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MAC Tournament Seeding
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wmubasketballsupporter Offline
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Post: #1
MAC Tournament Seeding
We are currently in 5 team tie for 2nd place! Just looked up the MAC tie-breaker procedure and for ties between 3 or more teams the best overall win/loss percentage is the tie breaker. In that case we end up with the 6th seed. We need to win Friday and hope some of the teams we are tied with choke and lose!

http://www.mac-sports.com/sports/2015/3/...50307.aspx

Friday games:

NIU vs BSU
Miami vs Ohio
Bowling Green vs Buffalo
Akron vs Kent State
CMU vs WMU

All 5 teams tied at 10-7 are the Home team!
(This post was last modified: 02-28-2017 10:32 PM by wmubasketballsupporter.)
02-28-2017 10:31 PM
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MRBUTTONS Offline
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Post: #2
RE: MAC Tournament Seeding
(02-28-2017 10:31 PM)wmubasketballsupporter Wrote:  We are currently in 5 team tie for 2nd place! Just looked up the MAC tie-breaker procedure and for ties between 3 or more teams the best overall win/loss percentage is the tie breaker. In that case we end up with the 6th seed. We need to win Friday and hope some of the teams we are tied with choke and lose!

http://www.mac-sports.com/sports/2015/3/...50307.aspx

Friday games:

NIU vs BSU
Miami vs Ohio
Bowling Green vs Buffalo
Akron vs Kent State
CMU vs WMU

All 5 teams tied at 10-7 are the Home team!
03-banghead Well that certainly stinks. It would be nice if we got at least as high as third unless N.I.U. PULLS OFF A BIG UPSET.
.
02-28-2017 11:25 PM
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BaldBuster Offline
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RE: MAC Tournament Seeding
The tie-breaker for 3 or more teams is the winning percentage between the tied teams. See below:

TIE-BREAKER PROCEDURE
Ties in winning percentage, and thus for tournament, seeding positions shall be broken as follows:

1. Between TWO teams:
A. Head-to-head competition

B. Division Record (10 games)^

C. Winning percentage* vs. ranked conference teams (top to bottom, regardless of division, vs. common opponents regardless of the number of times played)

D. Coin flip

2. For MULTIPLE (3 or more) team ties:
E. Total won-lost record/winning percentage* of games played among the tied teams

F. Two (2)-team tie-breaker procedure goes into effect (refer to A)
[NOTE: Once a three-team tie has been reduced to two teams, the two-team tiebreaker will go into effect.]

^ - For the purpose of determining the Division champion. This tiebreaker is ONLY used for seeding purposes if the two teams in question are tied for the Division lead. (Teams will still be considered co-divisional champions)

* - Winning percentage is used instead of record because of situations where teams do not play each other the same number of times. Therefore, a team that is 1-0 (1.000) would win the tiebreaker over a team that is 1-1 (.500).


So with Akron getting the top seed, as of today the 2 through 6 seeds would be:

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).

So if Western wins Friday, we'll be guaranteed no worse than a 4th seed and a pass to Cleveland and the quarter-finals on Thursday. With a Western win and a Buffalo and/or Ohio loss, we would move up in the seedings.

Assuming no upsets on Friday or Monday, we would face Kent on Thursday in the second game in the afternoon.

Fight-on
03-01-2017 12:36 PM
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GullLake Offline
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Post: #4
RE: MAC Tournament Seeding
(03-01-2017 12:36 PM)BaldBuster Wrote:  The tie-breaker for 3 or more teams is the winning percentage between the tied teams. See below:

TIE-BREAKER PROCEDURE
Ties in winning percentage, and thus for tournament, seeding positions shall be broken as follows:

1. Between TWO teams:
A. Head-to-head competition

B. Division Record (10 games)^

C. Winning percentage* vs. ranked conference teams (top to bottom, regardless of division, vs. common opponents regardless of the number of times played)

D. Coin flip

2. For MULTIPLE (3 or more) team ties:
E. Total won-lost record/winning percentage* of games played among the tied teams

F. Two (2)-team tie-breaker procedure goes into effect (refer to A)
[NOTE: Once a three-team tie has been reduced to two teams, the two-team tiebreaker will go into effect.]

