Wednesday, October 27, 2010

Oregon – Maybe Not the “Lock” You Think

While looking at the 4 remaining major conference unbeatens (Oregon, Missouri, Auburn and Michigan State) I wondered what chances Oregon has to run the table from here on out.

After glancing at their schedule, my thought was “pretty good”. After all, they have games –

At Cal
At Oregon State

Only Arizona is ranked. A pretty good chance to win out, no?

My next thought – can we put odds on it? Well, I think we can. And it isn’t the lock you might think.

Internet research led me to an invaluable find – a Phil Steele article laying out the odds that different Vegas spreads actually won a game.

Vegas, being Vegas, does a pretty fine job of predicting wins based on point spreads. The data from Steele’s article is below –

Favorite of
# of GMS
Lost Outright
31+ pts
3 or less

To apply this to Oregon’s chances, we would need Vegas’ lines on the Duck's remaining games. While we have the line for the game at USC (opened at Oregon by 6.5), we don’t have lines for future games.

We do, however, have something that claims to predict games between any teams – Jeff Sagarin’s Ratings, which are used (modified) in the BCS.

Using Sagarin, he predicts Oregon to be 7.8 points better than USC at a neutral site, which would give the Ducks a 4.8 point advantage at USC.

In our odds chart above, either 4.8 or 6.5 tells us that, historically, USC has a 34.1% chance of winning. This is based on a large sample size of 2588 games.

So if USC would win 34.1% of the time, the Ducks win 65.9%.

Since Sagarin’s spread was reasonably close to Vegas, here’s what he would have for the rest of Oregon’s games (as point spreads) –


Oregon St

The Cal one surprises me too, but Sagarin has them as his 8th best team per his Predictor, even with a 4-3 record.

Using these point spreads, we can estimate the chances of Oregon winning out. Including USC, here are Oregon’s chances to win each remaining individual game –





Oregon St


Just like you would multiply .50 times .50 to see the chance of getting two consecutive heads in a coin flip (.25 or 25% chance), you do the same here.

The results? 0.2261, or an only 22.6% chance that Oregon runs the table.

Less than 1 in 4.

Which begs the question – what are the similarly calculated chances for the other unbeatens? We will explore that in our next post.


TheCallahanKid said...

Interesting Mergz. Keep up the good work.

Anonymous said...

Interesting, indeed.


using your chart, the "average" chance of the favorite winning is .6295. When extrapolated out for a 12-game season, this mean that, by mathematical deduction, a team who is favored in all 12 games would have only a .004 chance of going undefeated; if a team is an underdog in one or more games, its chances would be systematically lower.

I do not have the exact figures at hand (too lazy...err, research right now), but I believe historically the ACTUAL % of undefeated teams is approx 1.5-2%--about 4 to 5 times higher than the "logical" mathematical approach you use would suggest should occur.

DA ;)

Mergz said...


First, I don't think you can merely "average" the chances because there are significantly more occurrences in the lower spreads.

I'd also have to do some more research, but there is what, 1 or 2 undefeated teams per year on average? (Some years there are none). 1 of 120 would equal a chance of 0.83%, or less than 1.

Also there are a lot of teams in the FBS. With 120 teams, even if all of them were equal, you would expect at least 1 to be undefeated after week 8 (120,60,30,"16",8,4,2,1). There would be a 50% chance of 1 after week 9.

But the chances aren't 50-50. If you have a team with a 70% chance in each game, that team would 1.3% chance to run the table. A "super team" with a 90% chance would have a 28.2% chance. With 120 teams of all different skills, we usually end up with 1. But even with your example of .004, that's an average of 0.48 teams a year, or 1/2 a team (once every other year).

This isn't perfect for sure. But I've done some more work and it offers some very interesting avenues to follow.

Anonymous said...

RE: "average"
The .6295 is a "weighted" average (u can do the math...or trust my math skillz) to account for the disparity in # of occurences.
Obviously no "system" is perfect; I just thought (probably wrongly)that that "predicted" # of undefeated teams per season by your method was disproportionately low compared to actual events. It certainly doesn't "invalidate" the data; I just think it's a point to be "taken under advisement."

Mergz said...

I didn't know you did a weighted average, so that would be correct - my bad (I'll trust your math skillz).

I guess we need to find out the actual (or best estimate) of number of undefeated teams per season. Plus, the 12 game season is relatively new, so older data might not apply. It appears there were 2 last year, and 1 the year before.

Mergz said...

I couldn't stand it, so I looked back to 2002 for undefeated teams -

2002 - 1
2003 - 0
2004 - 3
2005 - 1
2006 - 1
2007 - 0
2008 - 1
2009 -1

Even in some of the 1 no-loss teams years that team wasn't the BCS winner. Plus, Utah and Boise seem to have a disproportionate number of no-loss seasons (easy schedules?)

So we have 8 no-loss teams in 960 instances, or 0.008, twice the 0.004.

It bears further investigating for sure.

Mergz said...

One more note - you would need an average winning chance of about .6700 to get the 0.008.

Anonymous said...

In yout count of undefeated teams: is that at the end of the regular season, or after the bowl games?

For the purposes of this particular discussion, I move that we only consider the "regular" season, since the MNC game or other BCS bowls can often match undefeated teams. Since we are talking "average" odds, these arranged games would be more likely to have a lower point spread, therefore lowering the team's estimeted chance of winning said game (would bring their final % lower than the .004 previously calculated).

Mergz said...

The count was after the bowl games, so you are right. It should be regular season. After all, I'm more interested in who CAN make it, not who finishes the deal.

When I weighted the average chance of the favorite winning I got .7536. That would be a 0.034 chance of undefeated, or more like 4 teams a year. (8063 total sample size)