2015 ECAC Playoff Permutations

Short version first:

Possible ECAC tournament seedings (the number in parentheses is the best
seed the team can get with no help):

Quinnipiac     1      (1)
St. Lawrence   2      (2)
Yale           3-7    (3)
Harvard        3-7    (4)
Colgate        3-7    (5)
Dartmouth      3-8    (6)
Cornell        3-8    (7)
Clarkson       6-9    (8)
Rensselaer     8-11   (9)
Union          9-11   (10)
Brown          9-11   (11)
Princeton      12     (12)

And now the too-much-time-on-my-hands version:

Back and... well, at least not much worse than ever, it’s the ECAC Playoff
Permutation Spectacular!  Let’s get the easy ones out of the way first; no
matter what happens, Quinnipiac, St. Lawrence, and Princeton are guaranteed to
finish first, second, and twelfth respectively.  However, plenty of intrigue is
still left over for positions three through eleven.

Going into the final weekend of league play, here's a breakdown of where
each team in the ECAC could finish.  As always, I'm greatly indebted to
John Whelan's excellent playoff possibilities script at [www.elynah.com]

For each ECAC team, I've listed the following:

THIS WEEKEND:  The team's weekend games, its last two of the season.
ON THEIR OWN:  The highest the team could finish with no help from the
     competition.  Generally, this involves a weekend sweep.
BEST CASE:  The highest the team could finish if everything goes right.
WORST CASE:  The lowest the team could finish if everything goes wrong.
     This generally involves getting swept while teams nearby in the
     standings win.
TIEBREAKERS:  How the team would fare if they finished the season tied with
     some other team which is currently close (i.e. within 4 points) in the
     standings.  Note that there may be cases in which Team A "could win or
     lose" the tiebreaker against Team B, depending on whether there are
     more than just those two teams tied.  For instance, Dartmouth wins the
     head-to-head tiebreaker against Cornell with a 2-0 record; however, in a
     five-way tie (yes, I said five) involving these two, Yale, Harvard, and 
     Colgate, Dartmouth would actually be seeded lower than Cornell.

For two or more teams tied in the standings, the ECAC tiebreakers are:

1.   Head-to-head record in ECAC games (non-conference meetings, such as in
     tournaments, do not count).
2.   League wins.
3.   Record against the top four teams in the conference.
4.   Record against the top eight teams in the conference.
5.   Goal differential (net goals) head-to-head.
6.   Goal differential against the top four teams in the conference.
7.   Goal differential against the top eight teams in the conference.

Note that if the tie is among three or more teams, the tiebreaking steps are
used in order until a team, or multiple teams, is/are separated from the
"pack".  Once that happens, the process starts all over to break the remaining
ties.  For example, when the above steps are applied to a four-way tie, once
one team is separated out leaving a three-way tie, the procedure goes back to
the first step with the three remaining tied teams.


Without further ado, here's how things shape up:

Yale:
     THIS WEEKEND:  Colgate, Cornell.
     ON THEIR OWN:  Clinches third with a win over Colgate.
     BEST CASE:  Third.
     WORST CASE:  Would fall to seventh if they lose twice, Harvard and Dart-
     mouth get at least three points each, Cornell also beats Brown, and
     Colgate doesn’t lose to Brown.
     TIEBREAKERS:  Beats Harvard; loses to Cornell; could win or lose against
     Colgate and Dartmouth.

Harvard:
     THIS WEEKEND:  Quinnipiac, Princeton.
     ON THEIR OWN:  A sweep guarantees fourth place.
     BEST CASE:  Finishes third with a sweep if Yale gets no more than one
     point.
     WORST CASE:  Slides to seventh with two losses if Colgate gets at least one
     point and Dartmouth and Cornell pick up at least two points each.
     TIEBREAKERS:  Beats Colgate; loses to Yale; could win or lose against Dart-
     mouth and Cornell

Colgate:
     THIS WEEKEND:  At Yale, at Brown.
     ON THEIR OWN:  Wraps up fifth with three points.
     BEST CASE:  Climbs to third with a sweep if Harvard does not win twice.
     WORST CASE:   Drops to seventh if they lose twice and Dartmouth and Cornell
     get at least two points each.
     TIEBREAKERS:  Loses to Harvard and Cornell; could win or lose against Yale
     and Dartmouth.

