I didn't think the Bears would move up that much after beating 0-4 Carolina, but the Packers (3-->8), Saints (2-->9) and Texans (7-->13) all dropped past them, pushing them up.
This shows you how much parity there is in the NFL (and how little the "experts really know)! Here are the week one power rankings for 3 of the top 10 teams with current rankings in parentheses.
4. Cowboys (19)
5. Vikings (20)
8. Chargers (21)
And here are the 3 current top 10 teams that replaced them with week 1 rankings in parentheses.
1. Steelers (18)
7. Bears (21)
10. Chiefs (27)
It looks like there are very few really dominant teams this year. I Expect to see a lot more upsets.
Well, if parity is on the rise (or at a high level already) in the NFL, you really can't knock the "experts" for not getting the right pick every time (or some percentage thereof) or properly ranking the teams right off the bat.
Think about this: there is some definite, but unknown, probability X that team A will beat team B on any given day (or Sunday, for those of you that like wholly underrated football films). Likewise, there is some unknown probability Y that team B will beat team A. So then, in any given game (where one team has to win out over the other [we will exclude ties because they are so uncommon]), we can represent the probability of team B beating team A as 1-X=Y.
Let's now introduce Parity (P) which we will represent as the difference in X and Y: P=X-Y. Those of a quick mind will notice here that P can easily be negative in this sense, but since X and Y are both attributes of two nameless, talentless teams, we will assume that the lesser probability will be Y. Now, since we know that P=X-Y, and that Y=1-X, we can represent Y in the Parity equation as a function of X, and some quick algebra gives X as a function of P: X=(P+1)/2 (the same works if you use Y instead of X, but remember that we are assuming here that X is bigger than Y in most cases).
Now we must ask ourselves what
true parity is. In my mind, that would be when team A has just as good a chance of winning a game against team B as team B does winning out over team A. Given how P has been represented (a measure of how much bigger X is than Y), that would mean that P would have to be zero. Most of you can figure the rest out from here if you haven't already, but it's been awhile since I've stretched my calculus legs, so what we can do is take the limit of our function X (which is dependent on P) as P approaches true parity (zero). Doing this, we see that the closer and closer P gets to zero, the closer and closer X gets to 1/2 or 0.5 (if we take P to be 0.01, X is .505; 0.001, .5005; .000001, .500005; and so on).
This agrees with what we intuitively think about parity: since it is impossible to know for sure the outcome of the game beforehand, each team has some likelihood of emerging victorious, and with parity, that likelihood comes closer and closer to each team having a
true equal chance of winning the game.
It is clear though, that not every team has an equal chance as their opponent at winning a given game, even in today's NFL. However, with parity seemingly on the rise, those X's and Y's are creeping ever-closer to each other with every given game, and because of that, luck alone could explain away a seemingly inferior team rattling off a few wins against supposedly better opponents.
Those "experts" may actually know a lot about the given teams in a given game, and they may even correctly identify the strict "better" team, but when parity brings those teams closer together, luck starts to play a bigger and bigger role in the outcomes of those games, and it would be improper to discredit those "experts" for not accounting for the randomness of luck in their predictions.