Re: Hockey East - Who's in, who's out, who's home: by the numbers - 2009-10 edition
Doesn't top-down make more sense? That way you're rewarding the team that has done the best against the other teams they are tied with?
I think "more sense" in this case, as it so often is, is arbitrary. The case can be made for any approach.
For example:
They could just take the initial results of the first tie-break calculation and seed everyone there, all at once. Why throw anyone out and recalculate at all? Aren't they
all tied with
each other?
If teams within the group are tied (say, the middle two of a four-way tie), do an internal tiebreaker to split those and decide
those seeds (which one is 2nd of 4 and 3rd of 4), but not have it effect the others (first would still be first and fourth would still be fourth).
Or...
Cutting from the bottom up means that in the end you're comparing the two strongest teams head-to-head. Why should it matter how the two best teams did against the worst team in a five-way tie? Isn't it more important how they did against each other? (Of course, you risk skewing who is "best" every time you recalculate with a different set of members.)
Or...
Placing from the top down means that the best of the group is first and then the best of the re-compared remainders is next, and so on. (See above re: skewing.)
To demonstrate:
........ A ........ B ......... C ......... D ..... Total
A ..... x ...... 0-3-0 ... 3-0-0 ... 2-0-1 = 5-3-1
B .. 3-0-0 ...... x ...... 1-2-0 ... 1-2-0 = 5-4-0
C .. 0-3-0 ... 2-1-0 ...... x ...... 2-1-0 = 4-5-0
D .. 0-2-1 ... 2-1-0 ... 1-2-0 ...... x .. = 3-5-1
With the
all-at-once method, you're done. The seedings are A, B, C, D. That seems fair to me. They're all tied. Their mutual records are taken into consideration. We have seedings and can move on.
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Using
bottom-up, you'd toss out D. Then you'd re-compare the others with A at 3-3-0, B at 4-2-0, and C at 2-4-0. C gets tossed. Now you'd compare A and B and the sweep makes B the top seed and A second. I can see the viscerally-pleasing logic in that, where A and B were the top two - separated only by a half-game in the standings - and B swept A, so they get the nod. Final seedings: B A C D.
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But what about the fact that the
only series that B won was against A, yet now they're the top seed of the group? Whether you win a game by 3-0 or 2-1, it's still two points - so why does a sweep against one team take more precendence than losing two other series more narrowly? Couldn't the test be that A won two
series and C won two series? Couldn't the top H2H be A v C (with A getting the nod) and then one-series winners B and D match up, making the seeds A C D B? In that case, A, C, and D keep their relative seedings, but B moves from best to worst.
(What if C and D split 1-1-1 - making their overall records 3-5-1 and 3-4-2, respectively? If we go by
series wins, we'd then have A at 2-1-0, C and D at 1-1-1 and B at 1-2-0, breaking the C/D tie with league wins, I guess.)
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Working
top-down, you'd take A first and also take the three wins for B, and three losses for C, out of the mix. B is now 2-4-0, C is 4-2-0 and D is 3-3-0. So C is 2nd and D takes 3rd by virtue of their H2H win over B. Final seedings: A C D B. (See above re: best to worst.)
Does it make "more sense" that the second overall team, who swept the top overall team, is the bottom seed of the four? To Hockey East, this year - and apparently you - it does.
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What seems clear in this scenario is that A, C, and D should be seeded in that relative order. What is unclear is how to evaluate B. Are they best because they beat, even
swept, the otherwise best team (bottom-up)? Are they worst because they couldn't beat
either of the two otherwise worst teams (top-down & series)? Or should some weight be given to all sets of results and they're somewhere in the middle (all-at-once)?
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One of the things that seems counter-intuitive to me is the throwing out of teams and re-breaking the tie - which can lead to different criteria being used when the current criteria are sufficient. In the potential case of UMA/NU/UVM heading into last night, they would go to league wins to split the round-robin 3-3-0 tie, and then compare the remaining two of the teams H2H once a winner (or loser) was pulled from the three. I understand the mechanics of how that works, but once we've gone to league wins, we've determined that H2H is insufficient - so why go back up to H2H record when the teams are
not tied at the criterion you just used?
FWIW, in the case of UMA/NU/UVM, which tie-breaker is chosen decides whether NU would have been 1st (bottom-up), 2nd (all at once) or 3rd (top-down) of that group.
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Personally, I think all-at-once is the fairest method - and the most transparent to the casual observer. Since
all of the teams are contributing to the tie, shouldn't they
all be considered simultaneously to determine the outcome for
all of the teams? Given the choice between
only bottom-up and top-down, I'd pick top down because at least the top choice seems correct to me, but I think that's a false choice.
Ultimately, for the purpose of this thread, the concern of some (myself included) is more a matter of taking one method, whatever it is, and sticking with it or announcing the change.