Sports Asked by Ryan Ta on February 22, 2021
A follow-up to the question What is “Strength of Victory”?:
I was trying to find the theoretical maximum of a team’s strength of victory (SOV) after an NFL regular season is complete. Certainly, just like other won-loss-tied percentages, the SOV is a numerical value between 0 and 1.
With regard to how NFL match-ups are set for any given NFL team in any given season (e.g. 6 divisional games, 4 games against another division in the same conference, 4 games against a division in the opposite conference, etc.), is there a way to compute the maximum SOV of a given NFL team?
I am asking because it seems to be rare, in practice, that a team’s SOV is over 0.500. For instance, according to Reddit, the 2018 New England Patriots have an SOV of 0.539 and they say that is the highest in the NFL in that season.
The maximum value of SoV is 0.9375 (15/16), in the case where the team in question only wins one game while the opponent from that game wins every other game and goes 15-1.
Edit:
The highest possible SoV for a 16-0 team is 0.656. This stems from the following results (with an actually possible schedule of games for KC, the opponents are marked with an asterisk). I double-counted division rivals' records as there are two games, I couldn't find any info about if the NFL does it this way.
AFCW AFCE AFCN AFCS NFCE NFCN NFCS NFCW
KC 16-0 *NE 12-4 *BAL 14-2 *JAX 10-6 *DAL 12-4 DET 8-8 CAR 10-6 SF 6-10
*LV 14-2 *MIA 10-6 PIT 12-4 TEN 8-8 *PHI 10-6 MIN 6-10 NO 8-8 LAR 4-12
*DEN 12-4 *NYJ 8-8 CLE 10-6 HOU 6-10 *WSH 8-8 CHI 4-12 TB 6-10 ARI 2-14
*LAC 10-6 *BUF 6-10 CIN 8-8 IND 4-12 *NYG 6-10 GB 2-14 ATL 4-12 SEA 0-16
Correct answer by Daniel on February 22, 2021
Theoretically, this value could be very high and there is a way to calculate the maximum possible. It may not be a likely scenario, but imagine a scenario where the Detroit Lions have a perfect season: 16-0. This means that they beat each divisional opponent 2 times each, leaving them with a maximum record of 14-2. However, this is not a possible result as each other division opponent must also face each other twice. To make things easy, let's say they all split with each other -- each team takes a loss from each of the other two teams. This means that each divisional opponent is sitting at 12-4.
Lions 16 - 0
Packers 12 - 4
Bears 12 - 4
Vikings 12 - 4
From here, the Lions have 10 other opponents on the season. In a perfect world, 7 of those teams could come from different divisions and could each be 15-1. Three of these 10 teams would need to take an additional two losses from the division leaders and would end up at 13-3. If my math is correct, it would look something like this:
E(A) N(A) S(A) W(A) E(N) N(N) S(N) W(N)
NE 15-1 BAL 15-1 HOU 15-1 KC 15-1 DAL 15-1 DET 16-0 NO 15-1 SF 15-1
BUF 13-3 PIT 13-3 TEN 13-3 OAK 3-13 PHI 3-13 MIN 12-4 TB 3-13 SEA 6-10
NYJ 4-12 CLE 4-12 IND 4-12 DEN 3-13 WSH 3-13 CHI 12-4 CAR 3-13 LAR 6-10
MIA 2-14 CIN 3-13 JAX 2-14 LAC 2-14 NYG 3-13 GB 12-4 ATL 3-13 ARI 3-13
This gives us enough information to calculate the maximum possible Strength of Victory.
((12 / 16) * 6) ---- Divisional wins
((15 / 16) * 7) ---- Non-divisional wins against division leaders
((13 / 16) * 3) ---- Non-divisional wins against division runner ups
(((12 / 16) * 6) + ((15 / 16) * 7) + ((13 / 16) * 3)) / 16
( 4.5 + 6.5625 + 2.625 ) / 16
So the theoretical maximum for strength of victory is
0.84375
The likelihood of this happening in reality is slim to none, due not only to the incredible difficulty in finishing a season undefeated, but also adding the difficulty of beating 16 other VERY GOOD opponents. For example, the Patriots undefeated season in 2007 didn't break the .500 threshold due to games against teams that only had a single win.
Answered by Steve-o169 on February 22, 2021
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