Help! Three tables of 4 players, 7 games

[Jerry_G]

I'll be in a Mah Jong tournament in about 1.5 weeks. There will be a total of 12 people playing, and a total of 7 games.  There are no teams, just 12 individual players, simultaneously playing at 3 tables of 4 people.  Each person will just keep his or her individual score sheet throughout the 7 games. The idea is that at the end of the 7 games, everyone will have played with all of the 11 other people as equal a number of times as possible. I've never tired to figure this type of thing out before and nothing I try is working. It seems as if every one should end up playing with 10 of the other 11 people twice and 1 of the 11 other other people once.

Can anyone give me a schedule for this?  Thank you very much!

[Ian Wakeling]

Please have a look at this thread as it describes a schedule that you might consider, also it has a link to another thread that attempts to explain why it's not possible have the properties that you described - 10 opponents twice and the last opponent once.

If you must have 7 rounds of play in tables of 4, then I think the schedule below is the best you can do:

There are 66 possible pairs of players  (12*11)/2=66, in the schedule:

9 pairs occur once
54 pairs occur twice
3 pairs occur three times.

   7   4  11   5
   1   2   8   6
  10   3   9  12

   4   7   1   9
  10   8   2   5
   6   3  12  11

   1  12   5   6
   4   3   9   2
  11   7   8  10

   6   5  11   9
   3   8   1   7
  12   2  10   4

   9   2   6   7
   3  10   1   5
  12   4  11   8

  12   5   8   9
  11   3   1   2
  10   7   4   6

   1  11  10   9
   5   7   2  12
   6   3   4   8