The article that used post season games, based wins expected on net rushing differential and points allowed, admitted it was flawed
Yeah, it's in that article...
and is now a link that doesn't work?
All of the archived blogs aren't working.
By the way, if what you are saying is that you think that formula indicates the chances to win based on everything but QB play (which clearly it is horrible at) why would you add expected wins and actual wins to determine the quality of QB play?
Wouldn't you subtract wins expected regardless of QB from actual wins?
Anyway, since the link doesn't work, would you care to answer how you calculated this, or are you going to stick to avoiding the actual answer?
Again, it shows what the QB had to work with. And lol at me avoiding the answer. It was in thearticle, and I didn't know the link stopped working. Anyways, basically it's separated into different categories. If it's a gimme, the score was 2.00 or above, the QB is expected to win 100% of the time (haven't found one that's lost yet). An easy game is 1.00-1.99, the QB is expected to win 92% of the time. A neutral game, is -.99 - .99, and they're expected to win 48% of the time. -1.00 - -1.99 is a difficult game, and the QB is expected win 9% of the time. And -2.00 and up is an impossibe game, and the QB is expected to lose 100% (Tom has won two of these lol).
Now to get the numbers you basically look at the QB's defense and the rushing on both teams. I forgot excatly what he called one of the numbers for the scoring and why he got it, if the article was working I'd just get it from there, so you'll basically have to take my word that it made sense. One of the scoring numbers is an average margin of victory (forgot which one, think 11.4), the rushing was the average rushing differential (93.1). Anyways, here are the actual calculatoins of a game.
Let's take the 07 game vs. the Redskins for example:
The defense allowed 7 points so you subtract that from 18.9:
11.9.
Then you divide that by 11.4:
1.04.
Now take the rushing differantial (105) and divide it by 93.1:
1.13.
Now add both the numbers:
2.17.
It's a gimme game so that's a 100% win.
The next game they had the colts which was a Neutral game (-.25), so that's a 48% chance of Tom winning.
Out of those two games, Tom's win expectancy is 1.48 (100%x1 = 1. 48%x1 = .48. 1+.48 = 1.48.) Since Tom won two games, he has a score of .52 (Just for those two games).
But this is what I did, I took the win expectancy (1.48), and the actual wins (2) and added the, together (3.48). But over the whole season and postseason. Is the end all to be all, probably not. Is it intersting, yeah.
Happy now?
