NESCAC 2014

[Becks]

If Wes fails to beat Conn today, it will be the first time they have gone winless in NESCAC since 2003.

Winless NESCAC teams to date (reverse chronological order):
2009 Conn 0-9-0
2008 Conn 0-8-1
2007 Colby 0-7-2
2006 Conn 0-8-1
2006 Trinity 0-8-1
2003 Wesleyan 0-9-0
2002 Wesleyan 0-8-1
2000 Wesleyan 0-8-1

[jc2]

I suspect most people are predicting that Conn, Amherst and Bowdoin will all win their last games and end up in a 3-way tie for second place. Massey lists the chances of winning as: Amherst 72%, Conn 84% and Bowdoin 68%. The chance of all 3 winning is then .72 x .84 x .68 = .41.

Surprisingly, if the Massey odds are accurate, it's more likely that at least one of the teams will lose than all 3 will win.

Only time will tell.

Actually, looking back at this. You are correct that the chance of all 3 winning is .41. But that doesn't mean it is more likely that at least one of them will lose, because ties are also possible. Massey provides both teams' chances of winning, and using those figures, we can easily calculate the chances of a tie as well. Here's what Massey says are Conn's, Amherst's and Bowdoin's chances of winning, tying and losing.

Conn - 84%, 9%, 7%
Amherst - 72%, 13%, 15%
Bowdoin - 68%, 13%, 19%

So, if the likelihood of those teams getting a win or a tie is 93%, 85%, and 81%, respectively, the likelihood of all 3 getting a win or a tie is the product of those 3 or 64%, which means that the chances of at least one of the three losing is only 36%. So, while it is more likely that one of the 3 will lose or tie (59%) than it is that all 3 will win (41%), it is more likely that all 3 will win (41%) than that one of them will lose (36%).

Correct as usual. I used the term "lose" rather loosely. I meant to imply that failing to get 3 points would drop that team out of the three-way tie for 2nd place.

[jc2]

If Wes fails to beat Conn today, it will be the first time they have gone winless in NESCAC since 2003.

Winless NESCAC teams to date (reverse chronological order):
2009 Conn 0-9-0
2008 Conn 0-8-1
2007 Colby 0-7-2
2006 Conn 0-8-1
2006 Trinity 0-8-1
2003 Wesleyan 0-9-0
2002 Wesleyan 0-8-1
2000 Wesleyan 0-8-1

Add Wes to the list. 

[jc2]

Entering the last few days of NESCAC play, only Wesleyan is out of the playoffs. Looks like there is a good chance of a 3 way tie for second thru fourth place and the tie break rules will get put to good use. If Amherst, Conn and Bowdoin do indeed win their last game, how will they line up?

It would seem that if Amherst, Conn and Bowdoin win their last games and finish with the same point total, and assuming Williams wins v. Hamilton, the seeding order should go as follows by virtue of the tie-breaker rule (i.e., both Conn and Bowdoin beat Amherst and Bowdoin beat Conn during the regular season):

- #1 Williams
- #2 Bowdoin
- #3 Conn
- #4 Amherst

Can anyone validate?

Looks like this is going to be the final finish order.

[Ocean 1]

I suspect most people are predicting that Conn, Amherst and Bowdoin will all win their last games and end up in a 3-way tie for second place. Massey lists the chances of winning as: Amherst 72%, Conn 84% and Bowdoin 68%. The chance of all 3 winning is then .72 x .84 x .68 = .41.

Surprisingly, if the Massey odds are accurate, it's more likely that at least one of the teams will lose than all 3 will win.

Only time will tell.

Actually, looking back at this. You are correct that the chance of all 3 winning is .41. But that doesn't mean it is more likely that at least one of them will lose, because ties are also possible. Massey provides both teams' chances of winning, and using those figures, we can easily calculate the chances of a tie as well. Here's what Massey says are Conn's, Amherst's and Bowdoin's chances of winning, tying and losing.

Conn - 84%, 9%, 7%
Amherst - 72%, 13%, 15%
Bowdoin - 68%, 13%, 19%

So, if the likelihood of those teams getting a win or a tie is 93%, 85%, and 81%, respectively, the likelihood of all 3 getting a win or a tie is the product of those 3 or 64%, which means that the chances of at least one of the three losing is only 36%. So, while it is more likely that one of the 3 will lose or tie (59%) than it is that all 3 will win (41%), it is more likely that all 3 will win (41%) than that one of them will lose (36%).

And as we now know, and as was predicted accurately from the elegant amount of analysis and collaboration...all 3 in fact did win!!!

[Ocean 1]

Entering the last few days of NESCAC play, only Wesleyan is out of the playoffs. Looks like there is a good chance of a 3 way tie for second thru fourth place and the tie break rules will get put to good use. If Amherst, Conn and Bowdoin do indeed win their last game, how will they line up?

It would seem that if Amherst, Conn and Bowdoin win their last games and finish with the same point total, and assuming Williams wins v. Hamilton, the seeding order should go as follows by virtue of the tie-breaker rule (i.e., both Conn and Bowdoin beat Amherst and Bowdoin beat Conn during the regular season):

- #1 Williams
- #2 Bowdoin
- #3 Conn
- #4 Amherst

Can anyone validate?

Looks like this is going to be the final finish order.

As Hannibal would say, I love it when a plan comes together! 

[2xfaux]

My goodness, NESCAC folks citing John Hannibal Smith. What's next Rocky and Bullwinkle.  I hope so.

[Ocean 1]

Very nice play by Trinity tonight as they fought brilliantly to beat Amherst by a score of 1-0 in double OT. While Amherst seemed to have the better of the play in regulation, Trinity essentially dominated Amherst in that last OT period with some pretty good defense and great pace.

