Calculating park factors from batting lines instead of runs

I missed a post Tango wrote back in 2019 about park factors. In the comments, he said,

"That’s one place where we failed with our park factors, using actual runs instead of "component" runs. They should be based on Linear Weights or RC or wOBA, something like that.

"Using actual runs means introducing unnecessary random variation in the mix."

Yup. One of those bits of brilliance that's obvious in retrospect.

The idea is, there's a certain amount of luck involved in turning batting events into runs, which depends on the sequence -- in other words, "clutch hitting," which is thought to be mostly random. If teams wind up scoring, say, 20 runs above average in a certain park, it could be that the park lends itself to higher offense. But, it could also be that the park is neutral, and those 20 runs just came from better clutch hitting.

So if we calculated park factors from raw batting lines, instead of actual runs, we eliminate that luck, and should get better estimates. We can still convert to expected runs afterwards.

Let's do it. I'll start with using runs as usual. Then, I'll do it for wOBA, and we'll compare.

-------

I used team-seasons from 2000-2019, except Coors Field (because it`s so extreme an outlier). I included only parks that were used at least 16 of the 20 seasons. 

To get the observed park effects, I just took home scoring (both teams combined) and subtracted road scoring (both teams combined). 

For those 444 datapoints, I got

SD(observed) = 81.6 runs

To estimate luck, I used the rule of thumb that SD(runs) for a single team's games is about 3. (Tango uses the square root of total runs for both teams, but I didn't bother.)  

If SD(1 game) = 3, then SD(81 games) = 27. But we want both teams combined, so multiply by root 2. Then, we want (home - road), so multiply by root 2 again. That gives us 54.

SD(luck) = 54 runs

Since var(observed) = var(luck) + var(non-luck), we get*

SD(non-luck) = 61.2 runs

*"var" is variance, the square of SD. I'm using it instead of "SD^2" because it makes it much easier to read.

Now, what's this thing I called "non-luck"? It's a combination of the differences between parks, and season-to season differences within the same park -- weather, how well the players are suited to the park, the parks used by other teams in the division (because of the unbalanced schedule), the parks used by interleague opponents, the somewhat-random distribution of opposing pitchers ... stuff like that.

var(non-luck) = var(between parks) + var(within park)

To estimate SD(within park), I just looked at the observed SDs of the same park across the 16-20 seasons in the dataset. There were 23 parks in the sample, and I took the root-mean-square of those 23 individual SDs. I got

SD(different seasons of park) = 64.1

But ... that 64.1 includes luck, and we want only the non-luck portion. So let's remove luck:

var(diff. seas. of park)= var(luck) + var(within park)
64.1 squared = 54 squared  + var(within park)
SD(within park) = 34.5 runs

And now we can estimate SD(between parks):

var(non-luck) = var(between parks) + var(within park)
61.2 squared = var(between parks) + 34.5 squared
SD(between parks) = 50.5 runs

Summarizing:

81.6  runs total
---------------------------------
54    luck
50.5  between parks
34.5  within park between seasons

Park squared is only 38 percent of the total squared. That means that only 38 percent of the observed park effect is real, and you have to regress to the mean by 62 percent to get an unbiased estimate.

That's a lot. And it's one reason that most sites publish park factors based on more than one season, to give luck a chance to even out.

-------

Now, let's try Tango's suggestion to use wOBA instead, and see how much luck that squeezes out.

For the same individual parks, I calculated every year's observed park difference the same way as for runs -- home wOBA minus road wOBA, both teams combined.

For the sample, SD(observed) was 0.01524, against an average wOBA of .3248. That's a ratio of 4.7%. I did a regression and found runs-per-PA increase 1.8x as fast as wOBA (probably proportional to the 1.77th power, or something), so 4.7% in wOBA is 8.45% in runs.

In the full sample, there were .118875 runs per PA, and an average 6207 PA for each home park-season. That's about 738 runs. Taking 8.45 percent of that works out to an SD of 67.3 runs.

SD(observed) = 67.3 runs

The luck SD for wOBA for a single PA is .532 (as calculated from an average batting line APBA card). I'll spare you repeating the percentage calculations, but for 6207 PA,

SD(luck) = 41.9 runs

As before, var(observed) = var(luck) + var(non-luck), so

SD(non-luck) = 52.7 runs

Looking at the RMS between-season SD of the 23 teams in the sample, 

SD(different seasons of park) = 51.2 runs

Eliminating luck to get true season-to-season differences:

var(diff. seas. of park)= var(luck) + var(within park)
51.2 squared = 41.9 squared  + var(within park)
SD(within park) = 29.4 runs

And, finally,

var(non-luck) = var(between parks) + var(within park)
52.7 squared  = var(between parks) + 29.4 squared
SD(between parks) = 43.7 runs

The summary:

67.3  runs total
---------------------------------
41.9  luck
43.7  between park
29.4  within park between seasons

Here the "between park" variance is 42 percent of the total, up from 38 percent when we used runs. So we have, in fact, gotten more accurate estimates.

