Three of a crime Card game

Three of a crime Card game

[gattamaneni abhiram]

Three-of-a-crime is a simple logic game for up to 3 players. There are 7 different 
criminals. The computer randomly chooses three of these (the "perpetrators"), but 
doesn't tell the players which are chosen. The computer then puts down three random 
criminals. 0, 1, or 2 of these may be the actual perpetrators. The computer also tells 
the player how many (but not which) of the three criminals are perpetrators. The 
players may either guess which three criminals are the actual perpetrators or they may 
pass. If a player guesses wrong, she is out of the game and the other players continue. 
If no player chooses to guess, the computer puts down another three randomly chosen 
criminals (0, 1, or 2 of which may be actual perpetrators) and tells the players how 
many (but not which) of these are actual perpetrators. Players can again use logic to 
deduce the three actual criminals and may guess. Play continues until some player 
guesses correctly or until all players have guessed incorrectly. 

I am not able to deduce the logic. Can anyone please help me with the logic? 

Re: Three of a crime Card game

[Dennis Lee Bieber]

On Tue, 26 Mar 2013 16:18:10 -0700 (PDT), gattamaneni abhiram
<abhiram.gattamaneni@gmail.com> declaimed the following in
comp.lang.ada:

> Three-of-a-crime is a simple logic game for up to 3 players. There are 7 different 
> criminals. The computer randomly chooses three of these (the "perpetrators"), but 
> doesn't tell the players which are chosen. The computer then puts down three random 
> criminals. 0, 1, or 2 of these may be the actual perpetrators. The computer also tells 
> the player how many (but not which) of the three criminals are perpetrators. The 
> players may either guess which three criminals are the actual perpetrators or they may 
> pass. If a player guesses wrong, she is out of the game and the other players continue. 
> If no player chooses to guess, the computer puts down another three randomly chosen 
> criminals (0, 1, or 2 of which may be actual perpetrators) and tells the players how 
> many (but not which) of these are actual perpetrators. Players can again use logic to 
> deduce the three actual criminals and may guess. Play continues until some player 
> guesses correctly or until all players have guessed incorrectly. 
> 
> I am not able to deduce the logic. Can anyone please help me with the logic? 

	Given your description, this is NOT a "simple logic game"... There
is, in your description, NO LOGIC to apply. That is, there are NO clues
provided to the players.

	Consider:

7 candidate suspects
3 of the 7 are guilty (perpetrators)
3 of the 7 are listed to the players (these three may or may not be in
the list of guilty)
"computer also tells the player how many": OKAY, ONE CLUE... but if the
computer says "0", then there is nothing for a player to do. Similarly,
if the computer says "3", then all of the listed names are guilty --
again, nothing for a player to do.

Therefore, the only "play" occurs if the computer says 1 or 2 of the 3
names is in the guilty list.

With three names, and one candidate, the first two players are going on
sheer luck (if they guess wrong, the third player picks the third name
-- win). And with two candidates, you again have just three
possibilities: A & B, A & C, or B & C... So again the first two players
are just guessing, and if they don't guess right, the third player picks
the only remaining choice and wins.



-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

Re: Three of a crime Card game

[gattamaneni abhiram]

On Tuesday, 26 March 2013 23:10:31 UTC-4, Dennis Lee Bieber  wrote:
> On Tue, 26 Mar 2013 16:18:10 -0700 (PDT), gattamaneni abhiram
> 
> <abhiram.gattamaneni@gmail.com> declaimed the following in
> 
> comp.lang.ada:
> 
> 
> 
> > Three-of-a-crime is a simple logic game for up to 3 players. There are 7 different 
> 
> > criminals. The computer randomly chooses three of these (the "perpetrators"), but 
> 
> > doesn't tell the players which are chosen. The computer then puts down three random 
> 
> > criminals. 0, 1, or 2 of these may be the actual perpetrators. The computer also tells 
> 
> > the player how many (but not which) of the three criminals are perpetrators. The 
> 
> > players may either guess which three criminals are the actual perpetrators or they may 
> 
> > pass. If a player guesses wrong, she is out of the game and the other players continue. 
> 
> > If no player chooses to guess, the computer puts down another three randomly chosen 
> 
> > criminals (0, 1, or 2 of which may be actual perpetrators) and tells the players how 
> 
> > many (but not which) of these are actual perpetrators. Players can again use logic to 
> 
> > deduce the three actual criminals and may guess. Play continues until some player 
> 
> > guesses correctly or until all players have guessed incorrectly. 
> 
> > 
> 
> > I am not able to deduce the logic. Can anyone please help me with the logic? 
> 
> 
> 
> 	Given your description, this is NOT a "simple logic game"... There
> 
> is, in your description, NO LOGIC to apply. That is, there are NO clues
> 
> provided to the players.
> 
> 
> 
> 	Consider:
> 
> 
> 
> 7 candidate suspects
> 
> 3 of the 7 are guilty (perpetrators)
> 
> 3 of the 7 are listed to the players (these three may or may not be in
> 
> the list of guilty)
> 
> "computer also tells the player how many": OKAY, ONE CLUE... but if the
> 
> computer says "0", then there is nothing for a player to do. Similarly,
> 
> if the computer says "3", then all of the listed names are guilty --
> 
> again, nothing for a player to do.
> 
> 
> 
> Therefore, the only "play" occurs if the computer says 1 or 2 of the 3
> 
> names is in the guilty list.
> 
> 
> 
> With three names, and one candidate, the first two players are going on
> 
> sheer luck (if they guess wrong, the third player picks the third name
> 
> -- win). And with two candidates, you again have just three
> 
> possibilities: A & B, A & C, or B & C... So again the first two players
> 
> are just guessing, and if they don't guess right, the third player picks
> 
> the only remaining choice and wins.
> 
> 
> 
> 
> 
> 
> 
> -- 
> 
> 	Wulfraed                 Dennis Lee Bieber         AF6VN
> 
>         wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/


 Thank You for the help. Now I am onto the coding part!!