4 Player Games, 14 or 11 People

[smtk]

Hello!

I want to run a CPU Round Robin Tournament for a Mario Party game, as I recently finished doing one for Mario Party 7, which can be found here (and here is a reddit post explaining the spreadsheet).

For those who don't know, Mario Party is a game with 4 players. Each Mario Party game has a different number of characters, but the Mario Party game I'm interested in doing next is either Mario Party 6 or Mario Party 8, which have 11 and 14 characters respectively.

Of course, since neither game has a roster that is divisible by 4, that means that there's going to be people sitting out every round, but that's ok.

What I need to know is: For either game, how do you organize a Round Robin tournament so that every character encounters every other character an equal amount of times? And how many games would it take to get perfect balance?

I'm fine with playing as many games as needed, I just need to organize it all.

Also also, while Mario Party 6 or 8 are the games I'm leaning most towards, that doesn't mean I'm not considering the other Mario Parties either, so if you want to organize this same type of tournament for the other games, here's what you need to know:

Mario Party 1-2 has 6 characters
Mario Party 3-4 has 8 characters 
Mario Party 5 has 10 characters
As mentioned before, Mario Party 6 has 11 characters and Mario Party 8 has 14 characters, and I already did Mario Party 7.

[Ian Wakeling]

To work out how many rounds for perfect balance you need to count pairs in two different ways.  Let's take Mario Party 8, with 14 characters. First count the total number of possible character pairings, this is  (14*13)/2 = 91.   Now count the number of pairs that are covered by one round of play,  there will be 3 games with 4 players in each, and 2 players with a bye.  In one game of 4 players there are (4*3)/2 = 6 pairs covered, so one round of 3 games covers 18 pairs. This is actually optimally bad, as there are no common factors betwen 91 and 18.  So you will need 91 rounds of play, at which point it should be possible to arrange that every character meets every other character exactly 18 times.

The situation is similar for 11 players and 2 games per round - you will need 55 rounds of play.  I think I could make both schedules, but I am thinking this is simply too many rounds.

[smtk]

Ah, I see...thanks for the help!
Do the other Mario Parties make for a fewer amount of rounds, and if so, can you try making the schedules for them?

[Ian Wakeling]

Mario Park 3/4 with 8 characters is easy to deal with as there are no byes.   Use the whist schedule for 8 items and just ignore the pairs.  It will give 7 rounds and each possible pair of characters 3 times in a game.  Richard's generator is available on this site.

Mario Park 5 with 10 characters is more interesting.  A 15 round solution with pairs of characters 4 times in a game is possible. For example:


  ( 8  4  1  3) ( 5  7 10  6)    9  2
  ( 9  5  2  4) ( 1  8  6  7)   10  3
  (10  1  3  5) ( 2  9  7  8)    6  4
  ( 6  2  4  1) ( 3 10  8  9)    7  5
  ( 7  3  5  2) ( 4  6  9 10)    8  1

  ( 9  5 10  8) ( 7  3  4  1)    6  2
  (10  1  6  9) ( 8  4  5  2)    7  3
  ( 6  2  7 10) ( 9  5  1  3)    8  4
  ( 7  3  8  6) (10  1  2  4)    9  5
  ( 8  4  9  7) ( 6  2  3  5)   10  1

  ( 8 10  1  2) ( 6  4  3  9)    5  7
  ( 9  6  2  3) ( 7  5  4 10)    1  8
  (10  7  3  4) ( 8  1  5  6)    2  9
  ( 6  8  4  5) ( 9  2  1  7)    3 10
  ( 7  9  5  1) (10  3  2  8)    4  6


The byes are in the last two columns and are arrange so that everyone has one bye each in the first 5 rounds, the middle 5 rounds and the last 5 rounds.