------

But wait! The two methods really should give us the same estimate of the SD of the "between" and "within" park factors, since they're trying to measure the same thing. But they don't:

runs  wOBA
-----------------------------------------
81.6  67.3   runs total
-----------------------------------------
54    41.9   luck
50.5  43.7   between park
34.5  29.4   within park between seasons

(The "luck" SD is supposed to be different, since that was the whole purpose of using wOBA, to eliminate some of the random noise.)

I think the difference is due to the fact that the wOBA variances were all based on averages per PA, while the runs variances were based on averages per game (roughly, per 27 outs).

On average, the more runs you score, the more PA you'll have. So changing the denominator to PA reduces the high-scoring games relative to the low-scoring games, which compresses the differences, which reduces the SD. 

Although the differences in PA look small, they actually indicate large differences in scoring. Because, per season, every park gets roughly the same number of outs, which means roughly the same number of PA that are outs. So any "extra PA" are mostly baserunners, and very valuable in terms of runs.

If you switch from "observed runs per game" to "observed runs per 6207 PA," the observed SD drops from 81.6 to 72.7 runs.  That's an 11 percent drop. When I did the same for wOBA, the observed SD dropped by 13 percent. So, let's estimate that the difference between "per game" and "per PA" is 12 percent, and reduce everything in the runs column by 12 percent:

runs  wOBA
--------------------------------------------
71.8  67.3   runs total
--------------------------------------------
47.5  41.9   luck
44.4  43.7   between park
30.4  29.4   within park between seasons
--------------------------------------------

62%   58%    regression to long-term mean 

I'm not 100% sure this is legitimate, but it's probably pretty close. One thing I want to do to make the comparison better, is to use the same value for "between park" and "within park", since we expect the methods to produce the same estimate, and we expect that any difference is random (in things like wOBA to run conversion, or how PA vary between games, or the fact that the wOBA calculation omits factors like baserunning).

So after my manual adjustment, we have:

runs  wOBA
--------------------------------------------
71.4  67.8   runs total
--------------------------------------------
47.5  41.9   luck
44    44     between park
30    30     within park between seasons
--------------------------------------------
62%   58%    regression to long-term mean 

-------

That's still a fair bit you have to regress either way -- more than half -- but that would be reduced if you used more than one season in your sample. If we go to the average of four seasons, "luck" and "within park" both get cut in half (the square root of 1/4). 

I'll divide both of those by 2, and recalculate the top and bottom line:

runs  wOBA

--------------------------------------------
52.3  51.0   runs total
--------------------------------------------
24    21     luck
44    44     between park
15    15     within park between seasons
--------------------------------------------

29%   26%    regression to long-term mean 

So if we use a four-year park average, we should only have to regress 29 percent (for runs) or 26 percent (for wOBA). 

-------

Thanks to Tango for the wOBA data making this possible, and for other observations I'm saving for a future post.

My three previous posts on park factors are here:  one two three

Regression to higher ground

We know that if an MLB team wins 76 games in a particular season, it's probably a better team than its record indicates. To get its talent from its record, we have to regress to the mean.

Tango has often said that straightforward regression to the mean sometimes isn't right -- you have to regress to the *specific* mean you're concerned with. If Wade Boggs hits .280, you shouldn't regress him towards the league average of .260. You should regress him towards his own particular mean, which is more like .310 or something.

This came up when I was figuring regression to the mean for park factors. To oversimplify for purposes of this discussion: the distribution of hitters' parks in MLB is bimodal. There's Coors Field, and then everyone else. Roughly like this random pic I stole from the internet:

Now, suppose you have a season of Coors Field that comes in at 110. If you didn't know the distribution was bimodal, you might regress it back to the mean of 100, by moving it to the left. But if you *do* know that the distribution is bimodal, and you can see the 110 belongs to the hump on the right, you'd regress it to the Coors mean of 113, by moving it to the right.

But there are times when there is no obvious mean to regress to.

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You have a pair of perfectly fair 9-sided dice. You want to count the number of rolls it takes before you roll your first snake eyes (which has a 1 in 81 chance each roll). On average, you expect it to take 81 rolls, but that can vary a lot. 

You don't have a perfect count of how many rolls it took, though. Your counter is randomly inaccurate with an SD of 6.4 rolls (coincidentally the same as the SD of luck for team wins).

You start rolling. Eventually you get snake eyes, and your counter estimates that it took 76 rolls. The mean is 81. What's your best estimate of the actual number? 

