Change in NCAA selection criteria

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

David Harding

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Remember that this is a pairwise comparison. It is just a different weighting of the games against mutual opponents when you are being compared to School X. If you play 4 different schools 1 time each, winning them all, you only get credit if School X has played exactly those 4 opponents, too, getting the chance to tally the same 4 points. If you play 1 school 4 times, you can get anywhere from 0 to 1 points (in increments of 1/4) in the comparison, but it is only School X's record in N games against that 1 school that earns it points between 0 and 1 (in increments of 1/N).

Opponent    X     Y
A           1-0   4-0
B           1-1   0-1
C           0-0-1 0-1
D           2-0   3-0

Robb

David Harding

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Remember that this is a pairwise comparison. It is just a different weighting of the games against mutual opponents when you are being compared to School X. If you play 4 different schools 1 time each, winning them all, you only get credit if School X has played exactly those 4 opponents, too, getting the chance to tally the same 4 points. If you play 1 school 4 times, you can get anywhere from 0 to 1 points (in increments of 1/4) in the comparison, but it is only School X's record in N games against that 1 school that earns it points between 0 and 1 (in increments of 1/N).

Do you by any chance write bills for Congress?

The way I understand it (and I very easily could be wrong), if team X and team Y have 4 common opponents (A, B, C, and D) and their records against them are:

Opponent    X     Y
A           1-0   4-0
B           1-1   0-1
C           0-0-1 0-1
D           2-0   3-0

Under the old system, X was 4-1-1 (.750) and Y was 7-2 (.778), so team Y won the comparison. But what if team A and team D are the absolute worst teams in college hockey? Team Y loaded up their common opponent record with 7 wins against those cupcakes. Team X would have undoubtedly beaten them every game, too, but did not have the opportunity to play them so many times. Under the new system, X and Y have equal records (1.000) vs A and D, so those are a wash. Team X has a better record against B (.500 vs .000) and C (.500 vs .000), so overall X has a better record against 2 opponents and Y has a better record against 0 opponents, so X wins the common opponents comparison point.

Are we saying the same thing? Do I have it wrong?

ursusminor

Robb

David Harding

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Remember that this is a pairwise comparison. It is just a different weighting of the games against mutual opponents when you are being compared to School X. If you play 4 different schools 1 time each, winning them all, you only get credit if School X has played exactly those 4 opponents, too, getting the chance to tally the same 4 points. If you play 1 school 4 times, you can get anywhere from 0 to 1 points (in increments of 1/4) in the comparison, but it is only School X's record in N games against that 1 school that earns it points between 0 and 1 (in increments of 1/N).

Do you by any chance write bills for Congress?

The way I understand it (and I very easily could be wrong), if team X and team Y have 4 common opponents (A, B, C, and D) and their records against them are:

(table I screwed up and will not retype)

Under the old system, X was 4-1-1 (.750) and Y was 7-2 (.778), so team Y won the comparison. But what if team A and team D are the absolute worst teams in college hockey? Team Y loaded up their common opponent record with 7 wins against those cupcakes. Team X would have undoubtedly beaten them every game, too, but did not have the opportunity to play them so many times. Under the new system, X and Y have equal records (1.000) vs A and D, so those are a wash. Team X has a better record against B (.500 vs .000) and C (.500 vs .000), so overall X has a better record against 2 opponents and Y has a better record against 0 opponents, so X wins the common opponents comparison point.

Are we saying the same thing? Do I have it wrong?

It seems to me, at least from that article, that X's records are 1.000, .500, .500, and 1.000 against the four opponents, respectively, while Y's are 1.000, .000. .000, and 1.000, respectively. So X wins 3.000 to 2.000. This could turn out differently than your interpretation, but I leave it as an exercise for the reader.

Opponent    X     Y    %    
A           1-0   4-0  -
B           1-1   3-2  Y: .1
C           0-0-1 3-2  Y: .1
D           2-0   3-0  -
E           1-0   1-4  X: .8

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. I you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

RichH

Jeff Hopkins '82
It probably discourages scheduling a 2-game weekend OOC series as well.

It's this bit that I find intriguing (and agree with). The ECAC has the potential to benefit from western teams having a new reason to come east (two teams in one weekend that is far less practical out west). It's a lot harder to go the opposite direction. Teams within an easy drive of NYC and Boston are ideally situated to take advantage as cost plays a big role. Two western teams could pair up and play two eastern teams that are relatively easy to get to as happens somewhat sporadically now.

Another note is that this would seem to put an incentive on in-season tourney organizers to diversify the field as much as possible. May also result in more such tournaments hosted by the big boys in lieu of weekend series.

Can they finally ditch the Ivy Showcase?

