Could It Be?

In 2008, eight SEC team competed in bowl games. In 2009, 10 out of 12 played in the post season. This got me to thinking (by the way, I couldn't sleep thinking about this); what is the maximum number of SEC that can participate in a bowl game? Is it possible that all 12 teams can go at least go 6-6? After some thought, I can solve a greater question: What is the largest number such that every team has this many wins? If you're enticed, then you'll accompany me on a short (yet indeed), mathematical journey!

Bear with me for a moment while I lay some groundwork. I'm investigating this question using an area of mathematics (math and football are a very natural combination) called combinatorial graph theory. Basically, a graph consists of two parts: vertices to represent objects, and edges to represent relationships between two objects. A simple example could be that the vertices represent people at a party, and two people are connected by an edge if they're friends. Ever heard of the six degrees of separation? In graph theory, we would state it as follows: Conjecture- Let G be a graph where vertices represent people and edges represent acquaintance, then the minimum distance between any two people is less than or equal to six.

Anyway, let's look at our question. We'll assume everybody wins their non-conference games, so everybody starts at 4-0. Now fore each conference win, there is also a conference loss. We determine that there are 12(8)/2 = 48 games. Dividing the wins evenly, we see that each team could win 4 conference games. So it seems that each team could go 8-4 before post season. But is this possible?

We must keep in mind that within each division is a round robin setup, and so it's easiest to concentrate on each division first. Pay attention... We'll construct a graph with 12 vertices with 6 in one group and 6 in another. Each team is a pair of vertices: one vertex in the first group and the another vertex in the second group. The edges will join a vertex in one group (say A) to a vertex in the second group (say B) if the team in A beats the team in B. Does that make sense?

I'll cut out some of the details, but you can find the picture of the West's graph here. Three teams can win three games within the West, and three team can lose three games with the West. The additional edges sticking outward signify the games played with the other division. For instance, Ole Miss still has to beat one team in the East and lose to two teams in the East. Now the Eastern division can be represented the exact same way (just re-label the vertices).

To finish, we'll construct a graph on 24 vertices by taking two copies of the graph mentioned above. We only have to worry about the edges jetting outwards. All we do is kind of match them up. For instance, Ole Miss still has to win a game against the East, and Florida has to lose a game against the West. Then put an edge from Ole Miss to Florida. You have to be a little careful here, but its not too bad. I haven't drawn up a computer rendering of this, but I'll try if anyone's interested.

If you're not bored out of your mind yet, I'll explain what this means. The construction of this graph demonstrates that it is indeed possible that each team in the SEC could end up 8-4 at the end of the regular season (barring any schedule mandates of which I'm unaware). The funny thing is, due to the SEC championship game, there has to be at least one team that loses 5 conference games each year.

Anyway, in this somewhat socialist anomaly (you know… we're all the same and everybody wins), it could end up: Alabama (9-4), Florida (10-4)(win SEC game), Kentucky (9-4), Vanderbuilt (9-4), Georgia (9-4), South Carolina (9-4), Tennessee (9-4), Auburn (9-4), Arkansas (9-4), Ole Miss (9-4), LSU (9-4), MSU (9-5) (lose SEC game.. why not?)