This time, it should be LOWER than 76. You actually have to regress AWAY from the mean.

-------

Let's go back to the usual case for a second, where a team wins 76 games. Why do we expect its talent to be higher than 76? Because there are two possibilities:

(a) its talent was lower than 76, and it got lucky; or
(b) its talent was higher than 76, and it got unlucky.

But (b) is more likely than (a), because the true number will be higher than 76 more often than it'll be lower than 76. 

You can see that from this graph that represents distribution of team talent:

The blue bars are the times that talent was less than 76, and the team got lucky.  The pink bars are the times the talent was more than 76, and the team got unlucky.

The blue bars around 76 are shorter than the pink bars around 76. That means better teams getting unlucky are more common than worse teams getting lucky, so the average talent must be higher than 76.

But the dice case is different. Here's the distribution of when the first snake-eyes (1 in 81 chance) appears:

The mean is still 81, but, this time, the curve slopes down at 76, not up.

Which means: it's more likely that you rolled less than 76 times and counted too high, than that you rolled more than 76 times and counted too low. 

Which means that to estimate the actual number of rolls, you have to regress *down* from 76, which is *away* from the mean of 81.

--------

That logic --let's call it the "Dice Method" -- seems completely correct, right? 

But, the standard "Tango Method" contradicts it.

The SD of the distribution of the dice graph is around 80.5. The SD of the counting error is 6.4. So we can calculate:

SD(true)     = 80.5
SD(error)    =  6.4
--------------------
SD(observed) = 80.75

By the Tango method, we have to regress by (6.4/80.75)^2, which is less than 1% of the way to the mean. Small, but still towards the mean!

So we have two answers, that appear to contradict each other:

-- Dice Method: regress away from the mean
-- Tango Method: regress towards the mean

Which is correct?

They both are.

The Tango Method is correct on average. The Dice Method is correct in this particular case.

If you don't know how many rolls you counted, you use the Tango Method.

If you DO know that the count was 76 rolls, you use the Dice Method.

------

Side note:

The Tango Method's regression to the mean looked wrong to me, but I think I figured out where it comes from.

Looking at the graph at a quick glance, it looks like you should always regress to the left, because the left side of every point is always higher than the right side of every point. That means that if you're below the mean of 81, you regress away from the mean (left). If you're above the mean of 81, you regress toward the mean (still left).

But, there are a lot more datapoints to the left of 81 than to the right of 81 -- by a ratio of about 64 percent to 36 percent. So, overall, it looks like the average should be regressing away from the mean.

Except ... it's not true that the left is always higher than the right. Suppose your counter said "1". You know the correct count couldn't possibly have been zero or less, so you have to regress to the right. 

Even if your counter said "2" ... sure, a true count of 1 is more likely than a true count of 3. but 4, 5, and 6 are more likely than 0, -1, or -2. So again you have to regress to the right.

Maybe the zero/negative logic is a factor when you have, say, 8 tosses or less, just to give a gut estimate. Those might constitute, say, 10 percent of all snake eyes rolled. 

So, the overall "regress less than 1 percent towards the mean of 81" is the average of:

-- 36% regress left  towards the mean a bit (>81)
-- 54% regress left  away from the mean a bit (9-81)
-- 10% regress right towards the mean a lot (< 8)
   -----------------------------------------------------
-- Overall average: regress towards the mean a tiny bit.

--------

The "Tango Method" and the "Dice Method" are just consequences of Bayes' Theorem that are easier to implement than doing all the Bayesian calculations every time. The Tango Method is a mathematically precise consequence of Bayes Theorem, and the Dice Method is an heuristic from eyeballing. Tango's "regress to the specific mean" is another Bayes heuristic.

We can reduce the three methods into one by noting what they have in common -- they all move the estimate from lower on the curve to higher on the curve. So, instead of "regress to the mean," maybe we can say

"regress to higher ground."

That's sometimes how I think of Bayes' Theorem in my own mind. In fact, I think you can explain Bayes exactly, as a more formal method of figuring where the higher ground is, by explicitly calculating how much to weight the closer ground relative to the distant ground. 

Regressing Park Factors (Part III)

I previously calculated that to estimate the true park factor (BPF) for a particular season, you have to take the "standard" one and regress it to the mean by 38 percent.

That's the generic estimate, for all parks combined. If you take Coors Field out of the pool of parks, you have to regress even more.

I ran the same study as in my other post, but this time I left out all the Rockies. Now, instead of 38 percent, you have to regress 50 percent. (It was actually 49-point-something, but I'm calling it 50 percent for simplicity.)

In effect, the old 38 percent estimate comes from a combination of 

1. Coors Field, which needs to be regressed virtually zero, and
2. The other parks, which need to be regressed 50 percent.

