AFC Wildcard, Resting Starters, and 16-0 Teams

A few weeks ago, I wrote that the 2nd AFC wildcard playoff spot come down to the last week of the season between TEN and CLE. Their relative strengths of schedule clearly favored CLE, but the difference in opponent strength was primarily to due their final opponents of the regular season. CLE faces a weak SF while TEN faces an elite IND.

But with IND locked into the second seed, IND may not be so elite in week 17. I can't begrudge a team for resting its best players prior to the playoffs. It's legal and in their self-interest. But it is a shame that who gets into the playoffs can be determined by such a quirk.

If TEN played IND in any other week, they wouldn't have much of a chance to win. But against Jim Sorgi, they would probably be favored. IF TEN loses to a full-strength IND, Cleveland wouldn't even need to beat SF. They would keep the tiebreaker due to having a better conference record.

On a similar note, I estimated that NE had about a 52% chance of finishing the regular season undefeated. And it's no surprise that here are the unbeaten Patriots facing the Giants tonight for a chance at NFL immortality. It's also a shame that the Pats may not be facing an opponent at full-strength, as the Giants have clinched the 5th seed.

These unfortunate circumstances are really just due scheduling quirks. Had NE's or TEN's schedule been arranged any differently, their prospects for victory this weekend would be very different.

How rare is a 16-0 team?

Here is a quick back-of-the-envelope analysis. Let's say that every year there are two legitimate 13-3 teams. In other words, there are two teams in the NFL that have a fundamental .813 winning percentage against an average strength of schedule without luck of any kind. In a 16-game season, a team with an underlying .813 win probability would have a 2.8% chance of winning all 16 games. (0.813 ^ 16 = 0.0281).

Because we assumed there are 2 such teams each year, the chance of neither team going undefeated is (1-0.028)^2 = 0.944, or 94.4%. The chance of one or both teams being undefeated is therefore 1-0.944 = 5.6%. We should expect an undefeated team about every 1 out of 20 years with the current state of talent distribution and a 16 game schedule.

I'm not saying the 2007 Patriots are really a 13-3 team that's been really lucky. Assuming they beat the Giants, my hunch is (and their stats say) they are really a 15-1 team that dodged a bullet or two (namely the Ravens and Eagles).

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6 Responses to “AFC Wildcard, Resting Starters, and 16-0 Teams”

  1. Tommy_Grand says:

    "...but the Browns do hold the tie breaker. So if they can beat a resurgent SF, they're in."

    Really? I thought if Tenn. wins, the Browns are out.

  2. Brian Burke says:

    You're right. Thanks.

    My error was in the fact that the Browns own the tiebreaker right now. But they have a inter-conference game tomorrow, and the Titans have a conference game.

    Assuming both teams win, the Titans get the spot. Assuming both teams lose, the Browns get the spot.

    I'll edit the original post.

  3. Tarr says:

    I would argue that the Pats dodged at least four bullets this year - the two you mentioned, plus the games against the Manning brothers. Both games were close enough late enough that a flukey play could have swung the result.

    The .813^16 back-of-the-envelope calculation is not very accurate, because it implicitly assumes 16 average opponents on a neutral field. In reality a team faces a mix of good and bad teams, and sometimes has the bad luck of playing the good teams on the road. If we throw a mix of 65% games and 95% games in with the 82% games, the chance of a perfect season drops somewhat.

  4. Brian Burke says:

    You're probably right.

    But wouldn't there also be the possibility that they'd get to face the harder teams at home, increasing the likelihood of going undefeated? Wouldn't those two possibilities balance in the long run? I'm not sure...

    Regarding the mix of home/away games. Home field advantage is usually around a .07 advantage (the home teams wins 57% of the time). A 2-game season for a theoretically average team would look like this: 0.43 * 0.57 = 0.245 chance of going undefeated. Without considering HFA it's just 0.5 *0.5 = 0.250. So, yes, the alternating home/away consideration would reduce the likelihood of going undefeated, but not by a large amount. That difference compounded over a 16-game season adds up, but I'm guessing we'd still be in the same ballpark of around 20 yrs.

    The other thing I did not consider is that the 2 contending teams may have to play each other (possibly twice). Because both teams can't win, the chance of 'undefeatedness' is reduced. However, the 5.6% rate I mentioned includes the rare possibility that both teams are undefeated, so we'd still have about the same chance one team would be undefeated.

    Here's another method, just for fun. Say the best team in the league each year is good enough to outright beat 12 opponents. The other 4 games are toss ups, that are ultimately decided by luck. The chance of that team converting those 4 toss-ups into wins is 0.5 ^ 4 = 0.0625, about a 6% chance of going undefeated. I'm really just trying to get a feel for the order of magnitude of it.

    But a better analysis than mine can be found at: http://thehothand.blogspot.com/

  5. Tarr says:

    Thanks for the link. It seems you could actually run with Alan's approach and add some more solid numbers to it. That is, based on your stats, and looking back at the actual schedules played, what was the a posteriori probability of the Pats going 16-0? Or the probability of either the Pats OR the Colts going 16-0? I assume the probability of any other team is insignificant.

  6. Brian Burke says:

    Tarr-Good point. It was actually very easy to do. NE's probabilities of winning every game came to 0.040. IND's chance was 0.019 (partly because they had to play NE.)

    (Plus teams like DAL or GB could have very small chances at 16-0 too.)

    The probability of either going undefeated would be 1-((1-NE)*(1-IND)) = 0.059, or about 6%, or once out of 17 years. My gut is that NE is a truly a special team, so this year might not be representative, but then again, a "special" team will appear every so often.

    But the order of magnitude is about the same. I think it's in the 20-30 year ballpark. We could go back and do some previous years, but I think we'd see some similar numbers from the 14-win Colts and 15-win Steelers from recent years.

    I think Alan's individual game probabilities are too low.

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