From: cor@bigwpi.WPI.EDU (Cory Shimer)
Newsgroups: rec.sport.football.pro
Subject: Re: Predictions - 6 Games Remaining
Date: 17 Nov 1992 17:00:25 GMT
Organization: Worcester Polytechnic Institute
NNTP-Posting-Host: bigwpi.wpi.edu

In article <1992Nov17.142342.29559@porthos.cc.bellcore.com> danp@sabre.bellcore.com (Daniel Pomplun) writes:
>
>Here's a simple prediction method of the rest of the season:
>
>	Expected Chance the HomeTeam wins =
>	   0.6 + HomeTeam's season record - AwayTeam's season record,
>	Expected Chance the AwayTeam wins =
>	   0.4 + AwayTeam's season record - HomeTeam's season record,
>	each limited by 0 and 1.
>
>For example, Team A (4-6) hosts Team B (7-3):
>  Team A will be given 0.6 + 0.4 - 0.7 = 0.3 win and
>  Team B will be given 0.4 + 0.7 - 0.4 = 0.7 win.
>


Hmm. This sounds ok, but here's my own algortithm:

Teams ability = (winning pct + opponents winning pct) / 2

where opponenets wining pct means the combined records of all the teams
you have played to this point in the season.

for the Patriots it looks like this :

(pats are 1-9, opponents are 59-41)

Teams ability = (0.1 + 0.59) / 2  = .30



Now the ability to win a particular game is another thing...

Ability to win game = Your Ability - Their Ability + 50%.

Take the Pats aginst the Jets for next week. I'm going to assume
the Jet's ability to be about 50%.

Pat's ability to win game = .30 - .50 + .50 = .30

30% chance they can beat the Jets.


notice that .50 should designate an average team. so a .50
for the Jets indicates that they are an average team, but have a
losing record, which means they probably have a tough schedule.

So the ability of any given team to beat an average team
is equal to their own ability.

Now, Ability should not be the only factor do determine
the chance of winning a game.

should ability make 30% of the chance? 50% ? 80%?

I figured 50% was good.

leave 10% for Home field
leave 20% for injuries
leave 10% for matchup tendencies over last severl years
leave 10% for "other" intangibles

make it a 100 point system.

if you have home field, you get 10 pts toward your chance of winning
if you have .XX game winning ability, you get (.XX)* 50 pts 
if you have no injuries you get 20 points
if you dominate the matchup in recent years you get 10 pts
if you have other intangibles in your favor you get 10 points.

Max points = 100.

X points = X percent chance of winning.

You can apply this to the "rest of the season" to estimate how many more
games your team will win.

Just apply the above formula to each of the remaining six games
for your team.

or more simply you can make a quick guess by assuming
your team's chance to win is based solely on it's own ability
for each game.

i.e. for the patriots, they wouild have a 30% chance of
winning any of the remaining six games.

well, thats slightly under 1 in three, so you would figure
the Patriots would win 1 or 2 more games the rest of this season,
which is probably about right.


Does anyone think this algorith has any merit?

would someone mind figuring out what your team's ability is
and comment if you think it's correct?

or maybe calculate your team's game winning ability for their
upcoming game, and whether you think this number makes sense?

P.S. obviously my algorithm leaves room for subjective decisions
     on injuries, matchup history, and "other" intangibles.
     But predicting the chance is not an exact science.


	I came up with a 40-50% chance the patriots will win
	at home vs. the jets next week.