For the 50-percent estimate, the 93% confidence interval is (41, 58), which is very wide. But the theoretical method from last post, which I also repeated without Colorado, gave 51 percent, right in line with the observed number.

--------

I tried this method for the Rockies only, and it turns out that the point estimate is that you have to regress slightly *away* from the mean of 100. But with so few team-seasons, the confidence interval is so huge that I'd just take the park factors at face value and not regress them at all. 

The proper method would probably be to regress the Rockies' park factor to the Coors Field mean, which is about 113. You could probably crunch the numbers and figure out how much to regress. 

--------

The overall non-Coors value is 50 percent, but it turns out that every decade is different. *Very* different:

1960s:   regress 15 percent
1970s:   regress 27 percent
1980s:   regress 80 percent
1990s:   regress 84 percent
2000s:   regress 28 percent
2010-16: regress 28 percent 

Why do the values jump around so much? One possibility is that it's random variation on how teams are matched to parks. The method expects batters in hitters' parks to be equal to batters in pitchers' parks, but if (for instance) the Red Sox had a bad team in the 80s, this method would make the park effect appear smaller.

As soon as I wrote that, I realized I could check it. Here are the correlations between BPF and team talent in terms of RS-RA (per 162 games) for team-seasons, by decade. I'll include the regression-to-the-mean amount to make it easier to compare:

             r    RTM
---------------------
1960s:    +0.14   15% 
1970s:    +0.06   27%
1980s:    -0.14   80%
1990s:    +0.03   84%
2000s:    +0.16   28%
2010s:    +0.23   28%
---------------------
overall:  +0.05   50%

It does seem to work out that the more positive the correlation between hitting and BPF, the more you have to regress. The two lowest correlations were the ones with the two highest levels of regression to the mean.

(The 1990s does seem a little out of whack, though. Maybe it has something to do with the fact that we're leaving out the Rockies, so the NL BPFs are deflated for 1993-99, but the RS-RA are inflated because the Rockies were mediocre that decade. With the Rockies included, the 1990s correlation would turn negative.)

The "regress 50 percent to the mean" estimate seems to be associated with an overall correlation of +.05. If we want an estimate that assumes zero correlation, we should probably bump it up a bit -- maybe to 60 percent or something.

I'd have to think about whether I wanted to do that, though. My gut seems more comfortable with the actual observed value of 50 percent. I can't justify that.

Regressing Park Factors (Part II)

Note: math/stats post.

-------------

Last post, I figured the breakdown of variance for (three-year average) park effects (BPFs) from the Lahman database. It came out like this:

All Parks   [chart 1]
-------------------------
4.8 = SD(3-year observed)
-------------------------
4.3 = SD(3-year true) 
2.1 = SD(3-year luck)
-------------------------

Using the usual method, we would figure, theoretically, that you have to regress park factor by (2.1/4.8)^2, which is about 20 percent. 

But when we used empirical data to calculate the real-life amount of regression required, it turned out to be 38 percent.

Why the difference? Because the 20 percent figure is to regress the observed three-year BPF to its true three-year average. But the 38 percent is to regress the observed three-year BPF to a single-year BPF.

My first thought was: the 3-year true value is the average of three 1-year true values. If each of those were independent, we could just break the 3-year SD into three 1-year SDs by dividing by the square root of 3. 

But that wouldn't work. That's because when we split a 3-year BPF into three 1-year BPFs, those three are from the same park. So, we'd expect them to be closer to each other than if they were three random BPFs from different parks. (That fact is why we choose a three-year average instead of a single year -- we expect the three years to be very similar, which will increase our accuracy.)

Three years of the same park are similar, but not exactly the same. Parks do change a bit from year to year; more importantly, *other* parks change. (In their first season in MLB, the Rockies had a BPF of 118. All else being equal, the other 13 teams would see their BPF drop by about 1.4 points to keep the average at 100.)

So, we need to figure out the SD(true) for different seasons of the same park. 

--------

From the Lahman database, I took all ballparks (1960-2016) with the same name for at least 10 seasons. For each park, I calculated the SD of its BPF for those years. Then, I took the root mean square of those numbers. That came out to 3.1.

We already calculated that the SD of luck for the average of three seasons is 2.1. That means we can fill in SD(true)=2.3.

Same Park     [Chart 2]
------------------------------------
3.1 = SD(3-year observed, same park)
------------------------------------
2.3 = SD(3-year true, same park)
2.1 = SD(3-year luck, any parks)
------------------------------------

(That's the only number we will actually need from that chart.)

Now, from Chart 1, we found SD(true) was 4.3 for all park-years. That 4.3 is the combination of (a) variance of different years from the same park, and (b) variance between the different parks. We now know (a) is 2.3, so we can calculate (b) is root (4.3 squared minus 2.3 squared), which equals 3.6.

