Ted Abernathy, Our Inadvertent Hero
It’s that time of year, again, when a young man’s fancy turns to baseball, and old men argue about Strat-O-Matic rules and details. The discussion came up on Hondo’s Strat Forum about Ted Abernathy, a right handed pitcher who gave up no home runs to left handed batters in 143 batters faced in the 1967 NL. The question is should he be able to “control” these left handed hitters on their cards, the same way batters are controlled on pitcher cards by the W and N designations. I’m generally against this, as I am most things contrary to the model - but let’s wait a bit.
My usual objections to these kinds of adjustments, which are by the way very, very popular with replayers, are three-fold.
A) Allowing 0 HRA is being treated as a Superpower, that is, an ability to change another player’s hitting, usually using twenty-sided dice, by denying him home runs (adjusting his card). My contention is in most cases this 0 HRA pitcher’s scenario is mere random variation, that is, in any population of pitchers there will be a sub population of pitchers of normal talent who allow 0 Home runs for a more extended period, because, while it is unusual for this to happen, it’s not impossible.
- The Example I gave was the 1978 AL, a league I know pretty well. Two pitchers in the 1978 AL gave up 0 home runs on more than 143 BFP, including Victor Cruz of Toronto and Don Stanhouse of Baltimore, who made it to 352 batters faced, which is a pretty good number. But there’s this part of me that also thinks that if you sent Jim Rice up there, he of the 400 total bases in 1978, that you couldn’t say, “Hey, you can’t homer off me, I’m Don Stanhouse.” That’s perhaps a bridge too far.
It’s imposing a deterministic view- “this result is zero in real life, so it has to be zero in the game”- on a game construct that’s meant to model possibilities. If you’re allowing Justin Verlander to face Babe Ruth in a game, there’s no real life data for that. It’s all a model.
I’m also of the opinion that a pitcher with 1 home run allowed in say 100-150 innings would be more of a unicorn than 0 HRA in 143 batters faced, so the use of a 0 home runs allowed standard is actually a bit arbitrary.
B) The game’s general model does not allow one group of outcomes to cross over and affect another group of outcomes. What I mean by this is in real life the combination of Ozzie Smith and Terry Pendleton (and to some degree Tommy Herr) was absolute poison to right handed batters who tried to pull pitches and hit them on the ground in St Louis in 1985. In some baseball models this would be reflected in their hitting capabilities and not just (as it is in Strat) on small sub segments on the pitcher’s cards. (I think Sherco actually did this with range and positioning).
To me Ozzie’s fielding actually was a superpower, but it’s bounded very carefully in our game. A 0 home runs allowed pitcher who is allowed to negate batter’s home runs is not bounded at all, he impacts even a sixty-two home run hitting 2024 Aaron Judge.
C) The entire model assigning cause and blame for hits and outs to pitchers, batters, and fielders, is just a construct, it did not come down from On High from the Burning Bush. I’m not saying it’s bad. But there are constraints and compromises in any such model, and it’s against my nature to assign any one batting or pitching scenario to have any more weight than it needs in any simulation.
So - let’s do just that. Hondo actually did something interesting - he posed the question a different way, a way that made me less argumentative. He asked under what conditions would such a 1-20 twenty sided dice roll (adjustment roll) might be applicable, and what should it be? That’s an entirely different question; it could be restated as “at what point, in terms of batters faced, does this scenario, 0 home runs allowed, become significant? How do we apply relevant data to this? “
Now that question has a statistical model, you can use reliability analysis or AQL lot size type “0 defects” models, where the whole point is to sample for acceptance or denial based on the possible presence of a single “defect”. A defect, here would be a home run, and what you’re testing for (or against) is the base frequency with which these were allowed.
Let’s look at the 1967 National League:
(Note: I’m using pitcher’s BFP and hitter’s PA more or less equivalently; I know they are not exactly equivalent. BFP includes AB, W, H and reached on errors. Plate appearances include official AB, walks, hit by pitch, sac bunts and flies, and teaching base on fielder’s choices. They’re close, close enough for this estimate.)
The Data: 1967 NL
PA. HR
Rhb versus rhp 24970 485
Rhb versus lhp. 14690. 276
Rhb: 39660. 762 0.019213
LHB versus rhp. 17474. 282
LHB vs lhp. 3746. 56
LHB: 21220. 338. 0.015928
At the batter level, right handed batters have more power in the 1967 NL.
Okay let’s look at this at the pitcher level:
1967 NL
PA. HR
Rhb versus rhp 24970 485
LHB versus rhp. 17474. 282. 0.016138*
Right handers 42444. 767. 0.018071
Rhb versus lhp. 14690. 276
LHB vs lhp. 3746. 56
Left handers 18436. 333. 0.018062
choices.
So now you have two choices:
A) level of confidence (90 percent, 95, 98, 99), in which case if you selected an outcome at that confidence you could be wrong on the cause Les and less frequently. 90 pct would be 1 in ten times, 99 pct would be one in 100.
