Some observations, then, on the RPI:

• If you have a good RPI, and you beat a weak opponent, your RPI can go down. Conversely, if you have a bad RPI, and you lose to a good opponent, your RPI can go up. This is a shortcoming. As it happens, if you play most of your games against teams that about your strength, the RPI yields a decent approximation of your strength; unfortunately, The Committee uses it to try to select the top 50 or so teams in the country, and while the "bubble teams" often have played a fair number of games against teams at their level, its use for seeding purposes is really dubious.
• If you subtract off .500 from the wins/games of each team, you get a number whose average will be zero. While some teams have better records than others, when you average over all the opponents you get a number that is closer to zero, and, in fact, if you keep calculating not just your opponent's opponent's average, but their opponent's opponent's, and so on, these numbers get very close to zero very quickly.
• I subtract off .500 because, in fact, once I have done this, I can keep adding these averages over and over, and since they are going to zero instead of to .500, the total actually plateaus. Not only does it change more and more slowly, but it will, in fact, converge to a final answer.
• If I multiply by 2 after subtracting .500, I get a number that varies from -1 to 1, and, in fact, is (W-L)/(W+L). I like this better; it will simply result in a doubling of the RPI.

Mathematics

The mathematically inclined people who should be reading this will know that, in some sense, (1-C)(1+C+C^2+...)=(1+C+C^2+... -C-C^2...)=1 so long as C^n gets small when n gets big; by multiplying both sides of my equation above by (1-C), I get (1-C)r=w, or r=w+Cr, so that a team's RPI is its winning percentage plus the average of its opponents' RPIs, which seems itself to be a fine ab initio definition of our RPI. Thus, if we can invert the matrix 1-C by our usual laws of matrix inversion, we can just multiply that by w to get r.

Unfortunately, we can't.

Each row of C, you remember, adds up to 1, so that each row of 1-C will add to zero. The columns are linearly dependent, and 1-C is therefore not invertible. So we've reduced this problem to one of inverting a noninvertible matrix.

Consider a matrix all of whose entries are the same, say E. Er, then, would be equal to the average of r for all teams. Another upside to subtracting off .500 to put the average at zero; Er=0. (Incidentally, because different teams play different numbers of games, it may not work out exactly to zero; let me then not actually subtract off .500 exactly, but whatever I need to to make Er=0. Not only is this actually necessary to make the C^n actually go to zero, but it also gives the closest approximation in a least squares sense to w when we end up multiplying (1-C) by the solution r that this will give us; remember, since 1-C is singular, no matter what we choose for r (1-C)r will be restricted to a subspace that doesn't necessarily exactly include w, but it gets darn close, and this little hack we're doing gives us the point of closest approach.) Since Er=0, and (1-C)r=w, (1-C+E)r=w. So, by adding some number to each entry of 1-C, we get a matrix 1-C+E whose inverse, M, can be determined by standard methods.

In some sense we don't have to calculate M; all we really need to do is solve (1-C)r=w for r. There is a little value in calculating M:

1. Modulo early season tournaments, we know the schedule before the season, and can thus calculate M before the season begins, rather than having to wait until we know w.
2. Having an actual matrix M for which r=Mw can provide a certain amount of insight into how closely a team is tied to another. Some of you probably don't care, but I think this sort of thing is enlightening; a team might think it's doing well because it has some good wins over teams with good records, but its row in the matrix is dominated by entries against Towson St. and Yale.