Card game 13 players 1 v 1 v 1 v 1 v........

[spork]

Hi there,
I will be running a Wizard card game each week, there are no teams, and you play for yourself.  I wanted each player to play each other player once in the night. (Playing a person at the same table will constitute playing that person.)  Each night a winner (by accumulative points) will be awarded.  We can play 3, 4 or 5 to a table.  To further confuse things, each night the number of players will probably change.  This is the inaugural tournament and I expect 13 players but it could change next week.   I can use a ghost to make the numbers work if need be.
Just to get by this week, is there a solution for the 13?
Thanks in advance for any help.

[Ian Wakeling]

The closest balanced solution is when there are 15 players who play 3 to a table.  For example see the 7 round schedule here.  Use this as a starting point, and when 13 or 14 players turn up, simply delete one or two players from the schedule.  With 13 players, this will mostly leave two tables of two, which can be merged together into a table of 4, for the single round where it leaves a table of one, then this player can be added to any of the other tables of three.  With 14 players you will have to make tables of 5.

Hope that helps.

[spork]

Ian,
Thank you so much.  I apprecate your response.