Wednesday, January 30, 2008
Per Possession Win%
In response to a Phil-linked thread about figuring the chances of the Patriots winning if their number of possessions goes down, I said:
The win% in Brian’s data has an almost perfect match to this equation:
Wins divided by Losses
= 1.1 ^ possessionsSo, if you have 12 possessions, the above equation will give you 3.14. That is, 3.14 wins per loss, which is a win% of .758. Brian’s simulator said .755.
Here’s how Brian’s simulator compares to my above equation:
9 0.714 0.702
10 0.727 0.722
11 0.740 0.740
12 0.755 0.758
13 0.765 0.775So, a 12-possession game for the Pats has the equation at .758. If I extend back to only 6-possessions (half a game), I get a win% of .639. And that is pretty much the half-way point between .500 and .758.
Didn’t you post something on here a while back about “odds ratio”, which was a better way to determine the results of a matchup than just taking the simple average? Would that be appropriate in this situation, in trying to determine the Patriots’ chance of scoring versus the Giants’ chance of stopping (and vice versa)?