^ - For the purpose of determining the Division champion. This tiebreaker is ONLY used for seeding purposes if the two teams in question are tied for the Division lead. (Teams will still be considered co-divisional champions)

* - Winning percentage is used instead of record because of situations where teams do not play each other the same number of times. Therefore, a team that is 1-0 (1.000) would win the tiebreaker over a team that is 1-1 (.500).


So with Akron getting the top seed, as of today the 2 through 6 seeds would be:

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).

So if Western wins Friday, we'll be guaranteed no worse than a 4th seed and a pass to Cleveland and the quarter-finals on Thursday. With a Western win and a Buffalo and/or Ohio loss, we would move up in the seedings.

Assuming no upsets on Friday or Monday, we would face Kent on Thursday in the second game in the afternoon.

Fight-on

Interesting and helpful information.

Thank you!

04-cheers
03-01-2017 12:45 PM
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Potro Offline
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Post: #5
RE: MAC Tournament Seeding
I wondered how our 4th seed was determined given all the teams with identical records. Thanks for the information.
03-01-2017 12:59 PM
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BaldBuster Offline
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Post: #6
RE: MAC Tournament Seeding
(03-01-2017 12:59 PM)Potro Wrote:  I wondered how our 4th seed was determined given all the teams with identical records. Thanks for the information.

It was the five teams records against each other and then the two team tie-breakers applied.

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).
03-01-2017 01:16 PM
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wmubasketballsupporter Offline
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Post: #7
RE: MAC Tournament Seeding
(03-01-2017 12:36 PM)BaldBuster Wrote:  The tie-breaker for 3 or more teams is the winning percentage between the tied teams. See below:

TIE-BREAKER PROCEDURE
Ties in winning percentage, and thus for tournament, seeding positions shall be broken as follows:

1. Between TWO teams:
A. Head-to-head competition

B. Division Record (10 games)^

C. Winning percentage* vs. ranked conference teams (top to bottom, regardless of division, vs. common opponents regardless of the number of times played)

D. Coin flip

2. For MULTIPLE (3 or more) team ties:
E. Total won-lost record/winning percentage* of games played among the tied teams

F. Two (2)-team tie-breaker procedure goes into effect (refer to A)
[NOTE: Once a three-team tie has been reduced to two teams, the two-team tiebreaker will go into effect.]

^ - For the purpose of determining the Division champion. This tiebreaker is ONLY used for seeding purposes if the two teams in question are tied for the Division lead. (Teams will still be considered co-divisional champions)

* - Winning percentage is used instead of record because of situations where teams do not play each other the same number of times. Therefore, a team that is 1-0 (1.000) would win the tiebreaker over a team that is 1-1 (.500).


So with Akron getting the top seed, as of today the 2 through 6 seeds would be:

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).

So if Western wins Friday, we'll be guaranteed no worse than a 4th seed and a pass to Cleveland and the quarter-finals on Thursday. With a Western win and a Buffalo and/or Ohio loss, we would move up in the seedings.

Assuming no upsets on Friday or Monday, we would face Kent on Thursday in the second game in the afternoon.

Fight-on

Thanks for clarifying the tie-breaker scenario! I initially thought it was the total win/loss percentage for the entire season but after reading your post noticed it was the win/loss percentage against teams tied. Glad it means we get the bye to Cleveland with a win Friday!
03-01-2017 02:15 PM
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BrownandGoldforever Offline
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Post: #8
RE: MAC Tournament Seeding
(03-01-2017 02:15 PM)wmubasketballsupporter Wrote:  
(03-01-2017 12:36 PM)BaldBuster Wrote:  The tie-breaker for 3 or more teams is the winning percentage between the tied teams. See below:

TIE-BREAKER PROCEDURE
Ties in winning percentage, and thus for tournament, seeding positions shall be broken as follows:

1. Between TWO teams:
A. Head-to-head competition

B. Division Record (10 games)^

C. Winning percentage* vs. ranked conference teams (top to bottom, regardless of division, vs. common opponents regardless of the number of times played)

D. Coin flip

2. For MULTIPLE (3 or more) team ties:
E. Total won-lost record/winning percentage* of games played among the tied teams

F. Two (2)-team tie-breaker procedure goes into effect (refer to A)
[NOTE: Once a three-team tie has been reduced to two teams, the two-team tiebreaker will go into effect.]

