Thursday, September 11, 2008

The Value of Brady

Tom Brady is out for the season. You may have heard. How much worse does this really make the Patriots?

Before the season, the Vegas line indicated that New England's mean projection was 12.1 wins. Since any analysis of the Pats without Brady will be limited to their final 15 games, this needs to be adjusted. New England was favored against Kansas City by 16 points last week, with the total sitting at 43. This would indicate an average final score of 29.5-13.5. Plugging this into the football Pythag formula, we get an 86.4% chance of the Patriots winning. Subtracting 0.864 from the original 12.1 brings us to 11.24; with Brady, New England was expected to win 11.24 of their last 15 games, a winning percentage of .749.

At Matchbook, you can currently get the Patriots winning over 12.5 games at +250, and the under at -490. This indicates that there's somewhere between a 16.9% and 28.6% chance of them winning at least 13 games. That's an unfortunately large gap, but it's the best thing we have, so we're forced to take the average, 22.8%. When you're taking the over at this point, you are betting that they'll win at least 12 of their final 15 games. For a team to have a 22.8% chance of doing that, they'd have to have a "true" win % of .674.

With Tom Brady, New England was a .749 team against their Week 2-Week 17 schedule; without him, they're a .674 team. That's a W% difference of 0.075, or 1.2 wins over a full 16 game season.

For reference, we can compare this to baseball. Although the BBRAA may not agree, Albert Pujols is currently the best player in the league. Pujols' preseason VORP projection was 72.1, so offensively he's 7.2 wins above a replacement 1B. We can't ignore defense of course, where Dewan's +/- had him at +25 in '06, and +37 in '07. That needs to be regressed, so we'll put him at 9 wins above replacement overall. Over 162 games, Albert Pujols adds .056 to your W% over a replacement player.

So, according to this analysis, Tom Brady is 35% more valuable than Albert Pujols, for whatever that's worth. That makes some intuitive sense; it's not surprising that the best player in the NFL is worth more to his team than the best player in baseball, since in football one individual can seemingly have a much larger impact. One problem with this--and I'm certainly not saying this is the only potential problem--is that Matt Cassel isn't necessarily a replacement-level QB. I don't know how you'd determine whether he is or not, since he hasn't started a game in nine years, but it's certainly worth considering.

Photo: Squib Kick.

10 comments:

Sky said...

advancednflstats.com has done some working valuing players, mostly QBs. from what I remember, they had Brady between 2.5 and 3 wins above average, with other really good QBs maxing out at about 2 wins. bad QBs were about -2 wins. that would imply that the Pats without Brady are probably more like 3 wins worse, not just 1.2. and, for whatever it's worth, that's more in line with popular opinion.

Vegas Watch said...

3-4 games is what most people would say just off the top of their head, but that really is an unbelievably large difference. It's possible that it's closer to 2-3, but that's just not what the odds indicate. They're not necessarily correct, of course.

Vegas Watch said...

Also, the advancednflstats thing was probably looking at game data, rather than a projection. The former would obviously be more extreme, especially if dealing with '07 Brady.

Anonymous said...

Is that the: The Baseball Riters' Association of America? No wonder they can never get the voting right...

Vegas Watch said...

I'm not sure if you get the joke or not, but no, it's not a typo. See here.

Anonymous said...

Fair enough, I didn't get it. Still thought my joke was kinda funny though...

FWIW if you are going to reference some semi-obscure inside joke/campaign, I'd recommend linking it straight in the article for disambiguation...

QB said...

Putting aside that those Matchbook lines are not likely to be very efficient, when the favorite is at -490 and the dog is at +250, the "true odds" (or no-vig line) is actually +291/-291, so the number you were looking for is 25.6%, not 22.8%. This then changes the rest of your calculations. If you'd care to know how this was done, Ganchrow does a fine job explaining it here - http://forum.sbrforum.com/handicapper-think-tank/11994-introduction-betting-lines-percentages.html#post112449

Vegas Watch said...

After giving it a bit of thought, I think you're right, QB. I am very much aware of that calculation, and that is what I do for most things. For these lines on Matchbook, I was thinking of it differently, but I think I was wrong. 25.6% makes it so the EV of both bets is -10.4%; I think that's what we are going for here. There is no reason to assume that either of these gamblers is sharper than the other, which is what I was going for, but averaging the percentages isn't the way to get there.

Using 25.6%, the revised number is 1.0 wins, rather than 1.2. Doesn't exactly make it easier to comprehend. Thanks for the comment.

Vegas Watch said...

Actually, the current Matchbook price is now +240/-430, so we're down to 0.96 wins.

rob said...

this analysis is ridiculous and makes no sense

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