MAC Reset with permutations added
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So, we're reaching the nitty gritty. BG is scrambling right now for a home berth in the MAC tournament, and with that, they can always hope to get to Cleveland and from there who knows what happens. Ask OU.
Here are some possibilities, based on my best reading of the MAC tie-breaker procedures. At this point, we're reasonably likely to get a home berth, but it it is no sure thing. That loss to CMU certainly does not help tie-breaker wise.
If BG wins both they go 9-7, they have home court
If BG goes 8-8: (There are a number of ways that could happen).
Four way tie, BG, OU, Buffalo, CMU BG w/BG win over Buffalo
The tie breaker is the record among the four years, using winning percentage.
Among these four teams:
BG 3-2
Buffalo 2-3
OU 2-3
CMU 2-1
Therefore
6. CMU
7. BGSU
Four way tie, BG, OU, Buffalo, CMU BG w/Buffalo win over BG
Among these four teams:
BG 2-3
Buff 3-2
OU 2-3
CMU 2-1
6. CMU
7. Buffalo
8. BG (Depends on whether Miami or Kent finishes first, but probable due to BG's Miami win).
9. OU
3-way tie, Buffalo, BG, CMU w/BG win over Buffalo
Among these three teams:
BG 2-1
Buffalo 1-2
CMU 1-1
7. BG
8. CMU
9 Buffalo
3-way tie, Buffalo, BG, CMU w/Buffalo win over BG
BG 1-2
Buffalo 2-1
CMU 1-1
7. Buffalo
8. CMU
9. BG
BG loses out and finishes 7-9
If CMU loses both...
8. BG
9. CMU
If CMU loses one
8. CMU
9. BG
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