Is your favourite colour red?")

You now know everyone's favourite colour.

@robert If you still believe that your solution works in 5 questions and that the 6th question is unnecessary, I’ll ask you to address my response and see if you can clarify why I’m mistaken about the 6th question being necessary.

@NBAP i believe 6 questions are needed. I originally misread the question

@NBAP I believe I have the sketch of a rough proof that 5 questions aren't sufficient:

I) We absolutely need to know who Random is, because if we tried without, then in the worst case they would answer in such a way that we could never get information from them AND never be able to distinguish them from either True or False, so we would only be able to get information from either False or True respectively, which isn't enough since they do not know the favourite colour of the others.

II) If we ask Q1, then Q2, then Q3, there is no way we can know everyone's favourite colour in only 2 more questions, which provide only 2 bits of information while we need to distinguish between 6 different colour-combinations.

III) If we ask Q1, then Q3 to know who Random is, and try to know the favourite colours from there with the last 3 questions without aiming to know who is True/False:
I brute-forced all possible combinations, taking symmetries into account to go faster, and it is impossible.
Indeed, we have 12 combinations True-False and RGB to consider, but don't actually care about True-False order as long as we use the fact that they know the codename of each other. However, when trying every 3 bit True/False answers that are coherent with current knowledge of the favourite colour of the others, supposing they can hear the previous answers and infer from them, even then we always end up not being able to distinguish between all cases, either because the last person we ask might not have been able to deduce the favourite colour of the others, or that we still had 3 cases before asking that last question.
Just ONE example, because I'm not going to post all the truth tables here:
(suppose without loss of generality that A is random) Ask B: decipher(you are True and ¬Blue or you are False and Blue), then ask C: decipher(you are False and Blue or you are True and Green or {RandGreen;TrueBlue;FalseRed} or {RandBlue;FalseGreen;TrueRed}), then B knows C (and A)'s favourite colour but we end up with 3 possible combinations in the worst case and so 1 last question is not enough.

Maybe there is a smarter approach to III), but actually given the constraints there aren't so many questions to consider, so you can convince yourself pretty easily with a bit of time that it is impossible.

ANd finally IV) If we ask Q1, and then try to ask other questions that mix knowledge of colours with who Random is, I believe it's just losing time, because then you can't switch to use knowledge of the other non-Random person of their personal favourite colour.

I feel like for this one, it might be indeterminately many since it's reliant on the coin toss results due to the randomizer being both linguistically and colorimetrically unknown to the other participants.

This is how you solve it :
First find random by asking as many questions as it takes to make his answer change.

Then pose the question to both of the remaining people “If I asked you ‘Is your favorite color red?’, would you say ‘Ja’?”, do all the colors. You now know which color is their's because it's the only one to have gotten the different answer.

Finally, you now know the color of random through elimination.

Please resolve @NBAP.

I don't know if I will resolve before the market close date. Maybe if virtually everyone agrees that a particular answer has no counterexample while all solutions of greater question number do have counterexamples.

However, I will tell you (in my capacity as a fellow commenter, not the market creator) that there is definitely a guaranteed solution in 6 or fewer questions.

@Velaris The issue is, it is possible (even though with probability 0) that Random never changes their answer. In fact, it might be possible that Random acts indistinguishably from True (resp. from False), except when we try to ask a question involving their favorite colour, which I believe implies that you can only find False's (resp. True's) favorite colour.

@MindcraftMax Yeah. A helpful tip would be to imagine that Random is actually antagonistic and always answers in whatever way would most frustrate your efforts.

Nevertheless, I think there is a solution.

@NBAP You can't find a solution that will work every single time, since random can answer exactly like true or false. However, the odds of it happening are the same as one divided by infinity since you can ask again and again until he changes his answer.
If you flip a coin 10*500 times it will land on heads at least once.
On the human scale the odds of it not happening are equal to infinity.
If you were to resolve the market to those oods, I would be glad.

@Velaris I have a solution that works every single time in fewer than 7 questions.

Can you share it since you just said you have one? Or is this market on whether or not the traders will find it?

@Velaris I will share the solution, though I would like to give it some time to extract more value from doubters.

@Velaris Keep in mind, if you find the truth teller, you can ask them whether one of the others is random. Thus, you never need a favorable random outcome

There are 36 possible configurations (3! codename × 3! color), so we need at least log₂(36) ≈ 5.17 bits of information to guarantee a unique solution.

But with 6 yes/no questions:

• One person is Random, so ≥1 answer is noise ⇒ max 5 usable responses.

• Each person speaks an unknown language (Ja ≠ always Yes), so unless you resolve that first, each response gives <1 bit.

• Even assuming perfect question design, the expected info gain is <5.17 bits.

So: you can’t always extract enough reliable info from just 6 questions.

@Velaris Your goal is to determine the favourite colour of True, the favourite colour of False, and the favourite colour of Random – you don't need to find out who is who, it seems.

I believe there should be an anwser "no amount of questions can solve this riddle", which is different from ">6".

@MindcraftMax I think you can brute force it. You can find random easily enough with a lot of questions, then you can use and sometimes not use meta questions to find who lies, and then you can find out which languges they speak similiarly. It just takes a lot more then 6.