tag:blogger.com,1999:blog-38600807.post2720787169871290133..comments2018-06-02T14:19:34.554-04:00Comments on Advanced Football Analytics (formerly Advanced NFL Stats): AFC Wildcard, Resting Starters, and 16-0 TeamsUnknownnoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-38600807.post-62149781940807641002008-01-07T22:14:00.000-05:002008-01-07T22:14:00.000-05:00Tarr-Good point. It was actually very easy to do. ...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.) <BR/><BR/>(Plus teams like DAL or GB could have very small chances at 16-0 too.)<BR/><BR/>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. <BR/><BR/>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.<BR/><BR/>I think Alan's individual game probabilities are too low.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-22523067501523882472008-01-03T11:13:00.000-05:002008-01-03T11:13:00.000-05:00Thanks for the link. It seems you could actually ...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.Tarrhttps://www.blogger.com/profile/14368810359650066790noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-58994552396360316772008-01-02T19:45:00.000-05:002008-01-02T19:45:00.000-05:00You're probably right. But wouldn't there also be ...You're probably right. <BR/><BR/>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...<BR/><BR/>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.<BR/><BR/>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.<BR/><BR/>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.<BR/><BR/>But a better analysis than mine can be found at: http://thehothand.blogspot.com/Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-13416440940146354602008-01-02T18:56:00.000-05:002008-01-02T18:56:00.000-05:00I would argue that the Pats dodged at least four b...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.<BR/><BR/>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.Tarrhttps://www.blogger.com/profile/14368810359650066790noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-85512866418177458662007-12-29T20:00:00.000-05:002007-12-29T20:00:00.000-05:00You're right. Thanks.My error was in the fact that...You're right. Thanks.<BR/><BR/>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.<BR/><BR/>Assuming both teams win, the Titans get the spot. Assuming both teams lose, the Browns get the spot.<BR/><BR/>I'll edit the original post.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-2191659787298494172007-12-29T18:41:00.000-05:002007-12-29T18:41:00.000-05:00"...but the Browns do hold the tie breaker. So if ..."...but the Browns do hold the tie breaker. So if they can beat a resurgent SF, they're in."<BR/><BR/>Really? I thought if Tenn. wins, the Browns are out.Anonymousnoreply@blogger.com