[jc2]

I suspect most people are predicting that Conn, Amherst and Bowdoin will all win their last games and end up in a 3-way tie for second place. Massey lists the chances of winning as: Amherst 72%, Conn 84% and Bowdoin 68%. The chance of all 3 winning is then .72 x .84 x .68 = .41.

Surprisingly, if the Massey odds are accurate, it's more likely that at least one of the teams will lose than all 3 will win.

Only time will tell.

Actually, looking back at this. You are correct that the chance of all 3 winning is .41. But that doesn't mean it is more likely that at least one of them will lose, because ties are also possible. Massey provides both teams' chances of winning, and using those figures, we can easily calculate the chances of a tie as well. Here's what Massey says are Conn's, Amherst's and Bowdoin's chances of winning, tying and losing.

Conn - 84%, 9%, 7%
Amherst - 72%, 13%, 15%
Bowdoin - 68%, 13%, 19%

So, if the likelihood of those teams getting a win or a tie is 93%, 85%, and 81%, respectively, the likelihood of all 3 getting a win or a tie is the product of those 3 or 64%, which means that the chances of at least one of the three losing is only 36%. So, while it is more likely that one of the 3 will lose or tie (59%) than it is that all 3 will win (41%), it is more likely that all 3 will win (41%) than that one of them will lose (36%).

Actually, only two of the three won.

And as we now know, and as was predicted accurately from the elegant amount of analysis and collaboration...all 3 in fact did win!!!

[Becks]

If Wes fails to beat Conn today, it will be the first time they have gone winless in NESCAC since 2003.

Winless NESCAC teams to date (reverse chronological order):
2009 Conn 0-9-0
2008 Conn 0-8-1
2007 Colby 0-7-2
2006 Conn 0-8-1
2006 Trinity 0-8-1
2003 Wesleyan 0-9-0
2002 Wesleyan 0-8-1
2000 Wesleyan 0-8-1

Add Wes to the list. 

Again

[Becks]

I've been sifting through the ashes of the Wesleyan season to figure out what happened. Wes finished 7th in the league last year, but 11th this year.

Wes allowed only 9 goals last year (4th best  in the league), but allowed 21 this year (worst in the league). That's the most allowed by a Wes team since 2004 (when they went 1-8-0). Shots allowed were very similar. 2013 - 186, 2014 - 189. However, opponents' SOG% went up from .398 to .460 and saves percentage went down from a very high .895 to a mediocre .774. Since the starting keeper was the same as last year and was still regarded as perhaps the best player on the team, the stats suggest that the large increase in GA was the result of the team defense in front of the keeper allowing opponents significantly more high quality scoring opportunities than in 2013. Wes only graduated 2 starters from last year, but both were on defense and one (Kerry Doyle) was Wes's only all-NESCAC player and very fast. Wes also suffered injuries to several key defenders and defensive midfielders this year.

With the dramatic decline in the defense, Wes would have needed a substantial increase in goal scoring this year (say, up to the 13 goals it got in 2010 and 2011) in order to maintain the same record. Instead, the offense stayed steady at 7 goals. That offensive production was enough for Wes to make the NESCAC tournament last year when its defense was top 4 in the league, but was not enough to get results this year when the defense was substantially weaker.

[jc2]

One more NESCAC regular season game today that will decide the last playoff spot. Looks like Hamilton and Wes are out.

[Ocean 1]

I suspect most people are predicting that Conn, Amherst and Bowdoin will all win their last games and end up in a 3-way tie for second place. Massey lists the chances of winning as: Amherst 72%, Conn 84% and Bowdoin 68%. The chance of all 3 winning is then .72 x .84 x .68 = .41.

Surprisingly, if the Massey odds are accurate, it's more likely that at least one of the teams will lose than all 3 will win.

Only time will tell.

Actually, looking back at this. You are correct that the chance of all 3 winning is .41. But that doesn't mean it is more likely that at least one of them will lose, because ties are also possible. Massey provides both teams' chances of winning, and using those figures, we can easily calculate the chances of a tie as well. Here's what Massey says are Conn's, Amherst's and Bowdoin's chances of winning, tying and losing.

Conn - 84%, 9%, 7%
Amherst - 72%, 13%, 15%
Bowdoin - 68%, 13%, 19%

So, if the likelihood of those teams getting a win or a tie is 93%, 85%, and 81%, respectively, the likelihood of all 3 getting a win or a tie is the product of those 3 or 64%, which means that the chances of at least one of the three losing is only 36%. So, while it is more likely that one of the 3 will lose or tie (59%) than it is that all 3 will win (41%), it is more likely that all 3 will win (41%) than that one of them will lose (36%).

Actually, only two of the three won.

And as we now know, and as was predicted accurately from the elegant amount of analysis and collaboration...all 3 in fact did win!!!

Not sure what game I was watching...thanks for the correction!

[Ocean 1]

One more NESCAC regular season game today that will decide the last playoff spot. Looks like Hamilton and Wes are out.

Agree. The way I think the NESCAC Quarters line-up:

- Williams v. winner of Bates/Colby
- Bowdoin v. Tufts
- Conn v. Trinity
- Amherst v. Midd

[Becks]

One more NESCAC regular season game today that will decide the last playoff spot. Looks like Hamilton and Wes are out.

Agree. The way I think the NESCAC Quarters line-up:

- Williams v. winner of Bates/Colby
- Bowdoin v. Tufts
- Conn v. Trinity
- Amherst v. Midd

It's amazing how many times it comes down to Bates v Colby for the 8th seed. This year, 2013, 2011, 2008, . . . I guess it's because for most of the last 10 years they usually have both been near the bottom of the table and they always play each other in the last game of the season.