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I have been thinking about 8 vs 9 game IN conference scheduling. For teams at the top of a conference is seems to me that the fewer the In conference games the better. For the sake of this discussion let's suppose that all five P5 conferences are equal in strength and distribution (1-14)_. For the #1 team in a conference no matter how many games you play on AVERAGE the remaining conference foes will be #2-14 which means that on average your foes will be 8. For games outside the conference the foes will be #1-14 with an average of 7.5. Every time you replace an IN conference game with an OOC game your AVERAGE foes ranking will increase by 0.5. So imagine replacing an 8 in conference with a 2 or 3 from another conference and what that would mean to your SOS. I think one of the problems with the PAC is just that. They play 9 games in conference believing that will improve their SOS when in fact it decreases it. Replacing the ninth In conference game with another P5 team from a different conference would IMPROVE the SOS of all teams.
Next, how does one determine SOS. Suppose two teams ()A and B) play EXACTLY the same 12 opponents under the same conditions. Clearly their SOS must be equal. Now suppose team A elects to play a 13th opponent which is an FCS team. Now which of the original two teams has the higher SOS? Many of the SOS calculations would say the the team A now has the LOWER SOS. I don't understand this. Clearly the extra opponent could have won the extra game therefore team A had a higher chance of losing than team B (which didn't play) and therefore MUST have the higher SOS.
Finally, suppose you have the choice of 2 schedules:
A: where you play teams ranked 1-6 and 125-130 (average 65.5.
or B: where you play teams ranked 60-71 (average 65.5).
Suppose you are the #7 team. With schedule A you should be 6-0 vs teams 125-130 and if you are lucky you might win a couple of the games vs teams 1-6. We'll say 8-4.
With schedule B you probably would win all 12 games or maybe lose 1 or two but you certainly aren't going to lose 4 games vs teams 50+ position lower than you. Thus for higher positioned teams schedule B is easier than A.
Many of the SOS systems say the two are equal. Some argue that the sum of the positions are the same and therefore must be the same over all. Others argue that the average team (number 65.5) would likely end up 6-6 with each schedule and therefore the two are equal. The committee shouldn't be interested in what the #65.5 team would do but what teams say 1-6 would do with each schedule.
Next, how does one determine SOS. Suppose two teams ()A and B) play EXACTLY the same 12 opponents under the same conditions. Clearly their SOS must be equal. Now suppose team A elects to play a 13th opponent which is an FCS team. Now which of the original two teams has the higher SOS? Many of the SOS calculations would say the the team A now has the LOWER SOS. I don't understand this. Clearly the extra opponent could have won the extra game therefore team A had a higher chance of losing than team B (which didn't play) and therefore MUST have the higher SOS.
Finally, suppose you have the choice of 2 schedules:
A: where you play teams ranked 1-6 and 125-130 (average 65.5.
or B: where you play teams ranked 60-71 (average 65.5).
Suppose you are the #7 team. With schedule A you should be 6-0 vs teams 125-130 and if you are lucky you might win a couple of the games vs teams 1-6. We'll say 8-4.
With schedule B you probably would win all 12 games or maybe lose 1 or two but you certainly aren't going to lose 4 games vs teams 50+ position lower than you. Thus for higher positioned teams schedule B is easier than A.
Many of the SOS systems say the two are equal. Some argue that the sum of the positions are the same and therefore must be the same over all. Others argue that the average team (number 65.5) would likely end up 6-6 with each schedule and therefore the two are equal. The committee shouldn't be interested in what the #65.5 team would do but what teams say 1-6 would do with each schedule.
