Why Denver, and not San Diego or Oakland won the AFC West

We know now that Denver backed their way into winning the AFC West.  And those that care know it came down to a tie-breaker, that being a better record against common opponents.  But let's take a deeper look at why exactly that was and if it always would have been this way.

If we look at the records, specifically those you typically see in the standings, we see that each of Denver, San Diego, and Oakland were all 8-8 overall as well as 3-3 in the division.  We can also see that San Diego was 7-5 in the conference while Denver and Oakland were just 6-6.  So why did Denver win the division with a worse conference record than San Diego?

It did used to be (pre-2008) that the tie-breaking procedure had conference record with a higher precedence and with those rules, San Diego would have won the tie-breaker and the division.  But in 2008 with the addition of the Texans and 4 divisions, they tie-breaking rules were changed to have the record in games against common opponents used after the division record but before the conference record.  Thus, Denver's better record in those games against common opponents trumps San Diego's better conference record.

So who were these common opponents and how did each team do?  Well, the teams all played Kansas City as a common opponent twice, then the AFC West all played the NFC North as well as the AFC East, 4 games each, or a total of 10 games against common opponents.  Throw in the 4 games they played against each other and the only games not common were Denver's against Cincinnati and Tennessee, San Diego's against Jacksonville and Baltimore, and Oakland's against Houston and Cleveland.

In the 10 common games and each unique game as well, here is how the teams fared.  Records in the heading are against common opponents and why Denver wins the tie-breaker.

Opponent	Denver (6-4)	San Diego (5-5)	Oakland (5-5)
Common
Kansas City	W @17-10	W 20-17	L 28-0
Kansas City	L 7-3	L @23-20	W @16-13 OT
Green Bay	L @49-23	L 45-38	L @46-16
Detroit	L 45-10	L @38-10	L 28-27
Chicago	W 13-10 OT	L @31-20	W 25-20
Minnesota	W @35-32	W 24-17	W @27-21
New England	L 41-23	L @35-21	L 31-19
NY Jets	W 17-13	L @27-21	W 34-24
Miami	W @18-15 OT	W 26-16	L 34-14
Buffalo	L @40-14	W 37-10	L 38-35
Not Common
Cincinnati	W 24-22
Tennessee	L @17-14
Jacksonville		W @38-14
Baltimore		W 34-14
Houston			W @25-20
Cleveland			W 24-17

The differences for Denver and San Diego are Denver beat Chicago at home while SD lost on the road, Denver beat the Jets at home while San Diego lost on the road, and San Diego beat Buffalo at home while Denver lost on the road.  It would seem San Diego drew the bad break of having to play some of the tougher opponents on the road (Detroit, Chicago, New England, and the Jets) while Denver got all of those at home, while Denver's road games did include tough opponent Green Bay, but also included not so tough Minnesota, Miami, and Green Bay.  Had San Diego had another tough game at home in exchange for an easier one on the road, they may have won it and finished 9-7 winning the division.  Note that Oakland had the same home/road schedule as Denver.

Also, San Diego had the toughest not-common opponent in 12-4 Baltimore while Denver's toughest was 9-7 (Cincinnati and Jacksonville).  Oakland's toughest was 10-6 Houston.

So, San Diego had the slightly tougher road, due in part to finishing ahead of Denver and Oakland last year.  My computer actually says San Diego is the better team rating at about 82.7 versus Denver's and Oakland's 79.5, but it is the record and the NFL's tie-breakers that count.