Suggestion for Murderer/Sheriff selection

Discussion in 'MurderMystery General Discussion' started by justnrik, Jan 7, 2019.

  1. justnrik

    justnrik New Member

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    Here I have a pseudo-algorimth for Murderer/Sheriff selection:

    Let players vote if they want to be a Murderer, Sheriff or Innocent(default)

    If 0 Players want to be Murderer/Sheriff, it will be picked randomly as it currently is.
    If 1 Player wants to be Murderer/Sheriff, (s)he will be picked.
    If 2 or more Players want to be Murderer/Sheriff, it will be picked from them accordingly to how often they've been Murderer/Sheriff. The more often the lower chance.

    Example:

    5 Players want to be Murderer. And they've been murderer this amount of times
    (Player | Count):

    Player 1 | 5 times
    Player 2 | 3 times
    Player 3 | 3 times
    Player 4 | 4 times
    Player 5 | 2 times

    The sum of times they've all been murderer is 5+3+3+4+2 = 17 times.

    Player 1 has been Murderer most of times, hence, he has the lowest chance to be picked.
    Player 2 and 3 have the same chance to be picked.
    Player 4 has low chance to be picked
    Player 5 has the highest chance to be picked.

    totalChance = sum * (nPlayers - 1) = 17 * (5 - 1) = 17 * 4 = 68
    chancePerPlayer = (sum - count) / totalChance

    so each player wil have a chance of:

    Player1 | (17 - 5) / 68 = 12 / 68 = 17.65%
    Player2 | (17 - 3) / 68 = 14 / 68 = 20.59%
    Player3 | (17 - 3) / 68 = 14 / 68 = 20.59%
    Player4 | (17 - 4) / 68 = 13 / 68 = 19.12%
    Player5 | (17 - 2) / 68 = 15 / 68 = 22.06%

    Also, for any player that has never been murderer, the formula gets reduced to:

    1 / (nPlayers - 1) which in above's example would be 1 / 4 = 25%

    Condition: If totalChance = 0 (basically, all players who applied have never been Murderer/Sheriff) then they will be picked randomly from the list of users who applied.

    Curious things of this algorimth:

    1) If all players except 1 have never been Murderer/Sheriff, the player who has been murderer is basically excluded.

    2) This formula won't work if all players have never been Murderer/Sheriff because of Division by Zero. Hence, this condition has to be handled separately

    3) Generally, the probabily is very fair with all players, so if there are 2 players and one has been murderer 10 times and the other only 2 times, then the chance is:

    Player1 | (12 - 10) / 12 = 2 / 12 = 16.67%
    Player2 | (12 - 2) / 10 = 10 / 12 = 83.33%

    Another example assuming a player with 12 times, another with 8 times and another with 0:

    Player1 | (20 - 12) / 40 = 8 / 40 = 20%
    Player2 | (20 - 8) / 40 = 12 / 40 = 30%
    Player3 | (20 - 0) / 40 = 20 / 40 = 50%

    As you can see, the player that was never a Murderer/Sheriff has 50% of chance, while the others still have a 20 and 30% chance respectively which is good and fair.
     
  2. GingerlyEnderly

    GingerlyEnderly Notable Member

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  3. Phoe

    Phoe Notable Member Emerald Member

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    I wish this is how it worked, but it’s always 1/(number of entered players)
     
  4. dyingful

    dyingful Notable Member Emerald Member

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    lmao lowkey wish this was a thing.
     
  5. Bxnjamin

    Bxnjamin Notable Member Emerald Member

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    As cool as I think this would be, I kind of like it randomized.
     
  6. Firefox1318

    Firefox1318 Notable Member Emerald Member

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    i can't math on holidays xd
    i'd rather stick to the current version
     
  7. LightningFire96

    LightningFire96 Notable Member Gold Member

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    I don't like the idea
    But I applaud you for doing the math
     
  8. Crystal Ren

    Crystal Ren Notable Member

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    Great idea- I think it coupd be inspiring for any new games to come.
    However, MurderMystery is fun as it always is randomized. If it weren't, everyone would want to be Murderer; that would be too predictable and by then all the Innos will know who's Murd.
    Nice suggestion!
     

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