^ - For the purpose of determining the Division champion. This tiebreaker is ONLY used for seeding purposes if the two teams in question are tied for the Division lead. (Teams will still be considered co-divisional champions)

* - Winning percentage is used instead of record because of situations where teams do not play each other the same number of times. Therefore, a team that is 1-0 (1.000) would win the tiebreaker over a team that is 1-1 (.500).


So with Akron getting the top seed, as of today the 2 through 6 seeds would be:

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).

So if Western wins Friday, we'll be guaranteed no worse than a 4th seed and a pass to Cleveland and the quarter-finals on Thursday. With a Western win and a Buffalo and/or Ohio loss, we would move up in the seedings.

Assuming no upsets on Friday or Monday, we would face Kent on Thursday in the second game in the afternoon.

Fight-on

Thanks for clarifying the tie-breaker scenario! I initially thought it was the total win/loss percentage for the entire season but after reading your post noticed it was the win/loss percentage against teams tied. Glad it means we get the bye to Cleveland with a win Friday!

Good job on posting the tie-breaker scenarios. Broncos are really rolling now and have a great chance to make it eight wins in a row over a struggling CMU team Friday night. Chips defense is porous right now. They gave up another 100+ pts last night as EMU ripped them for 109.
Fight on, Broncos !!
03-01-2017 03:15 PM
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BaldBuster Offline
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Post: #9
RE: MAC Tournament Seeding
(03-01-2017 12:36 PM)BaldBuster Wrote:  The tie-breaker for 3 or more teams is the winning percentage between the tied teams. See below:

TIE-BREAKER PROCEDURE
Ties in winning percentage, and thus for tournament, seeding positions shall be broken as follows:

1. Between TWO teams:
A. Head-to-head competition

B. Division Record (10 games)^

C. Winning percentage* vs. ranked conference teams (top to bottom, regardless of division, vs. common opponents regardless of the number of times played)

D. Coin flip

2. For MULTIPLE (3 or more) team ties:
E. Total won-lost record/winning percentage* of games played among the tied teams

F. Two (2)-team tie-breaker procedure goes into effect (refer to A)
[NOTE: Once a three-team tie has been reduced to two teams, the two-team tiebreaker will go into effect.]

^ - For the purpose of determining the Division champion. This tiebreaker is ONLY used for seeding purposes if the two teams in question are tied for the Division lead. (Teams will still be considered co-divisional champions)

* - Winning percentage is used instead of record because of situations where teams do not play each other the same number of times. Therefore, a team that is 1-0 (1.000) would win the tiebreaker over a team that is 1-1 (.500).


So with Akron getting the top seed, as of today the 2 through 6 seeds would be:

2 - Ohio -- .571
3 - Buf -- .571
4 - WMU -- .500
5 - KS -- .500
6 - BS -- .333

Ohio beats Buffalo on a tie-breaker (Div record). Western beats Kent on tie-breaker (head-to-head).

So if Western wins Friday, we'll be guaranteed no worse than a 4th seed and a pass to Cleveland and the quarter-finals on Thursday. With a Western win and a Buffalo and/or Ohio loss, we would move up in the seedings.

Assuming no upsets on Friday or Monday, we would face Kent on Thursday in the second game in the afternoon.

Fight-on

CORRECTION

I inaccurately stated that if we win Friday we would be guaranteed a spot in the quarters. I didn't run through all the 4-team tie-breakers as a result of Friday nights play. As pointed out by Hawk at last nights Fast Break Club Meeting, if four of the five currently tied teams win, with Kent being the loosing team, we would be playing on Monday night. Without Kent in the 5-team tie-breaker mix, we end up tied with Testicle Tech and they beat us on the Div Record tie-breaker. Sorry for the inaccuracy.

Kent/Akron play the late game at 9:00 (ESPN2). So if all four of the other teams win their 7:00 games, we need to become Kent fans for a couple of hours. A BS, Ohio, or Buffalo loss would negate the need for Kent to win (I think). If we win and two of the other four teams lose -- we're in!
03-03-2017 11:04 AM
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stdatwmu Offline
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RE: MAC Tournament Seeding
[Image: C6C8unCU8AA-3Mo.jpg]
03-03-2017 11:10 PM
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