Paul,

There are 2 threesomes per round, for a total of 6 players. That means 24 players have been in threesomes over 4 rounds, so at least 6 players must be in threesomes twice. I don’t know if it is possible to achieve that number.

Here is a possible solution that’s not terrible. Start with the schedule here for 20 golfers over 4 rounds, http://mathcentral.uregina.ca/QQ/database/QQ.02.06/mona1.html. The teams that never play together are those in the same column of the array. So delete teams 16 and 20 from this schedule (and renumber) and you will get a schedule with two 3s and two 4s in which no pair or players is together more than once.

This schedule has 9 players in threesomes twice, and 3 players never in them. The number of players involved is too large to search all possibilities with a computer.

—Victoria