Dartmouth:
     THIS WEEKEND:  Princeton, Quinnipiac.
     ON THEIR OWN:  Clinches sixth with a pair of wins.
     BEST CASE:  Would rise to third with a sweep if Yale gets no more than one
     point and Harvard and Colgate each get no more than two points each.
     WORST CASE:  Ends up in eighth place with two losses if Clarkson sweeps,
     Cornell gets at least two points, and Harvard and Cornell do not both
     finish in the top four.
     TIEBREAKERS:  Could win or lose against Yale, Harvard, Colgate, Cornell,
     and Clarkson.

Cornell:
     THIS WEEKEND:  At Brown, at Yale.
     ON THEIR OWN:  One point will give Cornell seventh place.
     BEST CASE:  Would finish third with a sweep if Harvard and Dartmouth get no
     more than two points each and Colgate also beats Yale but loses to Brown.
     WORST CASE:  Falls to eighth if they lose twice and Clarkson sweeps.
     TIEBREAKERS:  Beats Yale, Colgate, and Clarkson; could win or lose against
     Harvard and Dartmouth.

Clarkson:
     THIS WEEKEND:  At Rensselaer, at Union.
     ON THEIR OWN:  Gets eighth with at least a tie against Rensselaer or a win
     over Union.
     BEST CASE:  Climbs to sixth with a sweep if Dartmouth and Cornell both lose
     twice.
     WORST CASE:  Finishes ninth if they lose twice and Rensselaer also beats
     St. Lawrence.
     TIEBREAKERS:  Loses to Cornell; could win or lose against Dartmouth and
     Rensselaer.

Rensselaer:
     THIS WEEKEND:  Clarkson, St. Lawrence.
     ON THEIR OWN:  Two points will wrap up ninth place.
     BEST CASE:  Clinches eighth with a sweep if Clarkson does not beat Union.
     WORST CASE:  Would drop to eleventh with two losses if Union and Brown get
     at least three points each.
     TIEBREAKERS:  Beats Union; loses to Brown; could win or lose against
     Clarkson.

Union:
     THIS WEEKEND:  St. Lawrence, Clarkson.
     ON THEIR OWN:  Guarantees tenth with a sweep.
     BEST CASE:  Finishes ninth with two wins if Rensselaer gets no more than
     one point.
     WORST CASE:  Falls to eleventh with two losses if Brown gets at least two
     points.
     TIEBREAKERS:  Loses to Rensselaer; could win or lose against Brown.

Brown:
     THIS WEEKEND:  Cornell, Colgate.
     ON THEIR OWN:  Has already wrapped up eleventh place and can do no better
     without help.
     BEST CASE:  Would climb to ninth if they sweep, Rensselaer gets no more
     than one point, and Union gets no more than two points.
     WORST CASE:  Eleventh.
     TIEBREAKERS:  Beats Rensselaer; could win or lose against Union.

Give My Regards
Short version first:

Possible ECAC tournament seedings (the number in parentheses is the best seed the team can get with no help):

Union          1      (1)
Colgate        2-4    (2)
Quinnipiac     2-6    (3)
Cornell        2-6    (4)
Yale           3-7    (5)
Clarkson       3-7    (6)
RPI            5-9    (7)
Brown          7-11   (7)
St. Lawrence   7-11   (9)
Dartmouth      8-11   (10)
Harvard        8-11   (11)
Princeton      12     (12)

I like it, but isn't this 2014?

Dammit, pasted the wrong thing. Fixed now.

Give My Regards
Dammit, pasted the wrong thing. Fixed now.

I thought maybe this was a pure season thing.

Thanks for the fix.

