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Sunday, September 23, 2007

Graphing Margin of Victory against Probability of Victory

As a cautionary note, this particular blog entry is not inherently interesting to the average college football fan. It is important, because it lays the groundwork for much of what I will be doing in later blog entries, but if you don't want to read it all I suggest that you skip to the implications--that's the fun part.

All teams have good games and bad games, but most will tend to fall somewhere in the middle. In fact, if a football team were able to play the same game over and over again, and if we could objectively measure the performance of the team and graph it, it would look something like this:

Two teams will have overlapping distributions, and the better team (which we will call A) will peak further to the right than the inferior team (which we will call B).
The more they overlap, the more likely it is that B’s performance will land higher on its distribution than A’s performance and B will win.

Point margin is useful because it provides more information than win/loss records.
If team A beats team B by 40 points and C beat D by 3 points, we can make the educated guess that, were they to play again, it is more likely that A will beat B than that C will beat D. In this way, the Margin of Victory (MOV) is related to the Probability of Victory (POV)

Trade Sports ( allows participants to buy stocks (that represent a prediction) that will pay out at $1 if the prediction is correct, so the price of the stock represents the probability the prediction will be correct.
For college football, Trade Sports allows participants to buy stock on the outcome of the game and probability that a team will cover the spread. Therefore, if there is a 50% chance that a team will cover a 20 point spread and an 85% chance that they will win the football game (based on the price of the relevant stock), we can assume that the teams, based on their "averages", are 20 points different. More importantly, we can assume that if two teams are 20 points different on average then one will win 85% of the time and the other 15%.

Back to our previous example, if A beat B by 20 in week 1 and then they play in week 2, all else being equal, then we would assume that A has an 85% chance of winning the second game based on their performance in the first game.

Using scores from 47 games on Trade Sports, I came up with the following equation:

POV = 1/(1+10^(-MOV/18.8))

This equation represents a simplified version of the line of best fit. It is notable that the over/under had no effect—a 7 point win is a 7 point win, whether that win is 100 – 93 or 7 – 0.

Our best guess, then, if A beat B by 20 in game 1, is that A has a 92% chance of winning the rematch.

On average, a team will be within in 12.7 points of their average performance 80% of the time.

.8 = 1-(2*(1/(1+10^((12.685-0)/(18.8*SQRT(2)))))

Also, a team has a one in five chance of making up a 12.6 point difference in ability if the opponent puts up an average performance.

A team that was favored by 20 (10 points per half), but losing by 14 points at halftime, still has a 38% chance of winning the game.


Addendum: I've approached this problem several ways since I first wrote this entry, and every time I've come up with the same conclusion.


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