The Patriots came into Indianapolis and beat the Colts in one of the most hyped mid-season games in NFL history. NE came into the game ranked #1 in efficiency rankings, accounting for strength of schedule. IND came into the game ranked just behind at #2.
This week the Colts will be ranked #1 ahead of the Pats. IND's generic win probability (GWP) -- the probability they would win a game against a league average team at a neutral site-- is 0.92, while NE's is 0.91.
Except for the score, the Colts actually outplayed the Patriots. Given their in-game statistics, it was surprising that NE won the game. One model says IND would have had a 77% chance of winning the game. The 4-point victory by the Patriots came down to one or two critical, high-leverage plays.
Most fans would say, so what? Who cares who was more efficient. The only thing that matters is the score. "You are what your record says you are," says Bill Parcells. That's true, but in-game efficiency stats are very often more indicative of future performance.
Unfortunately for IND, their hopes of an undefeated season are now dashed while NE's are still intact. Updating a previous post, here is NE's probability of finishing the regular season 16-0 accounting for their performance in yesterday's win.
Here are NE's upcoming games and the associated probabilities of winning each one.
Vprob | Visitor | Home | Hprob |
0.85 | NE | BUF | 0.15 |
0.08 | PHI | NE | 0.92 |
0.89 | NE | BAL | 0.11 |
0.16 | PIT | NE | 0.84 |
0.03 | NYJ | NE | 0.97 |
0.02 | MIA | NE | 0.98 |
0.82 | NE | NYG | 0.18 |
So assuming they don't rest starters or suffer a critical injury, NE's probability of finishing the season 16-0 is now:
0.85 * 0.92 * 0.89 * 0.84 * 0.97 * 0.98 * 0.82 = 0.45.
interesting post. I especially liked the layout of the Patriot future games and associated win percentages. Would it be possible to include this for the rest of the games on the schedule? Or could I figure it out on my own from your other published ratings? Thanks.
hey will, there are only 7 games left and I see all of them there.
I was referring to all the games on the NFL schedule, not just the Pats.
I can post them all, but it would just be one very long and ugly table. I'll try to find a good way to do that this week.
But in the meantime, you can calculate the odds of any game, even theoretical ones not on the schedule. (Warning: extremely nerdy math)
Take the GWP of both opponents and covert them to odds (odds = GWP/(1-GWP). So if the home team has a 0.75 GWP, that would be 0.75/0.25, which is simply 3. And if the visiting team has a 0.67 GWP, that would be 0.67/0.33 which is 2.
Take the natural log of each team's odds--ln(3) and ln(2) in our example.
Subtract the visiting team's log from the home team's log, then add 0.74 for home field advantage.
ln(3)-ln(2)+0.74 = 1.15
That is the "log odds ratio" of the home team beating the visiting team. So now we go backwards through the process.
The odds ratio would be e^(log odds ratio), which is:
e^1.15 = 3.14
So the odds of the home team winning are 3.14:1. Convert that into a probability by using some algebra:
Prob(Home team wins) = x/(1+x)
=3.14/(1+3.14)
=0.76