B) which of the above instances are the right percentages. Since we are looking at a right handed pitcher versus left handed batters I’ll choose 0.016138
Okay now let’s calculate the zero defects sample size, or the number of plate appearances with 0 HRA where we start to look at significance. Note that it’s about 4.2 PA per IP as a rough estimate; this is easily calculated from the league’s summary data.
90 percent = ln(1-.90) / ln(1-0.016138) or 142 (dimensionless)
(we don’t have a specific IP for Abernathy versus just left handers but if we did, this is about 34 IP with 0 HRA)
95 percent = ln(1 - 0.95) / ln (1-0.016138) or 184
98 percent = ln(1/0.98) / ln (1-0.016138) or 240
99 percent = lb(1/0.99) / ln (1-0.016138) or 283.
As you look at higher significance the pitcher needs to face more and more batters in order for the 0 HRA to be significant at any given home run rate.
Now let’s see what we get when we use my favorite toy to convert these findings to look at possible 20 sided die results for each level of significance. The favorite toy is an estimator that is going to place the probability of the adjustment at 50 percent for the first point that is significant at 95 percent confidence, and it can estimate the surrounding data based on that midpoint:
At 90 percent confidence:
Opp BFP with 0 HRA :
50 : (142 from above /50) minus 0.5 = 2.33 times 20 = 47
47 is greater than 20, no adjustment to batter’s cards
This would be only 12 IP with no HR
100: ((142 /100) - 0.5) = 0.915 times 20 or 18.
You would use 1-18 as an adjustment at 90 percent confidence for a pitcher with 100 homerless BFP. Thus is about 24 homerless IP.
143: ((142/143) -0.5) = 0.489 times 20 or 10.
Ted Abernathy would be a 1-10 adjustment at 90 percent confidence
200: ((142/200) - 0.5) = 0.208 times 20 or 4
A pitcher with 200 BFP would use a 1-4 adjustment
A pitcher with 250 BFP with 0 HRA would be 0.066 times 20 or 1. The adjustment would be 1 in 20 on a 20 sided die.
Let’s try 95 percent confidence, often used as a standard
50: (184/50 - 0.5) = 3.18 times 20 equals 64
No adjustment
100: (184/100 - 0.5) = 1.34 times 20 equals 27
No adjustment
143: (184/143 - 0.5) = 0.788 times 20 equals 16
Ted Abernathy versus LHB would use a 1-16 adjustment at 95 percent confidence. Note the adjustment is higher (more favorable to a batter) at 95 percent confidence, but the chance there is another special cause would be half what it was at 90 percent.
200: (184/200 - 0.5) = 0.421 times 20 = 8
A pitcher with 200 BFP versus lhp with 0 HRA would be 1-8 for an adjustment to batters
250: (184/250 - 0.5) = 0.236 or 5; 1-5 on a 20 sided die.
Let’s try 98 pct confidence:
50: (240/50 -0.5) = 4.31 times 20 = 86 no adjustment
100: 1.90 times 20 = 38 no adjustment
143: 1.18 times 20 = 24 no adjustment
200: 0.702 times 20 = 14 1-14 on a 20 sided die
250: 0.461 times 20 = 9 1-9 on a 20 sided die.
Using this very high bar Abernathy would have no correction.
99 pct: (Abernathy only)
283/143 - 0.5 = 1.48 times 20 no adjustment
So which one should you use? I think this is up to the user, but an interesting result happens if we use the 184 figure- this is the minimum level where significance could be presumed under this model.
At 90 percent
142// 184 - 0.5 equals .269, times 20 equals 5 1-5 on a 20 sided die for a batter home run- this is a pretty strong offset. It reduces batters card home runs by 75 percent.
At 95 percent:
184/ 184 - 0.5 or .5 times 20 = 10. What this is saying is a significant result should mean a 1-10 adjustment, a 50 percent reduction. This is the reference model for the Favorite Toy, the first significant point is a 50 percent reduction or 1-10. But it’s up to the user what their comfort level is.
At 98 percent:
240/ 184 - 0.5 or 0.806 times 20 = 16. The first level at which significance is observed results in a mild offset of a 20 percent reduction of the batter’s numbers (1-16).
At 99 percent:
283/184 - 0.5 or 1.04 times 20 = 21. 184 homerless BFP provides no adjustment but a slightly higher number would.
If it was me, I would calculate the 98 percent numbers and use them to calculate my offsets. This way you can be reasonably sure the significance limits are met. For the Abernathy case it would be no adjustment. But I could also see using the 1-16 adjustment of the 95 percent confidence level calculation.
Review- Steps:
Calculate the HR/PA fur the scenario you wish to adjust;
Select significance, 90, 95, 98 or 99
Calculate the sample level (ln 1- significance) / (ln 1- hr/pa)
Calculate the 1-20 twenty sided die adjustment
((Sample level / BFP with 0 HRA) - 0.5) times 20
Fred Bobberts 3/12/2025
