“If all three of the answers are incorrect, they pick one of them with probability one-third.”

Aren’t there more lies that they would be smart enough not to tell, though? Out of the possible 10 combinations

A:1 B:2 C:3 = must be 0 or 1

A:1 B:3 C:3 = must be 0 or 1

A:2 B:2 C:3 = must be 0 or 2

A:3 B:3 C:3 = must be 0 or 3

A:2 B:2 C:1 = 0,1 or 2

A:3 B:3 C:2 = must be zero

A:3 B:1 C:1 = must be zero

A:2 B:2 C:2 = must be zero

A:2 B:1 C:1 = must be zero

A:1 B:1 C:1 = must be zero

half will never be said by smart liars, so given A:2 B:2 C:1, depending on if there are really 0, 1, 2 or 3 there will be 4 different odds for their choice.

If there are 0, the odds they picked A:2 B:2 C:1 would be 1/5.

If there are 1, the odds they picked A:2 B:2 C:1 would be 1/3.

If there are 2, the odds they picked A:2 B:2 C:1 would be 1/2.

If there are 3, the odds they picked A:2 B:2 C:1 would be 0.

(I wonder what that would look like at 4 or 5 villagers, or if you asked “How many liars among you?”)

Given that the odds of meeting a truthteller are 1/2, the odds of meeting 0 is 1/8, 1 is 3/8, and 2 is 3/8(and 3 is also 1/8, but we know it’s not 3 for A:2 B:2 C:1).

From there, you can figure out the odds of having both met the necessary number of truthtellers and the remaining liars having constructed A:2 B:2 C:1

The odds you get A:2 B:2 C:1 after meeting 0 is 1/40, the odds you get it after meeting 1 is 1/8, and the odds you get it after meeting 2 is 3/16. Your guess is about 18.06% better than before asking. The odds went from the same that there are 1 or 2(3/8), to 50% more that there are 2 than the next best guess, 1 ((so 15/27 there are 2, 10/27 there are 1, and 2/27 there are 0)). A:2 B:2 C:3 is 7.5 times more likely to be 2 than 0, and the remaining combinations that allow for 1 are 5 times more like to be 1 than 0(same as the relationships for A:2 B:2 C:1). A:3 B:3 C:3 actually gives you the least information, giving you even odds on 0 or 3.

At least I think that’s the solution… Does the order they speak at/you met them matter? I’m not even sure it would change the math too much if it did matter,(you have to wait to have all the information to think about the problem, and the odds are the same regardless of the order; both meeting and listening are statistically independent. You would only be adding more random less ambiguous lies, making it easier to guess, if anything), but I clearly don’t know set theory, or what the hell I’m talking about. I just gave a presentation on the Monthy Hall Problem in high-school once and play a lot of poker… Sorry for all the scrolling…

]]>There is no overlap between your post and my post.

]]>http://hexxponies.wirehound.com/blog/2013/12/24/knights-knaves/ ]]>

I was planning to move in the directions you are suggesting.

]]>1) go to an island where inhabitants try to make the life of the interviewer as hard as possible – a) interviewees can group themselves as they wish, b) (independent feature) define if answers can be thought as a set or a sequence, for example, if C is the last to speak and is a liar then he/she speaks a text which makes the solution of the interview problem as hard as posssible or even impossible.

2) Try to find complexity measures of solutions for different cases and classify answer sets. ]]>

1) Are there answer sets which are impossible (do not admit a solution)? Find all of them.

2) Describe answer sets which lead to the same answer.

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