THE BOOK--Playing The Percentages In Baseball

Tom

That seems like a great idea (about the SDs). But I am just not sure how to proceed. I’m afraid I need a step-by-step recipe for what you suggest.

Cy

I don’t really understand well how SDs and z-scores are used to figure out if clutch hitting exists. Is there any way you could do some calculations using SDs and z-scores to show how hitting singles, for example, is a real skill as some kind of comparison? If there was some table comparing the level of skill in different types of hitter actions, it might be easier to understand how much less of a skill clutch hitting might be.

This is the process I went through to show how much pitcher skill there is to getting outs on BIP.  And batter’s on getting on base per PA.  Check out those threads, labelled “Spread in Talent”.

In summary the z-score process shows actual variation from expected variation.

Take a bunch of data, say OBP and work out the Z score for all hitters. We know that for a standardized normal (Z) distribution then the st dev is 1. Therefore if you work out the Z-scores and then work out the st dev of the z scores and it = 1 then the data fits the expected distribution. As Tango says if it is higher then skill variance is present.

John

I think I understand what is being suggested. I found the standard deviation of DIFF. That is the difference between actual PWV and predicted PWV. So it is the measure of clutch abiltiy (or performance). I just found the sample SD. That was .392 (so was the population SD). The mean was -0.02189 (meaning the average hitter chokes just a little?). So I calculated the z-score for Lance Johnson as (1.383- (-.02189))/.392 or 1.383 + .02189)/.392. That gave me 3.583. I did that for everyone. Then found the sample SD of those z-scores and it was 1.001. So does that mean their is some slight clutch hitting going on, but very slight?

Keep in mind that I did not do a weighted average or weighted SD. Players with longer careers did not have their DIFF weighted more heavily.

I downloaded all hitters from 2005 with greater than 300 ABs, which was 257 hitters. I calculated what % of PAs (AB+BB) were singles. Then I figured the StDev of the Singles% and then used that to find the Z-score for each player. The StDev of Z-scores for all the players was .597 and the mean was .801 for them. So I’m not sure, exactly, what this is telling me.

Chris,

I am not sure you have done your calculation correctly. I did it for 1B% over the last 3 years for all hitters who qualified for a batting title (met minimum AB). I got a st dev on the Z of 1.8. The st dev Z should be higher than 1 becuase you expect skill variance in the skill.

Cy,

When you find the st dev initially, this should be the EXPECTED st dev, not the obsever. This is what you use to calculate Z scores and then take the st dev of this. This gives you the variation from expected.

Presumably using 300+ ABs wouldn’t affect the result too much. Maybe I missed a step or two.

I calculated 1B% and found stdev. To calc z-score did abs(1B%-mean 1B%)/stdev. Then stdev of z-scores.

If you can figure out what I did wrong or if not then show how you did it, I would appreciate it. Thanks.

Chris,

Let me outline the approach I took.

1) Gather all data for last 3 seasons (I used STATS Inc for this)

2) Calculate 1b% as 1b/(AB+BB) for all players.

3) calculate mean = 0.165 (not weighted)

4) calculate expected st dev = 0.015 based on 541 AB+BB per season (I should say here that the 3 seasons I chose were 2005,5,6YTD, s0 2006 has less PA than it should do. No matter, I am just outlining the methodology

5) calculate z scores as (x - 0.165)/0.015 for each hitter

6) calc st dev of this. I get 1.88 as the st dev of Z

I think your problem is that you calculate abs(Z). Leave the sign in and see what you get. Remember to calculate EXPECTED st dev in the first step

John

What do you mean by the “expected” SD? What is the “observer” SD? Are these two like sample and population SD?

And yes, I did take the SD of the z-scores.

I was using the observed stdev of 1B% and abs for the z-scores. Using .015 for stdev and nonabs for z-scores I got 1.88 stdev of z scores for my smaller sample.

How do you figure out the expected SD?

Cy,

Sorry, I should have been clearer. Expected st dv is the st dev if everyone has the same skill level, which in the 1B% example would be the mean. This is calculated as srqt(P*(1-P)/N).

Then you work out the z scores using this and then calculate the (sample) std dev of the Z score (this is what I call observed) ... my language should probably have been clearer.

Chris,

You st dev Z is very similar to mine .... which is good!

John

I am not working with a binomial variable. P is probability, right? I am not working with something that is or is not a hit. My variable, if you can call it that, is based on what happens in all PAs. It is a win total. Is there some other formula I can use here? I don’t think the one you show me applies in this case.

Cy,

You are right.

I believe in your case what I call expected std dev should be based on your predicted PWV equation (as this is what we expect a player to hit). Use this in the initial Z calculation which becomes (Actual - Predicted)/Std dev(predicted)).

Then take the standard dev of the Zs ... I think that should give you what you are looking for. I think anyway!

I think that is what I have already done. At my site

“The Problem With “Total Clutch” Hitting Statistics”

which is at

http://www.geocities.com/cyrilmorong@sbcglobal.net/totalclutch1.htm

I reported that the standard error was about .39 (that is close to what I used above). Actually, it (the standard error) was .00056 per 700 PAs, which is exactly .392. Of course, it comes from the equation

PWV/PA = -.0246 + .000149*OBP + .000097*SLG

Forgot. Yes, I already took the SD of the z-scores and got 1.001

Thanks for the help John!

Cy,

Much ado about nothing! I just wasn’t clear exactly what you had done from your original post, but it appears you have done it correctly.

