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Are they considered to have the same records? One should just apply tie-breaking procedures?

NFL rules say:

If, at the end of the regular season, two or more clubs in the same division finish with identical won-lost-tied percentages

Not sure what the won-lost-tied percentage means. Above they mention that tie counts as half win, then why would you compute a "percentage" instead of simply ranking teams by their equivalent wins?

Bonus question: has this ever happened?

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2 Answers 2

up vote 9 down vote accepted

Won-Lost-Tied Percentages are applied as follows:

  • A win = 1.000
  • A loss = 0.000
  • A tie = 0.500

8-6-2: (8(1.000) + 6(0.000) + 2(0.500))/16 = (8 + 1)/16 = 9/16 = 0.563 (0.5625)

9-7-0: (9(1.000) + 7(0.000) + 0(0.500))/16 = 9/16 = 0.563 (0.5625)

*Divisor is the # of games played...in this case, we assume 16 games have been played.

Therefore, both teams W-L-T % is 0.563 and has the same record percentage.

The reason why a percentage is computed is because of the scenario above, equivalent wins (in the win column) doesn't equal equivalent record percentage (such as 9-6-1 and 9-7, 9-6-1 will have the better W-L-T %). In a extreme case, a team who is 0-0-16 has the same W-L-T % as a team who is 8-8-0 (or 7-7-2, 6-6-4, etc...).

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You say"A percentage is computed for the scenario above, equivalent wins doesn't equal equivalent records/percentage". What's the difference with saying the 8-6-2 team has 9 equivalent wins, just like the 9-7-0 team? Likewise, the 9-6-1 team has 9.5 wins clearly better than the 9-7-0 team. I just feel like the percentage is an unnecessary complication. –  FrenchKheldar Dec 9 '12 at 23:53
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A tie isn't a win. In terms of percentage, two ties for two games = .500 and one win for one game = 1.000. You compare two 16-game records...one has one less win and loss and two more ties. For all intents and purposes: 1-1-0 = 0-0-2 in terms of W-L-T %. –  edmastermind29 Dec 9 '12 at 23:57
    
I feel the percentage removes any doubt...I would be more concerned with head-to-head, division, conference, and similar schedule records to determine standings. –  edmastermind29 Dec 10 '12 at 0:01
    
The percentage can be particularly usefully mid season when not all teams have played the same number of games. While, of course, rankings mid season don't mean much, the information can still be interesting. –  kralco626 Nov 4 at 13:34
    
An interesting scenario took place with the AFC North recently. The Bengals were leading the division at 4-2-1. The Ravens and Steelers were second and third at 5-3, and the Browns were last at 4-3. –  edmastermind29 Nov 4 at 14:18

To answer the bonus question, yes, two or more ties used to happen fairly often prior to 1974 because the rules did not allow an overtime to happen at all prior to that. In 1974, the old overtime rules where the first team to score in the 15 minute overtime wins began. During this period tie games were much more infrequent than before. The new overtime rules that began in 2012 appear to have made ties slightly more prevalent than before since the opposing team now has a chance to answer if the team playing offense scores a field goal on the opening overtime drive. This is generally considered to be a fairer system than the old rules by most, but fairness comes at the cost of more ties. One thing to consider though is that tie games did not count for anything in the NFL prior to 1972. Prior to that, if two teams tied, their winning percentage was calculated as if the game never even happened.

Bonus bonus answer: The 1936 Chicago Bears hold the record for most ties in one season having 6 of them, 3 of them coming in back to back to back consecutive games. The last time any team had two or more ties in one season was the 1973 season when the Browns, Chiefs, Packers, and Broncos all had two a piece.

Data on tied NFL games can be found at http://en.wikipedia.org/wiki/List_of_NFL_tied_games

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The 1936 Bears did play in an era where the 0-0 tie was fairly common, of course... –  Joe Nov 10 at 16:03

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