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Sweeps and splits

Posted by Doug on December 12, 2008

When two NFL teams play each other twice in a season, how often should we expect to see a sweep, and how often should we expect a split?

If every game was a coin flip, we'd have four equally likely possibilities:

Team A wins both
A wins first, B wins second
B wins first, A wins second
B wins both

So 50% of the time it'd be a sweep, and 50% of the time a split.

But games aren't coin flips. When the 2008 Steelers play the 2008 Bengals, the chances of a sweep would seem to be greater. If we assume the Steelers have an 80% chance of winning each game, then we have:

Steelers win both: .8*.8 = 64% chance
Bengals win both: .2*.2 = 4% chance

So we've got a 68% chance of a sweep and a 32% chance of a split. So differences in team strengths have a tendency to cause more sweeps.

But home field advantage should have the reverse effect. Since each team gets one home game, that increases the chances of a split. If the teams are evenly-matched, then assuming the home team wins 60% of the time would yield the following chances of a split:

Home team wins both: .6*.6 = 36% chance.
Road team wins both: .4*.4 = 16% chance.

So it'd be 52% split and only 48% sweep.

So we've got two opposing forces at work. And I've also assumed independence of the games in the above calculations. Are they really independent, or do we in real life see things like one team having another team's number? Or do we see the reverse: the losing team in the first game having extra motivation in the second? Do you think that sweeps or splits happen more often in the actual NFL?

It turns out that sweeps are more common. Whether you look at 1970--present, 1978--present, or 2000--present, the percentages are very consistent. About 58% of regular-season two game home-and-home series are sweeps, and 42% are splits.

So now the question is: is this caused by natural variation in team strengths, or is there something else going on?

To find out, I looked at each home-and-home series from 1978 to 2006. I recorded the SRS difference between the two teams and used a logistic regression to translate that into a probability of each team winning, both at home and on the road. Then I used those probabilities to estimate the probability of a sweep. Here's an example:

In 2000, the Falcons and Saints met twice. According to SRS, the Saints were about 9.4 points better than Atlanta. According to my formula, that translates to roughly a 72% chance of a Saint victory in Atlanta and an 85% chance of a New Orleans win at home. So the sweep probability is .72*.85 + .28*.15, which is about 65%.

Now we look at all 1543 such series and use a calculation like the above to estimate how many sweeps we'd expect to see in total. Answer: if the games are independent, and if my formula for estimating probabilities is OK, we should have expected 911 sweeps in those 1543 series. What we actually saw was 902 sweeps, which is extremely close to the expected value.

So this turns out to be a boring post. Nothing to see here. We see sweeps almost exactly as often as we should expect to see them.

This entry was posted on Friday, December 12th, 2008 at 6:02 am and is filed under General. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.