Tournament Seeding question

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how do you seed these teams?

  • BCAD

    Votes: 5 17.2%
  • CABD

    Votes: 6 20.7%
  • ABCD

    Votes: 17 58.6%
  • CBDA

    Votes: 1 3.4%
  • Other (please post in thread)

    Votes: 0 0.0%

  • Total voters
    29
  • Poll closed .
Sep 29, 2014
2,421
113
At least it was not the dreaded A beat B and B beats C and C beat A...nobody is every happy with those

Sent from my XT1254 using Tapatalk
 
Jun 6, 2016
2,724
113
Chicago
I can understand the logic of tossing out head-to-head if all 4 teams didn't play each other, but I can't accept placing A ahead of C when A lost to C. That just doesn't make any sense.

I'm in the BCAD camp since the rules, perhaps poorly written, say H2H is the first tiebreaker. OK, well, all we solve there is C ahead of A. What's next? Fewest runs allowed. So we take whatever is left -- meaning C ahead of A is decided -- and use that to figure it out. B allowed fewer runs than C, so it gets ranked ahead of C (And A can't be placed ahead of C based on the first, more important tiebreaker). And D, as others all seem to agree, is last.

If the rules don't state H2H is only if there are two teams or if all teams involved have played each other, you still have to use it.
 

CoreSoftball20

Wilson = Evil Empire
DFP Vendor
Dec 27, 2012
6,235
113
Kunkletown, PA
head to head is out, since every team is 1-2 and they all didn't play each other. You cant
give whatever team higher seeding because they beat one of those 3 teams. Heck, one
of the other teams could have beaten a 2-1 team. You also cant go backwards and do runs against, then
go back to head to head if those teams played each other.

So you go to runs against and then run diff...which makes it ABCD.

But of course, I guess interpretation of the rule by the person running the tourney
is more important :)
 
Last edited:
Sep 14, 2010
10
1
I can understand the logic of tossing out head-to-head if all 4 teams didn't play each other, but I can't accept placing A ahead of C when A lost to C. That just doesn't make any sense.

I'm in the BCAD camp since the rules, perhaps poorly written, say H2H is the first tiebreaker. OK, well, all we solve there is C ahead of A. What's next? Fewest runs allowed. So we take whatever is left -- meaning C ahead of A is decided -- and use that to figure it out. B allowed fewer runs than C, so it gets ranked ahead of C (And A can't be placed ahead of C based on the first, more important tiebreaker). And D, as others all seem to agree, is last.

If the rules don't state H2H is only if there are two teams or if all teams involved have played each other, you still have to use it.

Thanks for the reply. I think most of us seem to think the rules are poorly written. But, it still is the rules that we see on most of the tournaments. And the issue that seems to mess up seedings more than anything else except perhaps an incorrect score.

Just to continue the discussion. If D had beaten B would you rank them DBCA or ACDB?
 

marriard

Not lost - just no idea where I am
Oct 2, 2011
4,319
113
Florida
Just to continue the discussion. If D had beaten B would you rank them DBCA or ACDB?

No because for head to head to count, all teams involved in a tie breaker must have played each other (so A must have played BCD, B played ACD) and so on. Also it is impossible for them to all have 1-2 records if they have all played each other.

So if nothing else had changed - just that D had beaten one of the 4 teams - it would still be ABCD

For those not getting this btw (mainly those putting C first) - once a tie breaker has been passed by, you never go back to it.
So if the first criteria (head to head in this case) doesn't make it clear for all teams positioning, you go to the NEXT criteria for the remaining teams left to seed, and you no longer consider the previously available criteria in your decision. That C beat A is completely irrelevant as head-to-head is not the seeding criteria that can be used to seed the 4 teams as the first tie breaker..
 
Last edited:

Greenmonsters

Wannabe Duck Boat Owner
Feb 21, 2009
6,165
38
New England
Except if another team in the pool also won all of their games, and then you still end up in tiebreaker land.

if you win ALL your games, it makes no difference who you play

IME after a team has won a tournament or two, they realize that a tourney win isn't really special unless they earned it by beating the best teams
 
Jun 5, 2012
38
8
Ontario, Canada
No because for head to head to count, all teams involved in a tie breaker must have played each other (so A must have played BCD, B played ACD) and so on. Also it is impossible for them to all have 1-2 records if they have all played each other.

So if nothing else had changed - just that D had beaten one of the 4 teams - it would still be ABCD

For those not getting this btw (mainly those putting C first) - once a tie breaker has been passed by, you never go back to it.
So if the first criteria (head to head in this case) doesn't make it clear for all teams positioning, you go to the NEXT criteria for the remaining teams left to seed, and you no longer consider the previously available criteria in your decision. That C beat A is completely irrelevant as head-to-head is not the seeding criteria that can be used to seed the 4 teams as the first tie breaker..

I agree with marriard 100% that this is how it should be but if I was team B or C I would certainly be offering a different interpretation. TD's please save everyone (including yourself) the grief and write it into the rules!
 

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