Hello Cindy,
For this type of problem it is a good idea to count the number of rounds you need to achieve a fair tournament. You have 6 players and want everyone to be partnered once so you have 6x5/2=15 possible partnerships. As 15 is an odd number and as each round involves 2 partnerships it will not be possible to balance the schedule for the partnered-one-time scenario. Doubling things up, that is allowing 15 rounds where everyone is partnered twice, will work.
So at the risk of turning your seasonal family get together into a major card playing tournament I suggest the following schedule:
Euchre Game Watching the little ones.
N S E W
(5 1) v (6 3) (2 4)
(1 4) v (2 5) (3 6)
(6 3) v (4 2) (1 5)
(6 4) v (1 5) (2 3)
(1 3) v (2 6) (4 5)
(4 2) v (5 3) (1 6)
(5 2) v (6 1) (3 4)
(3 4) v (5 6) (1 2)
(3 2) v (1 4) (5 6)
(2 6) v (4 5) (1 3)
(3 5) v (2 1) (4 6)
(1 6) v (4 3) (2 5)
(6 5) v (2 3) (1 4)
(1 2) v (4 6) (3 5)
(5 4) v (3 1) (2 6)
The schedule above has the following properties.
All players partner each other exactly twice.
All players oppose each other exactly four times.
The 4 players sitting at the table are never the same. There are 15 ways of choosing 4 players from 6 and each possibility is one of the rounds above.
All partnerships play once on the left (N S) and once on the right (E W).
No player plays in more than three consecutive rounds.
And to confuse the little ones...
The pair of players sitting out the round are never the same, all 15 ways of choosing 2 players from 6 occur.
No player watches the little ones twice in succession.
If you use the schedule then let me know how things turn out.
Best wishes for the season,
Ian.
Thursday, December 15 2005, 06:08 pm
