Mark,

In the archives there is a schedule where 16 golfers play with each other exactly once over 5 rounds. (Search the archives for golf 16.) One way to go is to repeat it three times. For a bit of variety (to mix up the foursomes), you could use the schedule once, renumber the players and then use it again, and then renumber yet again and use the schedule over. It is also possible to change the day order within the 5-round schedule and not hurt anything.

Have fun!
Victoria

Mark responded

Thank you for your answer. But I failed to specify in my question that this is a golf league where each golfer must play ounce against the other 15. Therefore, I cannot use your suggestion to use the 5 day schedule 3 times. I can’t figure out how to rearrange the names so that each golfer plays against each of the other 15 one time.
I do have a schedule that accomplishes the playing against each other player ounce but it does not provide an even distribution playing with each golfer.

For example, I will play in the same foursome with some of the players only ounce and others I play in the same foursome as many as 6 times.

Is there a way to calculate this?

Thank you

Mark

Mark,

Maybe I still don't understand. Is it that each player in a foursome has exactly one opponent in that foursome? Or is it that every pair of players is together in a foursome exactly once? Or something else?

The second outcome is impossible. The best that can be done is that every pair are together exactly three times. After 5 rounds, if no pair are together twice, each player has played with 3 x 5 = 15 others.

The first outcome is possible with the schedule that was suggested. If the foursome is a, b, c, d, then the first time they are together a plays b and c plays d, the second time a plays c and b plays d, and the last time a plays d and b plays c.

If neither of these is what you mean, please let try again and I will try to help.

--Victoria