So we'll break the "4.3" from Chart 1 into those two parts:

All Parks     [Chart 3]
-----------------------------------
4.8 = SD(3-year observed)
-----------------------------------
3.6 = SD(3-year true between parks)
2.3 = SD(3-year true within park)
2.1 = SD(3-year luck)
-----------------------------------

Now, let's assume that for a given park, the annual deviations from its overall average are independent from year to year. That's not absolutely true, since some changes are more permanent, like when Coors Field joins the league. But it's probably close enough.

With the assumption of independence, we can break the 3-year SD down into three 1-year SDs.  That converts the single 2.3 into three SDs of 1.3 (obtained by dividing 2.3 by the square root of 3):

All Parks     [Chart 4]
-----------------------------------
4.8 = SD(3-year observed)
-----------------------------------
3.6 = SD(3-year true between parks)
1.3 = SD(this year true for park)
1.3 = SD(next year true for park)
1.3 = SD(last year true for park)
2.1 = SD(3-year luck)
-----------------------------------

What we're interested in is the SD of this year's value. That's the combination of the first two numbers in the breakdown: the SD of the difference between parks, and the SD of this year's true value for the current park.

The bottom three numbers are different kinds of "luck," for what we're trying to measure. The actual luck in run scoring, and the "luck" in how the park changed in the other two years we're using in the smoothing for the current year. 

All Parks     [Chart 4a]
-----------------------------------
4.8 = SD(3-year observed)
-----------------------------------
3.6 = SD(3-year true between parks)
1.3 = SD(this year true for park)

1.3 = SD(next year true for park)
1.3 = SD(last year true for park)
2.1 = SD(3-year luck)
-----------------------------------

Combining the top three and bottom three, we get

All Parks      [Chart 5]
----------------------------------------------
4.8 = SD(3-year observed)
----------------------------------------------
3.8 = SD(true values impacting observed BPF)
2.8 = SD(random values impacting observed BPF)
----------------------------------------------

So we regress by (2.8/4.8) squared, which works out to 34 percent. That's pretty close to the actual figure of 38 percent.

We can do another attempt, with a slightly different assumption. Back in Chart 2, when we figured SD(three year true, same park) was 3.1 ... that estimate was based on parks with at least ten years of data. If I reduce the requirement to three years of data, SD(three year true, same park) goes up to 3.2, and the final result is ... 36 percent regression to the mean.

So there it is.  I think this is method is valid, but I'm not completely sure. The 95% confidence interval for the true value seems to be wide -- regression to the mean between 28 percent and 49 percent -- so it might just be coincidence that this calculation matches. 

If you see a problem, let me know.

Part III is here.

Regressing Park Factors (Part I)

I think park factors* are substantial overestimates of the effect of the park. 

At their core, park effects are basically calculated based on runs scored at home divided by runs scored on the road. But that figure is heavily subject to the effects of luck. One random 10-8 game at home can skew the park effect by more than half a point.

Because of this, most sabermetric sources calculate park effects based on more than one year of data. A three-year sample is common ... I think Fangraphs, Baseball Reference, and the Lahman database all use three years. 

That helps, but not enough. It looks like park factors are still too extreme, and need to be substantially regressed to the mean.

-------

* Here's a quick explanation of how park factors work, for those not familiar with them.

Park Factor is a number that tells us how relatively easy or difficult it is for a team to score runs because of the characteristics of its home park. The value "100" represents the average park. A number bigger than 100 means it's a hitters' park, where more runs tend to score, and smaller than 100 means it's a pitchers' park, where fewer runs tend to score.


Perhaps confusingly, the park factor averages the home park with an amalgam of road parks, in equal proportion. So if Chase Field has a park factor of 105, which is 5 percent more than average, that really means it's about 10 percent more runs at home, and about average on the road.

The point of park factor is that you can use it to adjust a hitter's stats to account for his home park. So if Edouardo Escobar creates 106 runs for the Diamondbacks, you divide by 1.05 and figure he'd have been good for about 101 runs if he had played in a neutral home park.

--------

For my large sample of batters (1960-2016, minimum 300 PA), I calculated their runs created per 500 PA (RC500), and their park-adjusted RC500. Then, I binned the players by batting park factor (BPF), and took the average for each bin. If park adjustment worked perfectly, you'd expect every bin to have the same level of performance. After all, there's no reason to think batters who play for the Red Sox or Rockies should be any better or worse overall than batters who are current Mets or former Astros.

(Because of small sample sizes, I grouped all parks 119+ into a single bin. The average BPF for those parks was 123.4.)

