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I came to the same conclusion as Neremanth, but via a more concise/elegant argument. Step 1: which door leads to golds Step 2: bad news


Ask this: If you hear "Yes", then the one you asked is the cursed one. If you hear "No", then the one you didn't ask is the cursed one. Original answer (invalidated by an edit to the question after this was written):


The worst case has the potential to reach any positive, finite number. I will demonstrate this with a counter-example using the 11-question strategy. Let's label the guards: T always tells the truth. F always lies. Y always says yes. N always says no. R gives a random answer. State 1: State 2: State 3: State 4: Now what? Since the worst possible case is ...


Turned out to be the same answer, but followed by a proof: Prove that a better answer isn't possible:


I think you can do this with 9 questions first irrespective of the outcome Now to figure the remaining people so the final number of questions you need to ask is


Truthteller: Liar: This depends on the fact that So the truthteller says an obvious truth and the liar an obvious lie.


The existing answer from Deusovi is good, but it can also be solved in a way that will, one third of the time, only need 2 questions, and the rest of the time still be solved with a third question. I'm not saying that's better, but it might be interesting anyway... Please excuse my formatting troubles, it's my first answer on here. Hint 1st Question: I've ...


This solution works no matter what happens in the case of paradoxes or questions with ambiguous answers. First, ask: With this, A note: this problem is actually easier than it seems, because

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