Give My Regards
And now the too-much-time-on-my-hands version:

Cornell:
     THIS WEEKEND:  At Brown, at Yale.
     ON THEIR OWN:  One point will give Cornell seventh place.
     BEST CASE:  Would finish third with a sweep if Harvard and Dartmouth get no
     more than two points each and Colgate also beats Yale but loses to Brown.
     WORST CASE:  Falls to eighth if they lose twice and Clarkson sweeps.
     TIEBREAKERS:  Beats Yale, Colgate, and Clarkson; could win or lose against
     Harvard and Dartmouth.

If H2H is the 1st tiebreaker, this is wrong.

Cornell beats Colgate (1-0-1) , Clarkson (2-0-0), Harvard (1-0-1); Loses to Dartmouth (0-2-0); Could win or lose against Yale (1-0-0).

Unless I'm missing something.

RichH

Give My Regards
And now the too-much-time-on-my-hands version:

Cornell:
     THIS WEEKEND:  At Brown, at Yale.
     ON THEIR OWN:  One point will give Cornell seventh place.
     BEST CASE:  Would finish third with a sweep if Harvard and Dartmouth get no
     more than two points each and Colgate also beats Yale but loses to Brown.
     WORST CASE:  Falls to eighth if they lose twice and Clarkson sweeps.
     TIEBREAKERS:  Beats Yale, Colgate, and Clarkson; could win or lose against
     Harvard and Dartmouth.

If H2H is the 1st tiebreaker, this is wrong.

Cornell beats Colgate (1-0-1) , Clarkson (2-0-0), Harvard (1-0-1); Loses to Dartmouth (0-2-0); Could win or lose against Yale (1-0-0).

Unless I'm missing something.

Cornell beats Yale: The only way Cornell and Yale can be tied in the standings is if Cornell sweeps and Yale also loses to Colgate. This gives Cornell a 2-0 head-to-head edge over Yale.

Cornell could win or lose against Harvard and Dartmouth: You are correct that Cornell beats Harvard head-to-head and loses to Dartmouth head-to-head -- if those situations are both two-team ties. However, if you throw other teams into the mix, things get messy. For example, if Cornell and Harvard finish tied with Dartmouth and Colgate (24 points each), Harvard gets the head-to-head tiebreaker against the other three with a 3-1-2 record compared to Cornell and Dartmouth at 2-2-2 each and Colgate at 1-3-2. So Harvard ends up with a higher seed than Cornell despite the Big Red's 1-0-1 head-to-head record against them.

Similarly, in the five-way tie (at 25 points each) among those same four teams plus Yale, Cornell would take the head-to-head tiebreaker among the five with a 4-2-2 record, giving the Big Red a higher seed than Dartmouth.

I didn't say any of this was likely, just possible

(BTW, throwing other teams in doesn't help Yale overcome Cornell's head-to-head edge)

Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

Give My Regards

RichH

Give My Regards
And now the too-much-time-on-my-hands version:

Cornell:
     THIS WEEKEND:  At Brown, at Yale.
     ON THEIR OWN:  One point will give Cornell seventh place.
     BEST CASE:  Would finish third with a sweep if Harvard and Dartmouth get no
     more than two points each and Colgate also beats Yale but loses to Brown.
     WORST CASE:  Falls to eighth if they lose twice and Clarkson sweeps.
     TIEBREAKERS:  Beats Yale, Colgate, and Clarkson; could win or lose against
     Harvard and Dartmouth.

If H2H is the 1st tiebreaker, this is wrong.

Cornell beats Colgate (1-0-1) , Clarkson (2-0-0), Harvard (1-0-1); Loses to Dartmouth (0-2-0); Could win or lose against Yale (1-0-0).

Unless I'm missing something.

Cornell beats Yale: The only way Cornell and Yale can be tied in the standings is if Cornell sweeps and Yale also loses to Colgate. This gives Cornell a 2-0 head-to-head edge over Yale.

Cornell could win or lose against Harvard and Dartmouth: You are correct that Cornell beats Harvard head-to-head and loses to Dartmouth head-to-head -- if those situations are both two-team ties. However, if you throw other teams into the mix, things get messy. For example, if Cornell and Harvard finish tied with Dartmouth and Colgate (24 points each), Harvard gets the head-to-head tiebreaker against the other three with a 3-1-2 record compared to Cornell and Dartmouth at 2-2-2 each and Colgate at 1-3-2. So Harvard ends up with a higher seed than Cornell despite the Big Red's 1-0-1 head-to-head record against them.