Here's the chart:


BPF      PA     Runs   Adj'd  Regressed
------------------------------------------
 88   8554   67.07   76.22   72.67
 89   2960   56.57   63.56   60.85
 90  43213   62.53   69.48   66.78
 91  61195   62.42   68.59   66.19
 92 121203   61.04   66.35   64.29
 93 195382   62.21   66.89   65.07
 94 241681   61.93   65.89   64.35
 95 304270   64.72   68.12   66.80
 96 325511   64.13   66.81   65.77
 97 463537   63.05   65.00   64.24
 98 520621   65.62   66.96   66.44
 99 712668   64.29   64.94   64.69
100 674090   64.86   64.86   64.86
101 589401   66.53   65.87   66.12
102 514724   66.19   64.89   65.39
103 440940   65.48   63.58   64.32
104 415243   66.07   63.53   64.51
105 319334   67.35   64.15   65.39
106 177680   66.15   62.41   63.86
107 138748   65.85   61.54   63.21
108 105850   67.58   62.57   64.51
109   25751   68.50   62.84   65.04
110   48127   65.46   59.51   61.82
111   34278   69.61   62.71   65.39
112   36977   65.12   58.14   60.85
113   23001   67.94   60.13   63.16
115   13778   74.55   64.83   68.60
116    7994   72.21   62.25   66.12
117   20901   73.62   62.92   67.07
123.4  39629   79.28   64.28   70.10
---------------------------------------
row/row diff    0.60%   0.39%   0.00%


Start with the third column, which is the raw RC500. As you'd expect, the higher the park factor, the higher the unadjusted runs. That's the effect we want BPF to remove.

So, I adjusted by BPF, and that's column 4. We now expect everything to even out, and the column to look uniform. But it doesn't -- now, it goes the other way. Batters in pitchers' parks now look like they're better hitters than the batters in hitters' parks.

That shows that we overadjusted. By how much? 

Take a look at the bottom row of the chart. Unadjusted, each row is about 0.6 percent higher than the row above it. We'd expect about 1 percent, if BPF worked perfectly. Adjusted, each row is about 0.4 percent lower. 

So BPF overestimates the true park factor by around 40 percent. Which means, if we regress park factors to the mean by 40 percent, we should remove the bias.

That's what the last column is. And the numbers look pretty level.

Actually, I didn't use 40 percent ... I used 38.8 percent. That's what gave the best flat fit. (Part of the difference is due to rounding, and the rest due to the fact that I ignored nonlinearity when I calculated the percentages.)

Just to be more rigorous and get a more accurate estimate, I ran a regression. Instead of binning the players, I just used all players separately, and did a "weighted regression" that effectively adjusts for the number of PA associated with each bin. Because of the weights, I was able to drop the minimum from 300 PA to 10 PA. Also, I included a dummy variable for year, just in case there were a lot of pitchers' parks in 1987, or something.

The result came out almost exactly the same -- regress by 38.3 percent.

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Could we have calculated this mathematically just from raw park factors? Yes, I think so -- but not quite in the usual way. 

I'll show the usual way here, and save the rest for the next post. If you don't care about the math, you can just stop here.

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If we used the usual technique for figuring how much to regress, we'd use

SD^2(observed) = SD^2(true) + SD^2(luck)

We can figure luck. The SD of team runs in a game is about 3. For two teams combined, multiply by root 2. Calculating the different from a road game, multiply by root 2 again. Then, for 81 games, divide by the square root of 81, which is 9. Finally, because we're using 3 years, divide by the square root of 3. 

You get 0.385 runs. 

That figure, 0.385 runs, is 4.27 percent of the usual 9 runs per game. To convert that to a park factor, take half. That's 2.13 points. I'll round to 2.1.

The observed SD, from the Lahman database, is 4.8 points. 

We can now calculate SD(true), since 4.8^2 = SD^2(true) + 2.1^2. It works out to 4.3 points.

SD(observed)= 4.8
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    SD(true)= 4.3
    SD(luck)= 2.1

So, to regress observed to true, we regress the park factor by (2.1/4.8)^2, which is about 19 percent.

Why isn't it 38.3 percent?

Because, in this case, the "true" value is the three-year average. For that, regress 19 percent. But, that's not what we really want when it comes to a single year's performance. For that, we want SD(true) and SD(luck) for just that one year's park factor, not the average of the three years.

It makes sense you need to regress more for one year than three, because there's more randomness: the first 19 percent is for the randomness in the three-year average, and the next 19 percent is for the randomness in the fact that the other two of the three years the park might have been different, so the three-year BPF might not be representative of the year you're looking at.

There's no obvious relationship between the 19 percent and the 38.3 percent -- it's just coincidence that it comes out double. 

But I think I've worked out how we could have calculated the 38.3 figure. I'll write that up for Part II.  

