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THE BOOK--Playing The Percentages In Baseball

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Tuesday, May 06, 2008

Pitchers WAR Aging Curves: 5 year forecast

By Tangotiger, 11:39 AM

I finally got around to doing an initial look.  This is what I did:


1. Took all pitchers born since 1918 (since Bob Feller)
2. Figured out runs above replacement, runs per win converter, wins above replacement (WAR)
3. For any season where a pitcher had at least 50 IP, figured out how many WAR they had in the following 5 years
4. Create a function using age and WAR in year X, to figure out total WAR in years x+1 through to years x+5

And so, here it is:

WAR5 = 2*WAR + 10 - Age/4

where WAR5 is the WAR for the next 5 years, WAR is the WAR for the most recent season, and Age is season minus birth year

Let’s take an example.  Someone is a .630 pitcher, with 225 IP, and is 28 years old.  What’s our expectation for the next 5 years? 

His WAR is (.630-.380) * 225/9 = 6.25

WAR5 = 2*6.25 + 10 - 28/4 = 15.5

So, we expect 15.5 WAR over the next 5 years.  And this is a top-flight pitcher.

If you have a $ per wins over the next 5 years of 4.8, 5.3, 5.8, 6.4, 7.0, that gives us a rough average of 5.8, remembering though that we want to weight the earlier seasons more).  That gives us a 90 million$ contract over 5 years.  And that is for the prime meat of pitchers.

Take a more reasonable high quality pitcher, say one who is a .580 pitcher, 207 innings, and 30 years old, and we get: 11.7 WAR over the next 5 years.  This gives us a 68MM$ deal over 5 years.

And what about a .500 pitcher who is 32 years old, with 180 IP?  WAR5 = 7, or a 40MM deal over 5 years.

Gotta run…

#1    Tangotiger      (see all posts) 2008/05/06 (Tue) @ 12:02

One quick thing.

If you figure the average pitcher being 28 years old, the function becomes WAR5 = 2*WAR + 3

Remember also that my really quick rule of thumb would be to lop off 0.5 wins per year.  So, if you have 2.5 WAR, then the next 5 years would be 2, 1.5, 1.0, 0.5, 0.  The total would be 5.0.

If you have 4.0 WAR, then the next 4 years would be 3.5, 3.0, 2.5, 2.0, 1.5, for a total of 12.5.

You can see pretty quickly how the quick rule of thumb becomes:
WAR5 = 5 * (WAR - 1.5)

A WAR of 3.5 gives us a WAR5 = 10, using either of the two equations.

So, we can see how the rule of thumb of simply chopping off 0.5 wins per year had some basis.

***

I’ll compare to my Rule of 10 (drop IP by 10% and win% by 10 basis points) to see how that looks too, next time.


#2    Mark      (see all posts) 2008/05/07 (Wed) @ 10:13

Is win % (.630 in the first example) based off traditional W-L or is there some other method you use?  If so, where can I find those numbers.


#3    Tangotiger      (see all posts) 2008/05/07 (Wed) @ 11:06

It doesn’t really matter what you use.  It’s just a stand-in for quality.  I say someone is a .600 pitcher, we all know what that means. 

The quickest way is to simply use ERA+, divide by 100, square that number, and divided by itself+1.  So, someone with an ERA+ of 141.4 would be a .667 pitcher.  Not exactly right, but close enough for something quick.

The best pitchers at their very peak are probably .750 pitchers.  Over their careers, the best are somewhere between .650 and .700.

As a practical matter, I’d probably top someone off at .667 unless you look at it more closely, just so that you don’t have a bunch of pitchers all exceeding that level (Santana, Oswalt, Peavy, Holliday, etc).


#4    Hyltzn      (see all posts) 2008/07/10 (Thu) @ 17:02

The quickest way is to simply use ERA+, divide by 100, square that number, and divided by itself+1.

Tom, could you explain why you would do that (why you square, divide by itself+1)? I just really like to know why it is that I’m doing certain things.

Thanks


#5    Hyltzn      (see all posts) 2008/07/10 (Thu) @ 17:41

I actually see. It’s the same steps as using Pytheganpat for ERA. Am I correct?


#6    tangotiger      (see all posts) 2008/07/10 (Thu) @ 20:28

It’s the classic Bill James Pythag equation, where we have a fixed exponent of 2.

PythagenPat floats that exponent based on the run environment.


#7    Hyltzn      (see all posts) 2008/07/10 (Thu) @ 21:07

Yeah, I know that. I just looked at your explanation on the original How To Calculate WAR Thread and saw that it was the same thing.

Thanks


#8    Hyltzn      (see all posts) 2008/07/10 (Thu) @ 21:29

I should say: same formula for the exception of the exponent.


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