Similarly, in the five-way tie (at 25 points each) among those same four teams plus Yale, Cornell would take the head-to-head tiebreaker among the five with a 4-2-2 record, giving the Big Red a higher seed than Dartmouth.

I didn't say any of this was likely, just possible

(BTW, throwing other teams in doesn't help Yale overcome Cornell's head-to-head edge)

Great. I was wondering if you had gone into 2nd degree geekery. Thank you for clarifying. Excellent work, as always.

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

KGR11
With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

You could also differentiate the middle team, or the penultimate team, or the nth team randomly chosen at the moment of the differentiation. In all these cases and the top or bottom team, choosing to differentiate just one team from a ranked order of a full set of teams is arbitrary.

KGR11

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

marty

KGR11

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

[www.collegehockeynews.com]

marty

KGR11

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

The infinite loop is actually caused by re-differentiating. If you just started from #1, resolving all ties by single pass, and allowing lower ties to stand provisionally (in John's cited example, allowing all the teams tied for 8th to be counted in record vs top 8) without nesting the tie resolutions, there would be no return arrow and no infinite loop.

Say you begin with a 3-way tie for 4th and a 3-way tie for 8th. To resolve the tie for 4th, you allow all three teams tied for 8th to remain in the "record vs top 8" set. After a one-pass resolution, you now have just one 4th place team, so when you get to the 8th place tie you can use a 4-team "record vs top 4" set in that one-pass resolution.

Trotsky

marty

KGR11

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

The infinite loop is actually caused by re-differentiating. If you just started from #1, resolving all ties by single pass, and allowing lower ties to stand provisionally (in John's cited example, allowing all the teams tied for 8th to be counted in record vs top 8) without nesting the tie resolutions, there would be no return arrow and no infinite loop.

Say you begin with a 3-way tie for 4th and a 3-way tie for 8th. To resolve the tie for 4th, you allow all three teams tied for 8th to remain in the "record vs top 8" set. After a one-pass resolution, you now have just one 4th place team, so when you get to the 8th place tie you can use a 4-team "record vs top 4" set in that one-pass resolution.

So to pick team 4 you use all 8th teams, but to pick the 8th you only use team 4? As you said, it's arbitrary.

Jim Hyla

Trotsky

marty

KGR11

Trotsky
Huh, I've never read your table that way before. I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on? I've never really liked that, preferring to resolve the whole n-tuple on the first go. Can anybody make a compelling case for why they do it their way? I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens. So you could end up with a top team(s) or a bottom team(s). I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

The infinite loop is actually caused by re-differentiating. If you just started from #1, resolving all ties by single pass, and allowing lower ties to stand provisionally (in John's cited example, allowing all the teams tied for 8th to be counted in record vs top 8) without nesting the tie resolutions, there would be no return arrow and no infinite loop.

Say you begin with a 3-way tie for 4th and a 3-way tie for 8th. To resolve the tie for 4th, you allow all three teams tied for 8th to remain in the "record vs top 8" set. After a one-pass resolution, you now have just one 4th place team, so when you get to the 8th place tie you can use a 4-team "record vs top 4" set in that one-pass resolution.

So to pick team 4 you use all 8th teams, but to pick the 8th you only use team 4? As you said, it's arbitrary.

But at least you are using the same rules to resolve all the teams tied for a given slot.

We can't go lower than 7, nor higher than 5. The keys are the Harvard and Colgate games. If we win and both lose, we get 5. If only 1 loses, 6. If both win, we're 7. In fact we're 7 win or lose to Yale. We're also 7 if we all lose. So you could say the overwhelming odds are for us to be 7. And I have to decide whether I want to drive to New Haven and back tomorrow.

Edit: and in 7 we'd play Union or RPI.

All permutations aside, I was impressed with the comeback tonight.