Twelve Golfers Golfing -- Optimal Pairings

One of the most common golf tournament size is twelve (12). Well, perhaps not considered a tournament; more like a match or competition. Twelve golfers at the local club that tend to get together on a fairly regular basis. They either play the same game each time, or, they like to mix it up and play different games (stroke play, match play, par points, alternate shot, etc.

Here's an example of a group of 12 Golfers playing multiple rounds together.

The one thing these twelve do want, is to play with a variety of the others over the course of the season. It's very simple to get variety for a few players, ensuring that they be placed in foursomes with different players each round. Unfortunately, this is at the expense of the others, who end up meeting many times just so that those few have optimum variety.

To mix all twelve players up so that they all experience a similar degree of variety is a very complex problem. In mathematics, the topic which describes the different mixings of players is called combinatorics. It has been utilized in golf for many years, centering around the infamous Social Golfer Problem (optimum mixing of players).

Ironically, other size groups (such as 16) which are larger are easier to mix. The problem with twelve is that there are only 3 foursomes to arrange, thus there is limited opportunity to separate players round after round. A 16 player group has another foursome and thus another opportunity of an equitable arrangement of players.

So why is twelve so difficult? Let's take a look at the combinatorics of the problem.
How many variations of 12 players are possible? By variation, I mean how many different ways can they be ordered?
1 2 3 4 -- 5 6 7 8 -- 9 10 11 12 -- and 12 11 10 9 -- 8 7 6 5 -- 4 3 2 1 are just two examples, but, you'll notice that they are the same 3 foursome groups. To order 12 objects, one must select them one by one. For example, say you're choosing 12 people for a team, which has 12 different roles. You need to choose one person for each role, one by one. So, for role #1, you have 12 choices. For role #2, you then have 11 choices. If you were choosing only 2 players from the 12, there would be 12x11 = 132 different pairs of people that you could choose for the 2 roles. In choosing 3 people from the 12, you'd have 12x11x10 = 1320 different triplets. Choosing all 12 people yields 12x11x10x9x .... x3x2x1 = 479 001 600 or about 479 million different arrangements. 

In combinatorics, we have a shorthand for representing this: it's 12! which is read 12 factorial (6! = 6 factorial = 6x5x4x3x2x1 = 720) As demonstrated with the two examples of 12 players above, many of these 479 million arrangements are the same 3 foursomes. Let's call the foursomes A, B and C. In the example above, A = 1 2 3 4 , B = 5 6 7 8 , C = 9 10 11 12 and then A = 9 10 11 12 , B = 5 6 7 8 , C = 1 2 3 4 So basically, I've just rearranged the order of the foursomes and not changed the foursome at all. ABC --> CBA How many different orders are there for the foursomes which would be the same foursomes? How many ways can one order 3 objects? Answer? 3! = 3x2x1 = 6

 In the 479 million arrangements, there are 6 repeats with the same foursomes (same golfers grouped together). So, we need to divide 479 million by 6. 479 001 600 / 6 = 79 833 600 or about 80 million But each foursome of players can be the same foursome but with a different order. Foursome A = 1 2 3 4 can also be A = 4 3 2 1 and A = 2 4 3 1, etc. There are 4! = 24 different arrangements of each of the four players in a foursome. So, the original 479 million also has 24 repeats of the same foursomes. We need to divide by another 24. 79 822 600 / 24 = 3 326 400 different arrangements of the 12 golfers that are different. But in those arrangements, there are foursomes which has players that met in many of the different foursomes of the different arrangements. For instance, the foursomes 1 2 3 4, 1 8 9 4, and 1 5 10 4 are all different but all contain players 1 and 4 so these players meet three times (say in successive rounds). What also needs to be considered is how many "pairings" of different players are possible? 

What we want is to minimize the number of pairings of golfers. For 12 golfers there are 66 different pairings: 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9, 1-10, 1-11, 1-12 (12 unique pairings with Player 1) 2-3, 2-4, 2-5, 2-6, 2-7, ........,2-12 (11 more unique pairings with Player 2 (1-2 repeat) 3-4, 3-5, 3-6, 3-7 ......., 3-12 (10 more cause 1-2, 2-3 already counted) this pattern continues until finally 11-12 (1 more unique pairing) Total = 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66. One can also calculate this using factorials. 66 = 12!/(10!2!) = 12 C 2 called a combination.

  Let's consider Round 1. Foursome 1 2 3 4 has 6 different pairings. The other two foursomes would also have 6 for a total 18 pairings in the first round: A: 1 2 3 4, B: 5 6 7 8, C: 9 10 11 12 To ensure as few repeats as possible in the second round, as many players as possible need to only play with players from a different foursome (1 plays with anyone from foursomes B and C). But if player 1 plays with a player in foursome A, then player 2 must play with a player from foursome B to avoid a repeat with player 1. As you can see, repeats are unavoidable. Round 1: 1 2 3 4, B: 5 6 7 8, C: 9 10 11 12 Round 2: 1 , 2 , 3 1 5 , 2 6 , 3 10 no repeat pairings 1 5 9 , 2 6 12 , 3 10 7 still no repeat pairings 1 5 9 4, 2 6 12 8, 3 10 7 11 3 repeat pairings unavoidable If you jumble the numbers above even a little bit, you'll end up with more than 3 repeat pairings. The minimum number of repeat pairings in successive days is 3. So the best one can achieve is 15 new pairings (18 - 3 repeats). In Round 3, the same would hold with a minimum of 3 repeats with Round 1 and 3 more repeats with Round 2, so 18 - 3 - 3 = 12 more unique pairings. Note that a minimum of 3 is harder to obtain in Round 3 than in Round 2. So, the total unique pairings possible in the first 4 rounds is 18 + 15 + 12 = 45 leaving another 11 to achieve (11 + 45 = 66). Round 4 would have a potential 9 unique pairings (if you chose the previous 3 rounds just right) which leaves at least 2 more pairings to achieve in Round 5. It ends up that it actually takes 6 rounds to achieve all 66 unique pairings, but at the expense of having multiple repeats with other pairings (twice and thrice). See if you can work it out. I've given you some information to get started above. Or, if you'd like to get your hands on this fully optimized solution, go to the one of these:

12 Golfers 3 Rounds 

 

 Round 1:     1,2,3,4     5,6,7,8,    9,10,11,12

 Round 2:     1,5,9,2     3,6,10,11,    4,5,12,7

 Round 3:     1,7,11,5     2,4,10,8,    3,16,12,9