As the title suggests, I have an NFL tiebreaker question. It is concerning best win loss percentage in common games. This is purely hypothetical but lets say we have two teams from different divisions that are tied: The Giants and the Saints. (again, this is hypothetical) Lets say one of their common opponents was the Eagles. Since the Eagles are in the Giants division, the Giants will obviously play them twice, while the Saints will only play them once. The question is: Do you count both Giants vs Eagles games, while only counting the one Saints vs Eagles game when comparing their win loss percentage in common games? How does this apply to the rule that there must be a minimum of 4 common games to use this tiebreaker?
Yes, you count both games for calculating the Giants’ won/lost percentage. The four game thing means that both the Giants and the Saints must have at least four games each to calculate the percentage. It’s ok if the Giants have five games and the Saints only four.
Right on. Thanks for the response by the way. It’s easy to pull up the NFL tiebreaker procedure, but finding more in-depth explanations for it has been pretty tough.
I was just studying this because I was wondering what would happen if the Chargers lost their last two games and the Bengals or Pats won their last two.
Chargers beat out the Bengals because of head to head.
Chargers never played the Patriots. Chargers and Patriots would have the same Conference record. Next is won/loss in common games but I didn’t have the time to figure out what that would be yet.
Did you find a website that explains this stuff in more detail?
I just Googled to find my answer. In the off chance that anyone cares, I will share it with you.
If the Chargers and Patriots both end up at 11-5 and the Bengals don’t, the Patriots would get the 2nd seed over the Chargers because of records against common opponents.
If the Chargers, Patriots and Bengals all end up at 11-5, it becomes more complicated because the three way tie rules are more complicated and the result can’t be projected at this point.
Not exactly. I found places where they explained specific situations and applied them to the general.
Given the current scheduling algorithm, here is what will happen if two teams from the same conference, but different divisions, end up tied:
They have three common opponents, totalling four games. This happens if they are not from divisions in the conference matched against each other for the season, and if they did not finish the same placing the year before in their division (e.g.: both second place in their respective divisions).
They have two common opponents, totalling two games. This happens if they are not from matched divisions, but they DID finish the same placing, since they will each play all the teams in the conference with that same placing. But note bene: in this case, they will have played each other, and this tiebreak will only be involved in the rare instance of a tied result between them.
They have six common opponents, totalling nine games. This happens if they are from matched divisions, but did not finish in the same placing in their division. In each case, the number of games with common opponents will total nine: each team will have played the common opponents within its division twice (six games) and the common opponents from the other division once (three games). Once again, in this situation the teams have had a head-to-head meeting.
They have eight common opponents, totalling 11 games. This happens if they are from matched divisions, and they finished the last season tied for division placing. They will have the same six common opponents from 3) above, but will also both have played the teams from the other divisions in the conference who placed last year at the same level in their own divisions. Again, however, in this case they will have played each other.
Examples of each, using the Green Bay Packers (3d last year in the NFC North):
Philadelphia Eagles (2d last year in the NFC Esst), whose common opponents with the Packers are the Dallas Cowboys (2games in division), the Tampa Bay Buccaneers (matched division), and the Chicago Bears (Packers’ division 2d place).
Dallas Cowboys (3d last year in the NFC East), whose common opponents with the Packers are the Tampa Bay Buccaneers (3d NFC South) and the Seattle Seahawks (3d NFC West).
San Francisco 49ers (2d last year in the matched NFC West), whose common opponents with the Packers are the Seahawks, Rams and Cardinals (2 games each), and the Bears, Vikings and Lions (one game each); the Packers of course have two each with the Bears, Vikes and Lambs, and one each with the other NFC West teams.
Seattle Seahawks (3d last year in the matched NFC West), whose common opponents are the same as the 49ers, plus the Tampa Bay Bucs and the Dallas Cowpokes.
So, unless the teams have tied head-to-head, the only scenario here that matters is category 1), in which case there are an identical number of games, just not the same number against all the teams involved.
OK, so what I’m getting out of this is that the way the schedule works, if common games need to be calculated, both teams will wind up with the same number of games played.
One nitpick though. I’m looking at the Vikings’ and Eagles’ schedules since they could wind up tied. If they do, it seems that no matter what, the Eagles win the tiebreak due to conference record, but for the hell of it, i looked at common opponents as well. From what I gathered, they will have 4 common opponents (Carolina, NYG, Chicago, and San Francisco) and 5 common games. Maybe I’m missing something here.
Oops, the first category should be modified to read:
They have four common opponents, totalling five games. This happens if they are not from divisions in the conference matched against each other for the season, and if they did not finish the same placing the year before in their division (e.g.: both second place in their respective divisions).
I forgot that both the “other” divisions will be involved. :eek: :o
Note that the Eagles play the Niners, another common opponent with the Packers, and consistent with this emendation.
OK, now it makes sense. Thanks for the detailed answer. This is one less tiebreaker procedure that I don’t get now.
Just to throw it out there, in three-way tiebreakers involving two teams in the same division, the first thing you do is use the divisional 2-way tiebreaker to knock one of the divisional teams out, then proceed to a normal two-way tiebreaker.
For example, let’s say the Falcons had clinched the #1 wildcard and the Packers, Cowboys and Giants were all fighting for the last spot and all had the same record. First you eliminate either the Cowboys or Giants based purely on division standing. So division record looms very large here even though division record normally doesn’t matter for wildcard seeding. Similarly, the two head to head games loom even larger despite the fact that head to head games normally get tossed out in three-way ties.
Once you eliminated either the Cowboys or Giants, the one left standing is now pitted against the Packers in two-way tiebreakers to see who wins the wildcard. The net result of this system is to make divisional play even more important than you’d normally think.
By extension, it also makes your games against the other conference more important than you’d think because they feature heavily in the “common games” divisional tiebreaker despite being completely irrelevant for playoff tiebreakers between teams from different divisions. (They are not conference games, and they can’t be common games either.)
Missed edit window:
So everything else being equal, if the Cowboys swept the AFC while the Giants got swept by the AFC, that means the Cowboys are ahead of the Giants for the three-way wildcard tiebreaker (by virtue of divisional tiebreaker) despite the Giants having a four-game better conference record.