Consider the game Dethy Mafia. The full rules can be found here, but this is all you need to know to solve this puzzle:

Each of the five roles "Sane Cop", "Insane Cop", "Paranoid Cop", "Naïve Cop", and "Mafia Goon" are given randomly to one of the five players. All players are told whether they're a cop or the Mafia Goon, but not their sanities. On the first night, the Mafia Goon does nothing, and each of the cops secretly chooses another player to investigate. The Sane Cop always gets the right answer, the Insane Cop always gets the wrong answer, the Paranoid Cop always gets a guilty result, and the Naïve Cop always gets an innocent result. In the morning, the players all share and discuss their results and try to deduce who's secretly the Mafia Goon (who, not having a real result, will have made up a fake one).

The puzzle: Alice, Bob, Carol, Dave, and Eve are all playing a game of Dethy Mafia. The roles were distributed as follows:

  • Alice - Naïve Cop
  • Bob - Insane Cop
  • Carol - Paranoid Cop
  • Dave - Sane Cop
  • Eve - Mafia Goon

Unfortunately for Eve, based on only the investigative results from the first night, the other players were all able to deduce for certain that she was the Mafia Goon on the first day. Can you come up with a set of investigative choices and results that would have led to this happening?

  • $\begingroup$ We just need to find one setting that works? $\endgroup$
    – justhalf
    Aug 21, 2022 at 6:47
  • $\begingroup$ @justhalf Correct. $\endgroup$ Aug 21, 2022 at 6:57

2 Answers 2


At any point:

only two players can truthfully claim the same result on the same player. If three do, the other two are innocent; since the conflicting results are on one of the two confirmed innocent players, they will confirm which of the Sane and Insane cop is within the conflict.

For example:

Alice (Naively) claims Carol is innocent.
Dave (Sanely) claims Carol is innocent.
Eve (as the Goon) claims Carol is innocent.

These results prove that

between Carol and Bob, there is an Insane and a Paranoid cop.
The Insane Cop can get an Innocent result only on the Goon; the Paranoid one will always have a guilty and their target is irrelevant.

Thus, to prove Eve the Goon:

Carol (Paranoid) claims anyone is guilty. Bob (Insane) claims Eve is innocent.

Similar logic can also be done in reverse:

since Carol is the sole Guilty result, she must be Paranoid (and therefore, not the Goon)
thus, since there are three innocent results on her, neither is Insane;
thus Bob's Insane result is actually damning for Eve.

Edit: changed a word to remove self-targeting cases, since the question was edited to disallow such.

  • $\begingroup$ Were these meant to be marked down as spoilers? $\endgroup$
    – Wyck
    Aug 21, 2022 at 18:16
  • $\begingroup$ "if the conflicting results are on one of the two confirmed innocent players" When is that ever not the case? $\endgroup$ Aug 21, 2022 at 18:48
  • 1
    $\begingroup$ J S-R M: Consider the first example again, except that Bob (Insane) claims Alice is guilty. In that scenario, the cops can't deduce which of Alice/Dave/Eve is the Goon (Alice can rule out herself but not Dave, and Dave can rule out himself but not Alice). $\endgroup$
    – Ed Murphy
    Aug 21, 2022 at 18:57
  • $\begingroup$ @EdMurphy While you would indeed not be able to prove who the Mafia Goon is in that situation, both the premise and conclusion of the sentence I'm referring to still seem to be true. $\endgroup$ Aug 21, 2022 at 19:28
  • 1
    $\begingroup$ Not if self-targeting is legal. (The question was edited: "a player" -> "another player".) If the Goon claims a self-target, the two are switched. $\endgroup$
    – Braegh
    Aug 21, 2022 at 20:13

Braegh's solution describes a situation in which the goon could be determined on the first day. However, it makes some assumptions about Eve's play:

a) Eve reveals her "result" first, with no knowledge of the other players' results, or
b) Eve chooses her result intentionally (or randomly) to allow town to identify her.

For my solution, I will assume that Eve chooses when to reveal her check, and that Eve will attempt to choose a result which does not incriminate her.

I contend that if Eve plays optimally, the situation described cannot exist, and the puzzle has no solution.

It is actually very simple for Eve to thwart all attempts to uniquely identify the goon. She may choose to reveal her fake check either third, fourth, or fifth. As soon as she sees two checks with the same result, she needs only to claim a check on the same person as one of the previous checks, with an opposite result.

This allows Eve to claim that she and the person she is contradicting are the sane and insane cops. Furthermore, it cannot be determined which of Eve and her contradictor are the sane cop, so no other information is gained from the result of their checks. Since Eve waited until there were two checks of the same type before claiming one came from the (in)sane cop, there will be at least one check of each type among the other three. Thus Eve claims the other three checks come from the naïve cop, paranoid cop, and mafia goon; these checks also give no information about their subjects.

  • $\begingroup$ This strategy doesn't always work. If all of the real cops checked Eve, she'd have to claim a check on herself to follow it, but real cops can't check themselves, so this would immediately out her as a liar and thus the Mafia. $\endgroup$ Aug 21, 2022 at 18:47
  • $\begingroup$ I do see where you're coming from, and you could patch your strategy to account for that case, but note that it generally wouldn't work in real games, since the assumption that Eve chooses when to reveal her check wouldn't hold. In real games, players generally demand that either everyone makes their claims simultaneously, or in a predetermined order, and treat refusal to go along with this as evidence of guilt. $\endgroup$ Aug 21, 2022 at 18:51
  • $\begingroup$ @JosephSible-ReinstateMonica In my experience players are usually allowed to self-target (even though it's typically suboptimal) unless explicitly told otherwise. Granted, I haven't played with cop sanities so perhaps self-targeting would be broken. Regarding your other point, I've found that it's generally quite difficult to get everyone to agree to a plan, especially day 1. And in any case, even if demanding to go last would be admission of guilt, such certainty would not be "based on only the (claimed) investigative results". $\endgroup$
    – ash4fun
    Aug 21, 2022 at 19:37

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