Interesting. Well, I'd love to see how you got those numbers if you ever get the time.
Alright, here's the process I used. I can't say that it's 100% the right way to do it, but it's 100% the best way I can think of doing it.Step 1: Go to
http://www.usatoday.com/sports/sagarin/nfl06.htm. Here are Jeff Sagarin's ratings for each team. The last column, "Pure Points", pretty much is a description of how good the team has played this year, or it's true ability (as shown on the field; a team that suffered a lot of injuries might have a good true ability but would score poorly in this system if they had in fact played poorly).
Step 2: Get the pure points rating for all 32 teams, and sort from 1-32. Then find the best fit line (standard linear regression) for the data, because we want to smooth out the bumps. This will make the difference between team 1 and 2 equal to the difference between teams 4 and 5, and between teams 24 and 25. While this arguably is not a perfect reflection of the NFL, I did this for a couple of reasons:
1) In my Team A and Team B hypotheticals, I was treating them this way. So when I ranked each team's schedule, and say the team would face the 16.5th or 19.5th best team on average, that included the number 1 team being the same amount better than the number 2 team as the number 15 team vs. the number 16 team.
2) Practically, it doesn't matter, for this example. The only problem with a linear model would be at the top and bottom, but in this case the schedules stipulated have the team playing the top five teams and the bottom 11 teams. So even if we bumped up number 1 and bumped down number 2, you're still playing both so there would be no practical difference.
Step 3: Figure out the likelihood of Team X beating Team Y.
If you go
here, you'll see a pretty neat formula for figuring out the chance of a team winning any game against any opponent:
Home team prob. of winning =~ 1 / (1 + e^(-.438 - .0826*diff))
If the home team is 7 points better than the road team, this model gives the home team a 73% chance of winning. If the home team is 7 points worse, this model gives the home team a 46% chance of winning.
Unfortunately, I only had the formula handy for the home team's chances of winning. So I ran through the schedules twice, one with each game being at home, and one with each game being on the road, and then averaged those numbers.Step 4: Insert the ratings from the linear model into the equation
All that stuff was pretty technical, but now we get the results: and it should jive with our intuition.
The model says that the 6th ranked team has a 49% chance of winning on the road against the 13th ranked team. The 6th ranked team has a 54% chance of winning at home against the 1st ranked team. The 6th ranked team has a 70% chance of beating the 21st ranked team in a neutral site (61% chance on the road, 79% at home).
Here is the full schedule for the 6th ranked team playing Team A's schedule:
OR = Opponent Rank
WinProb = Probability of winning
OR WinProb 1 43.0% 2 44.4% 3 45.8% 4 47.2% 5 48.6%22 71.1%23 72.3%24 73.4%25 74.5%26 75.6%27 76.6%28 77.6%29 78.6%30 79.5%31 80.5%32 81.3%#W 1070.1%And the 6th ranked team playing Team B's schedule:
Code:
OR WinProb 9 54.2%10 55.6%11 57.0%12 58.4%13 59.7%14 61.1%15 62.4%16 63.7%17 65.0%18 66.3%19 67.5%20 68.8%21 70.0%22 71.1%23 72.3%24 73.4%#W 1026.4%
Note that the 6th ranked team is given a 54% chance of beating the 9th ranked team, and a 46% chance of beating the 3rd ranked team. Those seem pretty realistic to me.Obviously the difference in expected wins is less than half a game here, so it's not incredibly significant in a practical sense. But it
is significant to the extent that focusing on the number of difficult games a team faces is not a perfect indicator of its actual strength of schedule.