re: A&M formally announcing departure from Big 12. SEC application to follow.

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how does it work then?

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do I need to repost what I typed on the last page?

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You will see that it's impossible.

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huh? Every team plays 8 conference games


somebody help me here, what basic math I am not getting across? Are you having some sort of home/away issue? Do you feel that if (for example) LSU plays at Kentucky in 2013 we MUST have Kentucky reciprocate that game in Tiger Stadium the next year? Because that's not the case at all. As long as everyone plays an equal amount of conference games, I couldnt give two shits about how the division/cross division games work out. It wont be an issue long enough for anyone to care

I really think A&M will benefit from exposure with these other SEC schools. From an on-field standpoint definitely, and from a fan experience standpoint. We can learn how to tailgate on a "professional" level. It's time to broaden our horizons a little bit and learn from others who seem to be doing it all very well.

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huh? Every team plays 8 conference games


somebody help me here, what basic math I am not getting across? Are you having some sort of home/away issue? Do you feel that if (for example) LSU plays at Kentucky in 2013 we MUST have Kentucky reciprocate that game in Tiger Stadium the next year? Because that's not the case at all. As long as everyone plays an equal amount of conference games, I couldnt give two shits about how the division/cross division games work out. It wont be an issue long enough for anyone to care

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Okay, every team plays 8 games. The west teams each have 6 intra-division games and 2 inter-division games. The east teams have 5 intra-division games and 3 inter-division games. That means that the west needs to play a total of 14 games against east teams, and east needs to play 18 games against west teams. Do you not see how that is impossible? Those numbers have to be equal.

Edit: typo corrected

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Those numbers have to be equal.

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ah, I see your point and disagree 1000%

why would they have to be equal? Every SEC team gets to play 8 games against a conference opponent. It'll be a little divisional-heavy in the west, but oh well.

those numbers do not have to be equal in the short term. Scheduling may be a little more tricky, but that's what they pay people to do and bye-weeks should alleviate any of that problem. I'm sure we'll all lose some scheduled non-conference opponents along the way

Am I the only one (besides chill) that noticed this at the bottom of that article?

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Correction: August 30, 2011

An earlier version incorrectly reported that the Texas A&M president had sent a letter to the Big 12 about its intent to leave the conference.

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I'm going to take the time to demonstrate how fricking dumb you are. Let's do the West, who each needs to have 2 games to make it an 8 game schedule. I'm going to add hypothetical opponents to prove this is impossible.

Texas A&M -- Florida, SC
LSU -- Florida, SC
Arkansas -- SC, Vandy
Ole Miss -- Vandy, Kentucky
MSU -- Kentucky, Georgia
Auburn -- Georgia, Tennessee
Alabama -- Tennessee, Florida

(count them up, it's 14 games)

Okay, so the West's conference slate is full. Based on this, the East's games are:

Florida -- A&M, LSU, Alabama
SC -- LSU, A&M, Arkansas
Vandy -- Ole Miss, MSU
Kentucky -- Arkansas, Ole Miss
Georgia -- MSU, Auburn
Tennessee -- Auburn, Alabama

(count them up, it's 14 games)

Okay, so only Florida and SC have all three interdivision games they need. Vandy, Kentucky, Georgia, and Tennessee only have 2 each, so they each need one more (4 more games + 14 they already have = 18). The problem is the West has all of their teams at 2 games each already. As I said, the east with 6 teams and 3 cross games each has to play the west 18 times.

There's no way for every west team to have the same number of games the other west teams do while all east teams have the same number of games as each other. It's physically not possible until you reach the lowest common denominator as I did with my hypothetical 2 and 3 team divisions playing a total of 6 cross games. The lowest common denominator of 6 and 7 is 42, meaning 42 total inter-division games would need to occur with each west team playing all 6 east ones (7x6=42) and each east team playing all the west teams (6x7=42).


If you don't understand this now, you never will.

I'm wondering if he's going to show up again... I'd bet not unless he comes back to say he still thinks he's right and there's a
magical formula that can make 14 = 18.

@TIGERSandFROGS

Thank you. I gave up trying to post logic with the rude dude. It's ok to be rude but not stupid. I'll bet Stu really looks like that in real life.

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@TIGERSandFROGS

Thank you. I gave up trying to post logic with the rude dude. It's ok to be rude but not stupid. I'll bet Stu really looks like that in real life.

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I feel kind of bad about being so insulting towards him. I've been trying to not hide behind my moniker on here and talk to people the way I would if I was speaking face-to-face with someone I didn't know. Regardless, he couldve had the marbles to say he was wrong at least, or thank me for teaching him something.

Oh well.