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2026 ACC Tiebreaker Unveiled at Media Days

CBurrr93 said:
So 8-0 automatically ahead of 8-1? I’m genuinely asking. In a meeting can’t read it in depth. That doesn’t seem like a great system since who is to say the 8-0 team wouldn’t be 8-1 if they played 9 games. Again I’m reading sparingly. May be missing a key point of your post/the ACC post
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Yes. The 8-0 team will be in over the 8-1 team, that's not considered a tie.
 
Looks good to me, can't see a problem...................


Crazy Eyes Looks Fine GIF by MOODMAN
<-------- every ACC executive
 
Zombie-Cane said:
Looks good to me, can't see a problem...................


Crazy Eyes Looks Fine GIF by MOODMAN
<-------- every ACC executive
Click to expand...

Absolute joke of a conference.

Had a dumb tie-breaker and an odd number of teams. Attempted to replace the dumb tie-breaker. In the same off-season, stupidly decided to increase the required conference games to (mostly) 9, even though it's mathematically impossible for every team to play 9 in-conference games. So the old dumb tie-breaker gets replaced with a new dumb tie-breaker.

And why does the ACC only have 17 members for football? Oh, yeah, stupidly allowed ND to be an ACC member in every sport but football (and men's hockey), and even threw them a life raft in 2020 just for fun.
 
No_Fly_Zone said:
Again, no to the bolded. An 8-0 team does NOT tie an 8-1 team. The 8-0 team has the better winning percentage, so they are not "tied teams" pursuant to Section 1. The section you are quoting from starts with "once the tied teams have been determined," making it inapplicable in your example. And if you are thinking Section 1.b. can be applied to have an 8-1 team "tie" an 8-0 team, you are misreading that section, which can only ever apply for the benefit of the team playing the alternate schedule (i.e., 8 games).

The part in italics is correct, though, but you're expressing it opposite of how I would. The 8-1 team has the better winning percentage than the 7-1 team. But, under Section 1.b., the 7-1 team is considered tied with the 8-1 team.
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This is the first section:

1. Defining Tied Teams:
a. Step 1: Identify the team or teams with the best Conference win percentage;
plus,
b. Step 2: Any team or teams which played an alternate number of Conference games and have either the same number of Conference wins or the same number of Conference losses as the team(s) identified in Step 1.
c. No other teams may be defined as Tied Teams.

Maybe I'm missing something but there's no way teams with dissimilar number of conference games are ever going to have the same win percentage with matching wins or losses, unless they're both undefeated. So why would the first part stand alone and not be in conjunction with part b?

I think the 8-0 team ties the 8-1 team the way it's written. You're going to punish a team that won the same number of games as the other but had to play an additional game? This is the, "why play a conference championship game, it can only hurt you", argument all over again. Imagine a scenario where Miami plays an identical conference schedule to another ACC team, they both win all those games going 8-0, but then Miami has to play a ranked Pitt team at the end of the season and losses and the other team doesn't have to play any additional games.
 
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hurricaneman said:
This is the first section:

1. Defining Tied Teams:
a. Step 1: Identify the team or teams with the best Conference win percentage;
plus,
b. Step 2: Any team or teams which played an alternate number of Conference games and have either the same number of Conference wins or the same number of Conference losses as the team(s) identified in Step 1.
c. No other teams may be defined as Tied Teams.

Maybe I'm missing something but there's no way teams with dissimilar number of conference games are ever going to have the same win percentage with matching wins or losses, unless they're both undefeated. So why would the first part stand alone and not be in conjunction with part b?

I think the 8-0 team ties the 8-1 team the way it's written. You're going to punish a team that won the same number of games as the other but had to play an additional game? This is the, "why play a conference championship game, it can only hurt you", argument all over again. Imagine a scenario where Miami plays an identical conference schedule to another ACC team, they both win all those games going 8-0, but then Miami has to play a ranked Pitt team at the end of the season and losses and the other team doesn't have to play any additional games.
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What I’m figuring too. But I admittedly didn’t care to read it very thoroughly. But alas. Hopefully we steamroll everyone and finally be freed of tie breaker scenarios this year
 
This was from CBS sport:

Defining a new tiebreaker procedure is particularly important for the ACC in 2026 as it moves to a new conference schedule in which some teams will play eight conference games while others will play nine. As such, the definition of "tied teams" includes not just squads with identical records but also those with the same number of conference wins or losses.

So if one team is 7-2 and another 7-1, they could be considered "tied" for the purposes of the tiebreaker process. This was done, Phillips explained, to both make sure that teams are not punished or helped by having more or fewer conference games than another.
 
No_Fly_Zone said:
Again, no to the bolded. An 8-0 team does NOT tie an 8-1 team. The 8-0 team has the better winning percentage, so they are not "tied teams" pursuant to Section 1. The section you are quoting from starts with "once the tied teams have been determined," making it inapplicable in your example. And if you are thinking Section 1.b. can be applied to have an 8-1 team "tie" an 8-0 team, you are misreading that section, which can only ever apply for the benefit of the team playing the alternate schedule (i.e., 8 games).

The part in italics is correct, though, but you're expressing it opposite of how I would. The 8-1 team has the better winning percentage than the 7-1 team. But, under Section 1.b., the 7-1 team is considered tied with the 8-1 team.
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I think separately a 8-0 and 8-1 team would be considered tied. Then separately a 8-1 and 7-1 team would be considered tied. But it would not be viewed as a 3 team tie with the 8-0, 8-1, and 7-1 team. And you also do the seeding top down….

So in effect they start with the top 2 team tie breaker between 8-0 and 8-1 team. And determine h2h and then sports source ranking for winner. If the 8-0 team is the winner of tie breaker they are in. And then you go to next tie breaker between the 8-1 team and the 7-1 team with same tie breaker method of h2h then sports source ranking.

If however the 8-1 team is winner of that original tie breaker, then they get the nod and THEN you would determine 2nd place between the 8-0 team and the 7-1 team, which means there is no tie to break, and thus the 8-0 team is in 2nd…

This imo makes sense actually. And given conference size and lack of comparable SOS, UNLESS you completely ignore W-L and go immediately to SportsSpurce/CFp rankings for the full ACC champ selections, this is really only decent way to do it
 
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