In a certain island, one-third of the native people are liars who always lie, one-third are knights who always tell the truth, and one-third are knaves that is, people who strictly alternate between speaking the truth and telling a lie, irrespective of order. The chances of encountering any one of the three natives on a road on the island are the same.

Four friends Art, Ben, Cal and Dan – who are natives of this island got the top four ranks in a certain quiz The following statements are made by each of Art, Ben, Cal and Dan.

Art
1. Exactly two of us are knights.
2. Ben got the first rank and Cal got the third rank.

Ben
1. Dan got the fourth rank.
2. Exactly one of us is a knave.

Cal
1. The absolute difference between Ben's rank and mine is 2.
2. Dan is not a liar.

Dan
1. Exactly one of us is a liar.
2. I am a knave.

Assuming no ties, determine the probability that:

(i) The absolute difference between Cal's rank and Dan’s rank is 1.
(ii) Exactly three of the four friends are liars.
(iii) Exactly two of the four friends are knaves.
(iv) At least one of the four friends is a knight.

*** For Art’s second statement - assume the entire statement is a lie if the whole statement or any of its parts thereof is false. For example- A2 is false if in reality Ben got the first rank and Cal got the fourth rank.

(In reply to re: One small step but still unclear. by Steve Herman)

Agreed Steve, and I think it may even be more complicated than that.  Before even considering the distribution of test score rankings, I'm still confusing myself with how to weight the 11 configurations of Knight-Knave-Liar.

It's sort of like the Monty Hall problem - at the outset each door has a probability of 1/3, but once you remove one door, the remaining two are not equiprobable. 

So for example, prior to the statements being made, there's a 1/3 chance that A is a Knight.  Once the statements eliminate 70 of the 81 configurations, how do we assign all that probability to the remaining 11?  In only 1 of the 11 is A a Knight, so does that row get the whole probability of 1/3 that was originally assigned to all the rows where A is a Knight?  Etc. 

Part of me thinks the remaining 11 should be equiprobable but another part of me thinks that's a mistake, and I can't convince myself which is right.

Posted by tomarken on 2014-07-16 12:06:54