9
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As in this puzzle, there are three different kinds of people: truth-tellers (always telling the truth), liars (always lying) and jokers (who may either tell the truth or lie). Based on these aspects of their personality, truth-tellers are considered more reliable than both jokers and liars, and jokers are considered more reliable than liars.

We meet Amelia, Bernard, Cicero, Daisy, Evelyn, and Frank. We know that among them there are exactly two truth-tellers, two jokers and two liars. Each of them says something about their kind.

  • Amelia: "I'm like Bernard"
  • Bernard: "I'm like Cicero"
  • Cicero: "I'm not like Amelia"
  • Daisy: "I'm more reliable than anyone else in our group, except for Evelyn"
  • Evelyn: "If my sentence is true, then Daisy's sentence is true"
  • Frank: "Amelia is a liar"

Which pairs of people are of the same kind?

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  • 5
    $\begingroup$ Actually, the liars are technically just as reliable as the truth-tellers --they are reliably wrong. It's the jokers who are unreliable... $\endgroup$ – Chris Sunami Jun 6 at 16:07
4
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I will assume that no one can make nonsense statements.

Examine E's statement. The sentence can be false only if the antecedent is true (the sentence is true) and the consequent is false (D's statement is false). This is contradictory. So, we will assume this statement is true.

This means that D's statement is true. This means that D is a Truth-teller.

This means that E is also a Truth-teller. We have found the two Truth-tellers.

1) Assume A is a Joker. If this is the case, then C cannot be a Liar or his statement would be true. A and C would be the Jokers, making B and F the Liars. Their statements would both be false, so this checks out.

2) Assume that A is a Liar. This means that B cannot be a Liar and must be a Joker. F's statement is true, so he must be the other Joker. C would then have to be the other Liar. His statement is false, so this seems to check out as well.

This leaves me with two solutions. Both give the same pairings: D and E are pairs (unambiguously Truth-tellers), A and C are pairs (either both Liars or both Jokers) and B and F are pairs (again either Liars or Jokers).

I have capitalized the terms here to make a distinction. A Joker can make false statements. If I hear X, who is a Joker, make a false statement, then I can say: "X is a liar." This is true even though X is not a Liar per se. If we assumed that F's statement means simply that A's statement is not true, then case 2) would be the only valid solution and we could tell everyone's orientation. I thought at first that this was the intention but then realized that in fact the two cases provide the one answer that was asked for. An extra twist that I certainly appreciated and should have called out in the first place.

Interesting puzzle!

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  • 1
    $\begingroup$ Great! You have two solutions, both correct, but the answer to the question is one. Can you please spell it out, so I'll accept your answer? $\endgroup$ – TheDude Jun 6 at 8:09
  • $\begingroup$ @TheDude I have updated it to spell out my thinking and the actual answer. Thanks! $\endgroup$ – Hugh Meyers Jun 6 at 8:29
3
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Solutions (all names here and below are abbreviated to initial letters):

4 different solutions, see below.

Explanation

Firstly, determine the pair of truth-tellers. A cannot tell the truth because if she did, she must be like B (so B is also a truth-teller), and B must be like C, so we had 3 truth-tellers instead of 2. So, if B is the truth-teller, so C must be too (first possible pair is BC). Now, note that D's sentence can be true if and only if D and E are both truth-tellers (the same can be said about E). So, the second possible pair is DE, and the remaining one (since D and E either are both truth-tellers or not) is CF.
Let's consider these variants one-by-one. If BC are truth-tellers, then A, D and E must all lie, and F's sentence can be either true or false. So, they can be liars or jokers in any combination, as long as we get 2 of each, except AF cannot be both liars, because F tells the truth in this case (2 solutions).
If DE are truth-tellers, then AB, BC, AF and CF cannot be both liars, because one of them should tell the truth in these cases. So, remaining pairs for liars are AC and BF (2 solutions).
Finally, if CF are truth-tellers, then A and B must be liars, and D and E both lie (and are jokers). But A is like B, and tells the truth in that case.

Verification

1) BC are truth-tellers, AD are liars, EF are jokers. A lies and is not like B, B tells the truth and is like C, C tells the truth and is not like A, D lies and is less reliable than all except A, E lies (her sentence is false, so D's can be any), F tells the truth and A is a liar. No contradictions.
2) BC are truth-tellers, AE are liars, DF are jokers. A lies and is not like B, B tells the truth and is like C, C tells the truth and is not like A, D lies and is less reliable than B and C, E lies (see above), F tells the truth and A is a liar. No contradictions.
3) DE are truth-tellers, AC are liars, BF are jokers. A lies and is not like B, B lies and is not like C, C lies and is like A, D tells the truth and is more reliable than all except E, E tells the truth (because D does), and F tells the truth as above. No contradictions.
4) DE are truth-tellers, BF are liars, AC are jokers. Same as above, except F lies and A is not a liar. No contradictions.

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  • $\begingroup$ You're close, but I'm sure you can rule out some of the solutions you've come up with so far! $\endgroup$ – TheDude Jun 6 at 7:29
  • $\begingroup$ @TheDude I've added the verifications. All 5 solutions seem to be correct and cannot be ruled out. Upd: Oh, yes. 1 of the 5 is indeed flawed $\endgroup$ – trolley813 Jun 6 at 7:52
  • $\begingroup$ Your 4 solutions may seem ok at first, but once you look at what people say, some of them are not applicable $\endgroup$ – TheDude Jun 6 at 8:13

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