2015 ECAC Playoff Permutations
Posted by Give My Regards
2015 ECAC Playoff Permutations
Posted by: Give My Regards (---.nys.biz.rr.com)
Date: February 23, 2015 04:32PM
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)
___________________________
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
Edited 1 time(s). Last edit at 02/23/2015 06:00PM by Give My Regards.
Re: 2015 ECAC Playoff Permutations
Posted by: Give My Regards (---.nys.biz.rr.com)
Date: February 23, 2015 04:34PM
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.
___________________________
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
Re: 2015 ECAC Playoff Permutations
Posted by: Jim Hyla (---.syrcny.fios.verizon.net)
Date: February 23, 2015 05:44PM
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?
___________________________
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
Re: 2015 ECAC Playoff Permutations
Posted by: Give My Regards (---.nys.biz.rr.com)
Date: February 23, 2015 06:01PM
Dammit, pasted the wrong thing. Fixed now.
___________________________
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
Re: 2015 ECAC Playoff Permutations
Posted by: marty (---.nycap.res.rr.com)
Date: February 23, 2015 07:14PM
Give My Regards
Dammit, pasted the wrong thing. Fixed now.
I thought maybe this was a pure season thing.
Thanks for the fix.
Re: 2015 ECAC Playoff Permutations
Posted by: RichH (134.223.230.---)
Date: February 23, 2015 07:40PM
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.
Edited 2 time(s). Last edit at 02/23/2015 07:43PM by RichH.
Re: 2015 ECAC Playoff Permutations
Posted by: Give My Regards (---.nys.biz.rr.com)
Date: February 23, 2015 10:31PM
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)
___________________________
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
If you lead a good life, go to Sunday school and church, and say your prayers every night, when you die, you'll go to LYNAH!
Re: 2015 ECAC Playoff Permutations
Posted by: Trotsky (---.dc.dc.cox.net)
Date: February 24, 2015 09:48AM
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.
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.
Re: 2015 ECAC Playoff Permutations
Posted by: RichH (---.hsd1.ct.comcast.net)
Date: February 24, 2015 11:05AM
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.
Re: 2015 ECAC Playoff Permutations
Posted by: KGR11 (---.stantec.com)
Date: February 24, 2015 01:30PM
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?
Re: 2015 ECAC Playoff Permutations
Posted by: Trotsky (---.dc.dc.cox.net)
Date: February 24, 2015 02:01PM
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
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?
Re: 2015 ECAC Playoff Permutations
Posted by: marty (---.sub-70-209-141.myvzw.com)
Date: February 24, 2015 03:05PM
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?
Re: 2015 ECAC Playoff Permutations
Posted by: RichH (134.223.230.---)
Date: February 24, 2015 03:21PM
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]
Re: 2015 ECAC Playoff Permutations
Posted by: Trotsky (---.dc.dc.cox.net)
Date: February 24, 2015 04:25PM
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.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?
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.
Edited 2 time(s). Last edit at 02/24/2015 04:39PM by Trotsky.
Re: 2015 ECAC Playoff Permutations
Posted by: Jim Hyla (---.twcny.res.rr.com)
Date: February 24, 2015 06:05PM
Trotsky
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.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?
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.
___________________________
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
Re: 2015 ECAC Playoff Permutations
Posted by: Trotsky (---.hsd1.md.comcast.net)
Date: February 24, 2015 07:12PM
But at least you are using the same rules to resolve all the teams tied for a given slot.Jim Hyla
Trotsky
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.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?
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.
Re: 2015 ECAC Playoff Permutations
Posted by: Jim Hyla (---.twcny.res.rr.com)
Date: February 27, 2015 10:42PM
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.
Edit: and in 7 we'd play Union or RPI.
___________________________
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
"Cornell Fans Made the Timbers Tremble", Boston Globe, March/1970
Cornell lawyers stopped the candy throwing. Jan/2005
Edited 1 time(s). Last edit at 02/27/2015 10:45PM by Jim Hyla.
Re: 2015 ECAC Playoff Permutations
Posted by: Icy (---.0377.apn.wlan.wireless-pennnet.upenn.edu)
Date: February 27, 2015 11:28PM
All permutations aside, I was impressed with the comeback tonight.
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