Steve,

The table you have is from Victoria's response to Evan. I am going to add a row to the table to indicate the golfers. I am naming the golfers S (that's you) and A, B, C, D, E, F and G.

Each day the golfers are divided into two foursomes called 0 and 1. Each player determines which foursome he or she belongs to by reading down his or her column. Steve, reading down your column you see all zeros so you are in foursome 0 every day. Look now, for example at golfer D. His column reads 1, 0, 0, 1, 1, 1, 0. Thus on Day 0 he is not in your foursome, on Day 1 and Day 2 he is in your foursome. On Days 3, 4, and 5 he is not in your foursome and on Day 6 he is in your foursome.

I hope this explains Victoria's table. Write back if you need more assistance.

Harley

In March of 2015 we received a further question from Gord.

Question from Gord:

I run an annual trip with 8 golfers over 7 days. With help from the schedule above.
I am nonetheless left with one issue:
The answer has every golfer paired with every other golfer once;
it has every golfer playing against every other golfer twice;
and it has every golfer in the same foursome with each other golfer 3 times.
Great.
BUT I would like to prevent any golfer being in the same foursome with any
other golfer on 3 SUCCESSIVE days. The model produces 2 of these
situations for each player.
Is there any way to avoid it. I've shuffled manually 'till my eyes bleed but can't do it.
Thanks
Gord

Gord,

It turns out that there is only one schedule, up to rearrangement of the days or renumbering of the players. The latter is irrelevant. So your only hope is to shuffle the days. Unfortunately, there are $7! = 5040$ possible ways to do that. If you have a friend who can program, then s/he could write a small program to look at all of the possible orderings of the days, count the number of “three in a rows”, and keep the best schedule.

Good luck.
—Victoria