 

___________________________
"Mark Mazzoleni looks like a guy whose dog just died out there..."

billhoward

It still will get us placed in the Gopher State bracket, against Minnesota, at the Minnesota rink, and all cowbells will be confiscated. Same as the old system.

ugarte

ursusminor

Robb

David Harding

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. If you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Remember that this is a pairwise comparison. It is just a different weighting of the games against mutual opponents when you are being compared to School X. If you play 4 different schools 1 time each, winning them all, you only get credit if School X has played exactly those 4 opponents, too, getting the chance to tally the same 4 points. If you play 1 school 4 times, you can get anywhere from 0 to 1 points (in increments of 1/4) in the comparison, but it is only School X's record in N games against that 1 school that earns it points between 0 and 1 (in increments of 1/N).

Do you by any chance write bills for Congress?

The way I understand it (and I very easily could be wrong), if team X and team Y have 4 common opponents (A, B, C, and D) and their records against them are:

(table I screwed up and will not retype)

Under the old system, X was 4-1-1 (.750) and Y was 7-2 (.778), so team Y won the comparison. But what if team A and team D are the absolute worst teams in college hockey? Team Y loaded up their common opponent record with 7 wins against those cupcakes. Team X would have undoubtedly beaten them every game, too, but did not have the opportunity to play them so many times. Under the new system, X and Y have equal records (1.000) vs A and D, so those are a wash. Team X has a better record against B (.500 vs .000) and C (.500 vs .000), so overall X has a better record against 2 opponents and Y has a better record against 0 opponents, so X wins the common opponents comparison point.

Are we saying the same thing? Do I have it wrong?

It seems to me, at least from that article, that X's records are 1.000, .500, .500, and 1.000 against the four opponents, respectively, while Y's are 1.000, .000. .000, and 1.000, respectively. So X wins 3.000 to 2.000. This could turn out differently than your interpretation, but I leave it as an exercise for the reader.

It isn't the same thing and I think ursusminor is correct. Take for example:

Opponent    X     Y    %    
A           1-0   4-0  -
B           1-1   3-2  Y: .1
C           0-0-1 3-2  Y: .1
D           2-0   3-0  -
E           1-0   1-4  X: .8

Y wins more H2H matchups, 2-1 with 2 ties, but X wins the comparison (X: 1 + .5 + .5 + 1 + 1 = 4;Y: 1 + .6 + .6 + 1 + .2 = 3.4). It isn't just number of comparisons won, it is also how decisively the comparisons were won.

David Harding

ugarte

ursusminor

Robb

David Harding

Jeff Hopkins '82
If I'm reading it right, I think this puts a premium on the number of different schools you play. If you play 4 games against 4 different schools and win all four, you get four points (4 x 1.000). But if you play one school 4 times and win all 4 you get 1 point.

Remember that this is a pairwise comparison. It is just a different weighting of the games against mutual opponents when you are being compared to School X. If you play 4 different schools 1 time each, winning them all, you only get credit if School X has played exactly those 4 opponents, too, getting the chance to tally the same 4 points. If you play 1 school 4 times, you can get anywhere from 0 to 1 points (in increments of 1/4) in the comparison, but it is only School X's record in N games against that 1 school that earns it points between 0 and 1 (in increments of 1/N).

Do you by any chance write bills for Congress?

The way I understand it (and I very easily could be wrong), if team X and team Y have 4 common opponents (A, B, C, and D) and their records against them are:

(table I screwed up and will not retype)

Under the old system, X was 4-1-1 (.750) and Y was 7-2 (.778), so team Y won the comparison. But what if team A and team D are the absolute worst teams in college hockey? Team Y loaded up their common opponent record with 7 wins against those cupcakes. Team X would have undoubtedly beaten them every game, too, but did not have the opportunity to play them so many times. Under the new system, X and Y have equal records (1.000) vs A and D, so those are a wash. Team X has a better record against B (.500 vs .000) and C (.500 vs .000), so overall X has a better record against 2 opponents and Y has a better record against 0 opponents, so X wins the common opponents comparison point.

Are we saying the same thing? Do I have it wrong?

It seems to me, at least from that article, that X's records are 1.000, .500, .500, and 1.000 against the four opponents, respectively, while Y's are 1.000, .000. .000, and 1.000, respectively. So X wins 3.000 to 2.000. This could turn out differently than your interpretation, but I leave it as an exercise for the reader.

It isn't the same thing and I think ursusminor is correct. Take for example:

Opponent    X     Y    %    
A           1-0   4-0  -
B           1-1   3-2  Y: .1
C           0-0-1 3-2  Y: .1
D           2-0   3-0  -
E           1-0   1-4  X: .8

Y wins more H2H matchups, 2-1 with 2 ties, but X wins the comparison (X: 1 + .5 + .5 + 1 + 1 = 4;Y: 1 + .6 + .6 + 1 + .2 = 3.4). It isn't just number of comparisons won, it is also how decisively the comparisons were won.

I agree with usrsaminor's calculation. The point I was trying to make in response to Jeff's comment ("I think this puts a premium on the number of different schools you play." ) was that you don't necessarily gain an advantage by playing lots of opponents. It does give you a better chance of having some overlap with any other school, but that school has the opportunity to gain the same number of points in the comparison as you do, regardless how many time you play any one school.

Jeff Hopkins '82
You're right, I missed the fact that it was a comparison of common opponents. I was applying it to the general TUC record.