10 runs = 1 win? (August 1, 2003)
I've written at length on this subject, but sometimes, a small sample makes things clearer. A team with 5 RS and 4 RA will win:
win% = .5 + (5-4)/10 = .600
Here are all the teams since 1946 who
: RS-RA = 1.00 +/- .05
: RS+RA = 9.00 +/- .50
year team win%
1948 BSN 0.595
1949 SLN 0.623
1950 CLE 0.597
1952 CLE 0.604
1957 ML1 0.617
1961 DET 0.623
1963 MIN 0.565
1978 ML4 0.574
1980 BAL 0.617
1980 NYA 0.636
1990 NYN 0.562
2002 SFN 0.59
The average win%? .600
--posted by TangoTiger at 10:49 AM EDT
Posted 4:44 p.m.,
August 7, 2003
(#1) -
bob mong
This is interesting...but what is the context? Where did the formula
WPCT = 1/2 + (RS-RA)/10
come from?
And what is the meaning of the title, "10 runs = 1 win?"
I suppose I am just out of the loop here - but I would like to be in the loop!
Posted 4:57 p.m.,
August 7, 2003
(#2) -
tangotiger
(homepage)
Every 10 marginal wins will add 1 marginal win.
So, you start with a team of 4.5 RS and 4.5 RA. What's their win%? .500.
Now, you have a team with 4.6 RS and 4.4 RA. What's their win%? If you use PythagaPat, it estimates .521. You can also try running the Tango Distribution (see above site), and you get .520.
Try 4.8 v 4.2, and you get .563 with PythagaPat and .561 with TD.
Try 5 v 4, and you get .604 with PP, .601 with TD, and .600 in real-life.
The equation: W% = .500 + (RS-RA)/10 will lead to a similar answer to the above.
It is a nice shorthand, but for more rigorous, I suggest the PP or TD processes.