7 golfers playing 4 rounds

[Macca]

We are going on a golf tour with 7 players and will be playing 4 rounds. What is the optimal combination of 4 ball and 3 ball per round to try and ensure that every person plays with each other person the same (or mostly the same) rounds with each other?

[Ian Wakeling]

Although every pair of players does not play together, I think the following schedule would be the most useful:

   (6  4  5  1)  (3  7  2)
   (1  4  7  2)  (6  3  5)
   (7  3  1  6)  (5  4  2)
   (3  2  5  1)  (6  4  7)

Player 1 is the nominated tour organiser, who gets to play against all other players exactly twice, but in order to do this they always play in the foursome.  The other 6 players, play exactly twice each in a foursome and twice in a threesome, they play 5 of the other players exactly twice, and one player not at all.  The pairs that don't occur are (2 6), (3 4) & (5 7).

It is possible for everyone to play together at least once - for example:

   (2  3  6  1)  (4  7  5)
   (3  4  1  7)  (5  2  6)
   (3  5  4  2)  (7  6  1)
   (1  5  7  2)  (3  4  6)

however, 9 pairs of players meet only once, and 3 pairs of players meet 3 times.  Also the number of times a player plays in a threesome varies from once (players 1 to 3) to three times (player 6).

I hope one of the schedules above is useful.

[Macca]

thanks for this ! - both are very useful and it gives us options - much appreciated.
 
we may play 5 rounds .... are you able to flex it to account for this?

thanks

[Ian Wakeling]

For 5 rounds there are still many of the same problems - 3 pairs only meet once, 6 pairs meet three times.

   (1  7  2  6)  (4  3  5)
   (5  2  3  7)  (1  4  6)
   (7  4  3  1)  (5  2  6)
   (6  7  4  5)  (1  2  3)
   (5  1  6  3)  (2  7  4)


Things will not get better until there are 7 rounds and then you can use Table 1 here. Ignore player 7 and use players 0 through to 6.  Then it will be perfect, every pair of players will meet once in a threesome and twice in a foursome.

[Macca]

Ian,

This is a great forum. Was raving about it at the golf club over the weekend.

A member of my weekend 4 ball is also going to be going on a golf tour - very similar to ours except with 8 players.

They plan to select 2 teams of 4 and play a Ryder cup style

4 rounds of golf will be played so each player from team A will play each member of Team B. The tricky part is to try and even out the number of times everyone (both same team and opposing team) play with each other.

The members of the same team will play with 1 person twice and the other 2 players once.

How can he best schedule the matches aginst the other team (or will it agian be the case that some will play together 3 times and others only once)?

Hope you can help with this one?

thanks again for a great forum

[Ian Wakeling]

I don't think there is any good solution to this, some opposing pairs will occur once, others three times.  For example

A1 A2 v B3 B4   &  A3 A4 v B1 B2
A1 A3 v B2 B4   &  A2 A4 v B1 B3
A1 A4 v B2 B3   &  A2 A3 v B1 B4
A1 A2 v B1 B2   &  A3 A4 v B3 B4

[Macca]

Thanks again