There is a school of thought that we should add two rounds to the NCAA
basketball tournament to include almost every school in the tournament. The
only reason I've ever heard for this is Because We Can, but I'm going to
here propose something like this for the football program; take 108 football
teams and have them all in a six-round tournament.
Many of you know, from the college basketball situation, that you can only
fit 64 teams in a six-round tournament that results deterministically in a
single champion; this six-round playoff, however, would work by judiciously
pairing good teams with a loss against bad teams that are still undefeated
in the hopes of thereby eliminating more teams from the ranks of the
undefeated than simply half of those left at the beginning of a round.
(Some of you will object to a probabilistic playoff system. Good for you.
Shut up.)
| Teams | round 1 | round 2 | round 3 | round 4 |
| 1-27 | 3 | 3 | 2 | 1 |
| 28-36 | 2 | 2 | 2 | 1 |
| 37-54 | 2 | 2 | 1 | 1 |
| 55-72 | 1 | 1 | 1 | 0 |
| 73-81 | 1 | 1 | 0 | 0 |
| 82-108 | 0 | 0 | 0 | 0 |
- Seed 108 teams from 1 to 108. In each round each team has a "pairing score" given by
- the number of games that team has won
plus - the number given in the table.
Teams with
the same pairing score are paired against each other; in the case of an odd number of teams with a given score, teams near the bottom of a score group play
teams near the top of the next one down. (For example, in the first
round the top 27 teams have three points, so they play each other. In
the next round, the 13 or 14 teams that lose will still have 3 points,
as will the 13 or 14 teams that have one win but are ranked between 28
and 54.) (This will be familiar to
chess players with some knowledge of the Swiss system.)
- At least 13 of the top 27 teams will lose in the first round; these teams
are now in the same score group as the at-least-13 teams in the 28-54
range who won. By pairing the teams that won with the teams that lost, one
expects that the odds of more than 6 teams coming out of these games undefeated
are small; this is the point, then, is that if we simply matched undefeated
teams we would have half of the teams coming out undefeated each round.
- Even if a team has no chance of winning the championship after an early loss, that team will keep playing, at least until the team has more losses than
the number given in the table. For example, the 10th team is likely to lose one of its first two or three games, but will play four games. (If it loses two of its first three, it's no longer needed after the third round.) These teams
could be kept in, but they wouldn't be necessary.
- These "pairing scores", as you can see, eventually relax to the point where, by the fifth round, teams are playing other teams with the same number of
wins. Thus a team with three wins against weaker teams will have to play other teams with three wins, if we decide to keep them in the tournament, and at
the end of six rounds inhomogeneities caused by early discrepancies in pairing
will get ironed out. Most teams, however, at the bottom will have mostly
played other teams near the bottom.
- Any team can, in fact, win the championship, but the probability of enough teams
upsetting enough other teams to generate more than one undefeated team by the
end of six rounds is slim to none. (In fact, I expect that the tournament, as given, would be more likely to suffer the opposite problem that there would only
be one undefeated team after five rounds, and, if they lose in the sixth round,
there would be no undefeated teams. It might be best, if a bit suspect, to
just say that the champion is crowned when only one undefeated team remains.)