- Sep 29, 2014
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At least it was not the dreaded A beat B and B beats C and C beat A...nobody is every happy with those
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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.
Just to continue the discussion. If D had beaten B would you rank them DBCA or ACDB?
Win all your games and you never have to worry about seeding
Except if another team in the pool also won all of their games, and then you still end up in tiebreaker land.
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..