(P.S.  You might have noticed that the last column of the chart was fairly level, except for the extreme hitters' parks.  I'll talk about that in part III.)


Update, 3/24/20: Part II is here.

Park adjusting a player's batting line has to depend on who the player is

Suppose home runs are scarce in the Astrodome, so that only half as many home runs are hit there than in any other park. One year, Astros outfielder Joe Slugger hits 15 HR at the Astrodome. How do you convert that to be park-neutral? 

It seems like you should adjust it to 30. Take the 15 HR, double it, and there you go. 

But I don't think that works. I think if you do it that way, you overestimate what Joe would have done in a normal park. I think you need to adjust Joe to something substantially less than 30.

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One reason is that Joe might not necessarily be hurt by the park as much as other players. Maybe the park hurts weaker hitters more, the kind who hit mostly 310-foot home runs. Maybe Joe is the kind who hits most of his fly balls 430 feet, so when the indoor dead air shortens them by 15 feet, they still have enough carry to make it over the fence.

It's almost certain that some players should have different park factors than others. Many parks are asymmetrical, so lefties and righties will hit to different outfield depths. Some parks may have more impact on players more who hit more line drive HRs, and less impact on towering fly balls. And so on.

I suspect that's actually a big issue, but I'm going to assume it away for now. I'll continue as if every player is affected by the park the same way, and I'll assume that Joe hit exactly 15 HR at home and 30 HR on the road, exactly in line with expectations.

Also, to keep things simple, two more assumptions. First I'll assume that the park factor is caused by distance to the outfield fence -- that the Astrodome is, say, 10 percent deeper than the average park. Second, I'll assume that in the alternative universe where Joe played in a different home park, he would have hit every ball with exactly the same trajectory and distance that he did in the Astrodome.

My argument is that with these assumptions, the Astros overall would have hit twice as many HR at home as they actually did. But Joe Slugger would have hit *fewer* than twice as many.

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Let's start by defining two classes of deep fly balls:

A: fly balls deep enough to be HR in any park, including the Astrodome; 
B: fly balls deep enough to be HR in any park *except* the Astrodome.

We know that, overall, class A is exactly equal in size to class B, since (A+B) is exactly twice A.

That's why, when we saw 15 HR in class A, we immediately assumed that implies 15 HR in class B. And so we assumed that Joe would have hit an extra 15 HR in any other park.

That seems like it should work, but it doesn't. Here's a series of analogies that shows why.

1. You have a pair of fair dice. You expect them to come up snake eyes (1-1) exactly as often as box cars (6-6). You roll the dice 360 times, and find that 1-1 came up 15 times. 

Since 6-6 comes up as often as 1-1, should you estimate that 6-6 also came up 15 times? You should not. Since the dice are fair, you expect 6-6 to have come up 1 time in 36, or 10 times.* The fact that 1-1 got lucky, and came up more often, doesn't mean that 6-6 must have come up more often.

(*Actually, you should expect that 6-6 came up only 9.86 times, since there are 5 fewer tosses left for 6-6 after taking out the successful 1-1s. But never mind for now.)

2. You have a pair of fair dice, and an APBA card. On that card, 1-1 is a home run, and 6-6 represents a home run anywhere except the Astrodome.

You roll the dice 360 times, and 1-1 comes up 15 times. Do you also expect that 6-6 came up 15 times? Same answer: you expect it came up only 10 times. The fact that 1-1 got lucky doesn't mean that 6-6 must also have gotten lucky.

3. You have a simulation game, with some number of fair dice, and a card for Joe Slugger. You know the probability of Joe hitting an Astrodome HR is equal to the probability of Joe hitting an "anywhere but Astrodome" HR.  But that probability -- Joe's talent level -- isn't necessarily 1 in 36.

You play a season's worth of Joe's home games, and he hit 15 HR. Can you assume that he also hit 15 "anywhere but Astrodome" HR? 

Well, in one special case, you can. If the 15 HR was Joe's actual expectation, based on his talent -- that is, his card -- then, yes, you can assume 15 near-HR. 

But, in all other cases, you can't. If Joe's 15 HR total was lucky, based on his talent, you should assume fewer than 15 near-HR. And if the 15 HR was unlucky, you should assume more than 15 near-HR.

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So I think you can't park adjust players via the standard method of multiplying their performance by their park factor. The park adjustment has to be based on their *expected* performance, not their observed performance.

Suppose Joe Slugger, at the beginning of the season, was projected by the Marcel system to hit 10 HR at home. That means that he was expected to hit 10 HR at the Astrodome, and 10 "almost HR" at the Astrodome.

Instead, he winds up hitting 15 HR there. But we still estimate that he hit only 10 "almost HR". So, instead of bumping his 15 HR total to 30, we bump it only to 25.