A st dev on Z of 1.001 shows almost no variation vs random. If you multiply 1.001 by the predicted PWV then you get the true PWV distribution, which will be very small. Effectively what you are showing is that there is no clutch skill in your study.

The fact that your DIFF has a mean less than zero is interesting. If you weight it by PA do you get zero? If not then this could be because in clutch situations hitters generally face better pitchers.

Tango has show in The Book that relievers have a better ERA / wOBA / [insert relevent pitching stat] than starters.

Very interesting work as ever Cy. To me it feels like a precursor to some of the WPA work that is doing the rounds at the moment.

John

The weighted mean diff was -.011. So closer to zero and pretty small. The mean seems unlikely to be exactly zero. We are talking about 1/100 of a win here. Also, I got just about 1 (.999 something) with a weighted SD of the z-scores (which was based on the weighted average and a weighted SD of DIFF)

John,

How would the expected StDev be calculated for the basic LWTS (using this formula):

Runs = (.47)1B + (.78)2B + (1.09)3B + (1.40)HR + (.33)BB - (.25)(AB-H)

Since it would average near 0 (I could change the .25 so LWTS would be 0), it would be somewhat similar to the clutch stats and not be like it would be for the hitting % stats, if what I’m understanding of the above posts is correct.

Chris,

I am not 100% sure what you are trying to do with this. Anyway ... here are my thoughts.

1)The formula as you have it is a count formula, first of all you need to change it into a rate formula (ie, divide by PA). Otherwise the st dev won’t work—a player with 100 PA will have less LWTS RUNS than a player with 600 PA.

2)The simplest thing to do is treat it as a multinomial statistic. I don’t know if you have bought a copy of The Book, if you haven’t then you should, but on pg 370 in talks about how to calcuate expected st dev of multinomials (in this case wOBA). To do it properly isn’t straightforward and would take a bit of explanation so I’ll assume you have The Book!

3)One other thing I’d do is to develop the equation without the out, otherwise you have dependendency in the outcomes. Note: for simplicity just add the value of the out (0.25) to each of othe other coefficients. This also gets rid of the zero issue.

So pulling it all together. As a first step what you could do is calculate RUNS/PA and then scale it to, say, batting average (add a constant if you must) and then use a deviation of the binomial std dev formula like std dev = sqrt(P*(1.1-P)/N). This is the formula given in The Book for wOBA. The 1.1 is because there is more variance in the data (particularly for power hitters) than a pure binomial stat. Use this as your expected st dev. Calcualte the observed st dev as before and you are on your way.

Good luck. Where I currently live it is v. late so need to go. But if you have any other questions I am sure others more wise than I can answer them ... I look forward to checking back in tomorrow

Keen to hear if anyone else has different ideas ....

John

I didn’t know it would be so much more complicated for a league relative stat (where the average for the league is zero) as compared to a probability? stat (like the % chance of hitting a single). I should have mentioned that I meant LWTS/PA. I was looking to compare that with 1B%/PA for hitters in 2005 to see if there was much difference in the stdev of the hitters z-scores between the two. I’ll look at the different things you mentioned and see if I can understand it better.

Thanks again for the help.

Chris

I might be on thin ice here .... but I don’t think the zero matters. There will still be variance in the data.

What matters for the approach of comparing standard deviation is to calcuate expected and observed distributions. The difference between is the skill factor.

However in your LWTS example the out (which has a negative co-efficient) means you can’t treat the stat like a probability. Beacuse for a true multinomial outcome the some of all the probabilities should equal 1. Hence why I suggested getting rid of the out.

I’m not sure what looking at the st dev of 1b% and LWTS/PA will tell you. The LWTS/PA will have a higher coefficient of variation, I’m sure ...

My question mainly had to do with your formula for expected StDev, srqt(P*(1-P)/N), because if P = LWTS/PA = 0, then obviously that formula couldn’t be used to calculate expected StDev like it could be for the 1B% example.

Let’s say instead of using 1B%, I used 1B% versus the league average. So if the league average was .165 per PA, then instead of having .198, .217, .142 etc., I had +.033, +.014, -.023 etc. In the first case I use P = .165 to calculate the expected StDev, but in the second case the mean would be 0, so P = 0, and I couldn’t use the formula to calculate the expected StDev.

So basically I was just wondering if there would be an easy way to figure out the expected StDev if I used stats relative to the league. I hope I made this a little clearer.

Chris,

That formula is for binomial statistics only (and can be adapted for multinomial statistics). In other words if you add together the probability of all the outcomes you get 1. Obviously in your formula the out is negative so you although if you add all the probabilities together you get 1 (by definition), when you include the weights you don’t.

If you make the adjustment I mentioned earlier (with the out) then you can use the modified shortcut forumla that is in The Book (with a 1.1 factor), otherwise you can’t.

In the example you present you’d need to know the 1B% (.165) and calculate the expected standard deviation. When you adjust to league average the expected std dev doesn’t change ....

If you are comparing say LWTS to say RC ... to see if LWTS has as much variation as RC what you do then is

1) Find all Z as (RC - LWTS)/std(LWTS) for each batter. Here RC is observed, and LWTS is predicted (in this example)

2) Work out the std dev for all of these Z

Why you’d want to do that I’m not too sure. You have to compare two things ... whether it is expected and observed data (as with the binomial) or whether it is how well a regression equation fits with reality, or if LWTS explains all the variance in RC.

I think to answer your question you can’t do what you want to do unless you translate the LWTS stat into something akin to wOBA in The Book, unless you have something very specific and similar you are trying to compare it to.