Suggested Tourney Trees

By Richard the Poor of Ely

This is an attempt at determining the optimal structure for a tourney given the number of entrants. “Optimizing” for these purposes includes several objectives:

  • Minimize the number of byes in the tourney.
  • Get rid of all byes as early as possible.
  • Avoid an excessively long tournament.
  • Avoid very short tournaments.
  • Avoid having an entrant participate in consecutive bouts.

There are three basic types of tourneys: single elimination, double elimination, and round robin.
In a single elimination, each bout removes one entrant from the tourney. So the total number of bouts is N-1, where N is the total number of entrants.
In a double elimination, each entrant must lose twice in order to be eliminated. So     the total number of bouts is 2N-1.
For a round robin, each entrant competes against every other entrant. Whoever wins the most bouts wins the tournament. The total number of bouts is Nx(N-1)/2. In case of ties, one can have the two entrants participate in a “winner take all” bout.

Here, then, are my proposals for optimal tourneys.

Less than four entrants:
Don’t bother.

A round robin. Suggested pairings are (for entrants A, B, C, and D):

AB    CD    AD    BC    AC    BD

Since time will most likely allow it, one can do this twice. You will more likely have a definite winner than, anyway.

A round robin. Suggested pairings are:


A round robin. Suggested pairings are:

Round 1    Round 2    Round 3    Round 4    Round 5
AB              AC               BC             CE               BE
CD              BF              AE             BD               AD
EF              DE             DF              AF                CF

Add one bye, and do a double elimination.

A double elimination.

Now we come to one of the most difficult situations – the case where you have one more entrant than the number for a perfect double elimination: 2X+1.  There are three options:

1. Persuade one of the entrants to drop out and do a double elimination.
2. Persuade another person to enter, and go to the tree for the next highest number.
3. Do a modified round robin.

For the case of nine entrants, to do the modified round robin, set up a grid like the one below. For the first part, do round robins with each of the rows. Then do round robins for each column.

A    B    C
D    E    F
G    H    I

The pairings for the first part look like this:

AB    DE    GH    BC    EF    HI    AC    DF    GI

For the second part, the pairings look like this:

AD    BE    CF    DG    EH    FI    AG    BH    CI

Each entrant is in four bouts, and there are a total of eighteen bouts. In case of a tie, you have a number of options.

If time permits, you can redo the tourney, but rearrange the grid as shown so that no pairing is repeated. Or, you can do a “playoff” with the tied entrants.

A    F    H
E    G    C
I    B    D

Split the entrants into two “divisions” of five each, and do round robins with each division. Then do a playoff with the division winners.

Sometimes, it’s better to split the entrants into divisions. This is only recommended when you can split the entrants into divisions of equal size. This way, each entrant in the tourney will be involved in the same number of bouts (before any playoffs, of course).
In the case of ten entrants, if you did a straight double elimination, you would have to add six byes, and the total number of bouts is 19 ((2 x 10) – 1).  With divisions, the number of bouts from the round robin divisional play is 20 (2 x (5×4/2)). Not much difference at this level, but with divisions, you don’t have to worry about byes.
At higher numbers, it gets more difficult. For example, a normal double elimination with 24 entrants will require eight byes and a total of 47 bouts. Split it into three divisions of eight and do double eliminations (no byes) , and there will be 45 bouts in the divisional play. The key with using double eliminations in divisions is to have the losers from one division go into the Losers’ Bracket of a different division. This is to keep two entrants from meeting each other in bouts more than once (until the playoffs).
With regards to the Playoffs, you will wind up with two to four “finalists”. With two, do a best two-out-of-three playoff. With three, do a round robin until there is a clear winner. With four, do a single elimination playoff.

Another really difficult case is where the number of entrants is a prime number that would need too many byes to bring it up to an easy double elimination, such as 11.

Two divisions of six entrants, and round robins in the divisions.

Thirteen to Fifteen:
Add byes to make a sixteen-entrant double elimination.

Double elimination.

This is one of the toughest (see Eleven). If you do a double elimination, you will need 15 byes to bring it up to the next straight double elimination number (32). The way the tree works out then, you will essentially wind up picking two entrants at random and having them face each other in a bout to see who does not get to participate in a double elimination with sixteen entrants.
Now how much fun is that?
You could split the entrants into three divisions of six, six, and five entrants each and do round robins, but the entrants in the small division will be in fewer bouts than the other entrants. And you’ll wind up doing more total bouts than in a straight double elimination (40 to 33).
It’s your choice.

A double elimination will require 14 byes and 35 total bouts. Splitting the entrants into three divisions of six each and doing round robins will require 45 bouts before any playoffs. Two divisions of nine each (and following the suggestion above for nine entrants) will require 36 bouts before playoffs.
Pick the one you’re most comfortable with.

Aarrggh! Another prime number!

A double elimination will need 12 byes and 39 total bouts. Four divisions of five each, with round robins, will need 40 total bouts before the playoffs. I recommend the latter.

Twenty-one through Twenty-four:
Three divisions, with enough byes to bring each division up to a double elimination of eight each.

Twenty-five through Thirty-two:
Add byes to bring it up to a double elimination with 32 entrants.

Thirty-three through Fourty-Eight:
The simplest (though sometimes not necessarily the best) method here is to split the entrants into three divisions of equal numbers (or as equal as possible) and add enough byes to bring the total number in each division to sixteen.
Do a standard double elimination in each division, then a round robin playoff with the divisional winners.

Forty-Eight through Sixty-Four:
Just add as many byes as needed to bring the total up to 64. Anything else introduces too much complexity.

More than Sixty-Four:
Hopefully, this will never happen outside of Crown Tourney. By the time the numbers get this large, you will need more than one list in order to handle the number of bouts. Even so, the tourney will still drag on inteminably.
Dare I suggest putting a limit of 64 entrants on all tourneys (save Crown, perhaps)?

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