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I was surprised by this, that there's no way to convert the Astrodome to a normal park that doesn't require you to estimate the player's talent. 

But here's what surprised me even more, when I worked it out: you only need to know the player's talent when you're adjusting from a pitchers' park. When you adjust from a hitters' park, one formula works for everyone!

Let's take it the other way, and suppose that Fenway affords twice as many home runs as any other park. And, suppose Joe Slugger, now with the Red Sox, hits 40 at Fenway and 20 on the road.

How many would he have hit if none of his games were at Fenway?

Well, on average, half of his 40 HR would have been HR on the road. So, that's 20. End of calculation. 

It doesn't matter who the batter is, or what his talent is -- as long as we stick to the assumption that every player's expectation is twice as many HR at Fenway, the expectation is that half his Fenway HR would also have been HR on the road.

(In reality, it might have been more, or it might have been less, since the breakdown of the 40 HR is random, like 40 coin tosses. But the point is, it doesn't depend on the player.)

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If you're not convinced, here's a coin toss analogy that might make it clearer.

We ask MLB players to do a coin toss experiment. We give them a fair coin. We tell them, take your day of birth, multiply it by 10, toss the coin that many times, and count the heads. Then, toss the coin that many times again, but this time, count the number of tails.

For the Fenway analogy: heads are "Fenway only" HR. Tails are "any park" HR.

We ask each player to come back and tell us H+T, the total number of Fenway HR. We then try to estimate the heads, the number of "Fenway only" HR.

That's easy: we just assume half the number. Mathematically, the expectation for any player, no matter who, is that H will be half of (H+T). That's because no matter how lucky or unlucky he was, there's no reason to expect he was luckier in H than T, or vice-versa.

Now, for the Astrodome analogy. Heads are "Any park including Astrodome" HR. Tails are "other park only" HR.

We ask each player to come back and tell us only the number of heads, which is the the Astrodome HR total. We'll try to estimate tails, the non-Astrodome HR total.

Rob Picciolo comes back and says he got 15 heads. Naively, we might estimate that he also tossed 15 tails, since the probabilities are equal. But that would be wrong. Because, we would check Baseball Reference, and we would see that Picciolo was born on the 4th of the month, not the 3rd. Which means he actually had 40 tosses, not 30, and was unlucky in heads.

In his 40 tosses for tails, there's no reason to expect he'd have been similarly unlucky, so we estimate that Picciolo tossed 20 tails, not 15.

On the other hand, Barry Bonds comes back and says he got 130 heads. On average, players who toss 130 heads would also have averaged about 130 tails. But Barry Bonds was born on the 24th of July. We should estimate that he tossed only 120 tails, not 130.

For Fenway, when we know the total number of heads and tails, the player's birthday doesn't factor into our estimate of tails. For the Astrodome, when we know only the total number of heads, the player's birthday *does* factor into our estimate of tails.

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So, when Joe Slugger plays 81 games at the Astrodome and tosses 15 home run "heads," we can't just expect him to have also tossed 15 long fly ball "tails". We have to look up his home run talent "date of birth". If he was only born on the 2nd of the month, so that we'd have only expected him to hit 10 HR "heads" and 10 near-HR "tails" in the first place, then we estimate he'd have hit only 10 neutral-park HR, not 15. 

If we don't do that -- if we don't look at his "date of birth" talent and just double his actual Astrodome HR -- our estimates will be too high for players who were lucky, and too low for players who were unlucky. 

Obviously, players who were lucky will have higher totals. That means that if we park-adjust the numbers for the Astros every year, the players who have the best seasons will tend to be the ones we overadjust the most. In other words, when a player was both good and lucky, we're going to make his good seasons look great, his great seasons look spectacular, and his spectacular seasons look like insane outliers. When a player is bad and unlucky, his bad seasons will look even worse.

But if we park-adjust the Red Sox every year ... there's no such effect, and everything should work reasonably well.

My gut still doesn't want to believe it, but my brain thinks it's correct. 

Well, my gut *didn't* want to believe it, when I wrote that sentence originally. Now, I realize that the effect is pretty small. When a player gets lucky by, say, 20 runs, with a season park factor of 95 ... well, that's only 1 run total. My gut is more comfortable with a 1-run effect.

But, suppose you're adjusting a Met superstar, trying to figure out what he'd hit in Colorado. Runs are about 60 percent more abundant in Coors Field than Citi Field, which means the park factor is around 30 percent higher. If the player was 20 runs lucky in that particular season, you'd wind up overestimating him by 6 runs, which is now worth worrying about.




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(Note: After writing this, but before final edit, I discovered that Tom Tango made a similar argument years ago. His analysis dealt with the specific case where the player's observed performance matches his expectation, and for that instance I have reinvented his wheel, 15 years later.)