I was working on a project for which I needed to state the results of a single game as a winning percentage.   The larger project will follow, but here’s how that can be done. 

            Of course, every game results in a winning percentage either of .000 or 1.000, but a 7-6 game is different than a 10-0 game.   The 10-0 victory is much more decisive, and thus represents a higher level of performance.  

            In the majors this year (last year. . ..2008) there were two games in which teams scored 20 runs, and those two teams were 2-0.    Here is a chart of the won-lost records at each run level: 

            Let us assume that the game is 50% offense, 50% defense.  

Let us take a game that is won 7-6.   From an offensive perspective, your winning percentage is .804.   Offense is half the game, so your offensive contribution to winning percentage is .402.   

From the defensive side, you have performed in such a manner that your winning percentage would be .291.  (When you score 6 runs in a game, your winning percentage is .709.   Therefore, when you allow 6 runs in a game, your winning percentage has to be .291)  Half of that is .1455.   Adding them together and using one more decimal than I’m showing you, your winning percentage for the game would be .548--.402 + .146. 

A 10-0 game, on the other hand, is .462 from the offensive side (.924 divided by 2), and .500 from the defensive side (1.000 divided by 2).  Thus, a 10-0 game has a “Game Winning Percentage” of .962.  

We have madcap data in the small groups, of course. . .one can’t really assume that a team that scores 17 runs in a game should lose two-thirds of the time, even though they did.   I fixed that by evening out the curve at the upper boundaries: 

             Except that algebra abhors zeroes, so you might want to use .999 for 20 runs and .001 for a shutout. Fortunately in 2008 there were no 20-0 games. …in fact, I don’t know if there has ever been a 20-0 game in the majors.   Anyway, you probably don’t want zeroes in your system; they cause problems.         

            This simple approach can be used to figure the “winning percentage” for a team in any game. . .for example, 6-5 if .536, 4-3 is .557, 11-3 is .792, 5-1 is .778, 11-7 is .574, 7-0 is .902, etc. 

            This system has the following virtues:

            1)  That every winning score has a winning percentage over .500,

            2)  That every game has a winning percentage between .000 and 1.000,

            3)  That the results seem reasonable, and

            4)  That it seems kind of cool to be able to state the score of a game as a winning percentage.

            Usually the winning percentage of a game is similar to what you would get if you just divided the runs by the runs allowed.   A 7-5 game, for example, gives a .583 winning percentage—the same winning percentage a pitcher would have if he was 7-5.  

            But a problem with that is that that would make a 1-0 game the same as a 9-0 game—1.000.    A 1-0 game is not a dominating performance.  A 1-0 game in this system is a winning percentage of .541--.041 for the offense, .500 for the defense.    A 9-0 game is .946.   These two should not be the same, and they are not.  

            A problem with our little system, however, is that the season’s winning percentage is not the average of the game winning percentages.   A .600 team will come in with a game average of .550, more or less; a .550 team will come in around .525.  

            Maybe this isn’t a problem, I don’t know.  Your season’s ERA is not the average of your ERAs for each start; your batting average is not the average of your batting averages from each game.

            You can probably figure out as well as I can why it doesn’t work that way.   The average game winning percentage tends to track the ratio of runs scored to runs allowed—as the individual games do.  The ratio of runs scored to allowed is not the winning percentage; the winning percentage is a ratio of squares.  

            Working on the game level. . .let’s take a 6-1 game.   Teams that score six runs in a game win 70.9% of the time, which means that scoring six runs in a game increases the win expectation by 20.9%--not 10.4%.   When we divide it by two, we get 10.4%.   Teams that allow only one run in a game win 91.8 percentage of the time, meaning there is a “positive win contribution” of +41.8—not +20.9.

            We are studying the combined impact of scoring six runs in a game and of allowing only one run in a game, as if these were separate and unrelated events.    It happens that they’re in the same game, so we put them together in one game.  But what it really should be, for a 9-0 game,  is .709 + .918 = 1.627 minus .500, which is 1.127.    But who wants a system which tells you that the winning percentage for a team in a game is 1.127?   It’s not the way winning percentages work. 

            But the winning percentage for one game part of it.  ..that works pretty well.   I don’t know how to make it adjust for the home field advantage; I guess I would need two sets of charts, one for home games and one for road games.  The highest winning percentage for any team in a game last season was .9825, for games that were 13-0; I think there were a bunch of games that were 13-0.   The lowest winning percentage for any team that actually won a game was .5025, for Colorado’s 18-17 ass whuppin’ of Florida on the Fourth of July.     I don’t know how to adjust it for park effects, either. 

Alternative Method

            I needed a method to do this, so I tried some other methods first, one of which is worth describing.

            I started by squaring the runs scored; a 3-1 game becomes 9 to 1, which is .900.   1 to 0 is 1.000.   Obviously that doesn’t work; everything is kept within zero to 1.000 but most everything comes out under .200 or over .800, and 1-0 is the same as 20-0.   Also, you get a lot of zeroes, and algebra despises zeroes. 

            Let’s try adding one run as “ballast” before we look at the ratio of squares.

            Still doesn’t work.   A 3-1 game becomes 16-4, or .800; 1-0 if 4-1, also .800.   

            I tried adding two, three, four.    Four seems to work.   A 3-1 score become 7-5 which becomes 49 to 25, or .662.    By the other method we’re at .643.     1-0 becomes 25-16, which is .610; 9-0 becomes 169 – 16, which is .914. 

            That method is, as a practical matter, about the same as this one, and the average game winning percentage is too close to .500, same as in the other method.   But in this method it is pretty easy to fold in the home field advantage.   Since the home field advantage is one-fourth of a run, you just add one-eighth of a run to the road team, take away one-eighth of a run from the visiting team, before you square the runs.   A 4-1 win becomes .760 if you’re on the road, .729 if you’re at home.  

It works OK, but I just prefer the other method.