Last night in the Cubs/Cardinals game, the Cardinals skipper took his starter, Lackey, out in the 8th inning of a 1-run game with one out, no one on base and lefty Chris Coghlan coming to the plate. Coghlan is mostly a platoon player. He has faced almost four times as many righties in his career than lefties. His career wOBA against righties is a respectable .342. Against lefties it is an anemic .288. I have him with a projected platoon split of 27 points, less than his actual splits, which is to be expected as platoon splits in general get heavily regressed toward the mean, because they tend to be laden with noise for two reasons: One, the samples are rarely large because you are comparing performance against righties to performance against lefties and the smaller of the two tends to dominate the effective sample size – in Coghlan’s case, he has faced only 540 lefties in his entire 7-year career, less than the number of PA a typical full-time batter gets in one season. Two, there is not much of a spread in platoon talent among both batters and pitchers. The less spread in talent for any statistic, the more the differences you see among players, especially in small samples, are noise. Sort of like DIPS for pitchers.
Anyway, even with a heavy regression, we think that Coghlan has a larger than average platoon split for a lefty and the average lefty split tends to be large. You typically would not want him facing a lefty in that situation. That is especially true when you have a very good and fairly powerful right-handed bat on the bench – Jorge Soler. Soler has a reverse career platoon split, but with only 114 PA versus lefties, that number is almost meaningless. I estimate his actual platoon split to be 23 points, a little less than the average righty. For RHB, there is always a heavy regression of actual platoon splits, regardless of the sample size (while the greater the sample of actual PA versus lefties, the less you regress, it might be a 95% regression for small samples and an 80% regression for large samples – either way, large) simply because there is not a very large spread of talent among RHB. If we look at the actual splits for all RHB over many, many PA, we see a narrow range of results. In fact, there is virtually no such thing as a RHB with true reverse platoon splits.
Soler seems to be the obvious choice, so of course that’s what Maddon did – he pinch hit for Coghlan with Soler, right? This is also a perfect opportunity since Matheny cannot counter with a RHP – Siegrest has to pitch to at least one batter after entering the game. Maddon let Coghlan hit and he was easily dispatched by Siegrest 4 pitches later. Not that the result has anything to do with the decision by Matheny or Maddon. It doesn’t. Matheny’s decision to bring in Siegrest at that point in time was rather curious too, if you think about it. Surely he must have assumed that Maddon would bring in a RH pinch hitter. So he had to decide whether to pitch Lackey against Coghlan or Siegrest against a right handed hitter, probably Soler. Plus, the next batter, Russell, is another righty. It looks like he got extraordinarily lucky when Maddon did what he did – or didn’t do – in letting Coghlan bat. But that’s not the whole story…
Siegrest may or may not be your ordinary left-handed pitcher. What if Siegrest actually has reverse splits? What if we expect him to pitch better against right handed batters and worse against left-handed batters? In that case, Coghlan might actually be the better choice than Soler even though he doesn’t often face lefty pitchers. When a pitcher has reverse splits – true reverse splits – we treat him exactly like a pitcher of the opposite hand. It would be exactly like Coghlan or Soler were facing a RHP. Or maybe Siegrest has no splits – i.e. RH and LH batters of equal overall talent perform about the same. Or very small platoon splits compared to the average left-hander? So maybe hitting Coghlan or Soler is a coin flip.
It might also have been correct for Matheny to bring in Siegrest no matter who he was going to face, simply because Lackey, who is arguably a good but not great pitcher, was about to face a good lefty hitter for the third time – not a great matchup. And if Siegrest does indeed have very small splits either positive or negative, or no splits at all, that is a perfect opportunity to bring him in, and not care whether Maddon leaves Coghlan in or pinch hits Soler. At the same time, if Maddon things that Siegrest has significant reverse splits, he can leave in Coghlan, and if he thinks that the lefty pitcher has somewhere around a neutral platoon split, he can still leave Coghlan in and save Soler for another pinch hit opportunity. Of course, if he thinks that Siegrest is like your typical lefty pitcher, with a 30 point platoon split, then using Coghlan is a big mistake.
So how do managers determine what a pitcher’s true or expected (the same thing) platoon split is? The typical troglodyte will use batting average against during the season in question. After all, that’s what you hear ad-nauseam from the talking heads on TV, most of them ex-players or even ex-managers. Even the slightly informed fan knows that batting average against for a pitcher is worthless stat in and of itself (what, walks don’t count, and a HR is the same as a single?), especially in light of DIPS. The slightly more informed fan also knows that one season splits for a batter or pitcher are not very useful for the reasons I explained above.
If you look at Siegrest’s BA against splits for 2015, you will see .163 versus RHB and .269 versus LHB. Cue the TV commentators: “Siegrest is much better against right-handed batters than left-handed ones.” Of course, is and was are very different things in this context and with respect to making decisions like Matheny and Maddon did. The other day David Price was a pretty mediocre to poor pitcher. He is a great pitcher and you would certainly be taking your life into your hands if you treated him like a mediocre to poor pitcher in the present. Kershaw was a poor pitcher in the playoffs…well, you get the idea. Of course, sometimes, was is very similar to is. It depends on what we are talking about and how long the was was, and what the was actually is.
Given that Matheny is not considered to be such an astute manager when it comes to data-driven decisions, it may be is surprising that he would bring in Siegrest to pitch to Coghlan knowing that Siegrest has an enormous reverse BA against split in 2015. Maybe he was just trying to bring in a fresh arm – Siegrest is a very good pitcher overall. He also knows that the lefty is going to have to pitch to the next batter, Russell, a RHB.
What about Maddon? Surely he knows better than to look at such a garbage stat for one season to inform a decision like that. Let’s use a much better stat like wOBA and look at Siegrest’s career rather than just one season. Granted, a pitcher’s true platoon splits may change from season to season as he changes his pitch repertoire, perhaps even arm angle, position on the rubber, etc. Given that, we can certainly give more weight to the current season if we like. For his career, Siegrest has a .304 wOBA against versus LHB and .257 versus RHB. Wait, let me double check that. That can’t be right. Yup, it’s right. He has a career reverse wOBA split of 47 points! All hail Joe Maddon for leaving Coghlan in to face essentially a RHP with large platoon splits! Maybe.
Remember how in the first few paragraphs I talked about how we have to regress actual platoon splits a lot for pitchers and batters, because we normally don’t have a huge sample and because there is not a great deal of spread among pitchers with respect to true platoon split talent? Also remember that what we, and Maddon and Matheny, are desperately trying to do is estimate Siegrest’s true, real-life honest-to-goodness platoon split in order to make the best decision we can regarding the batter/pitcher matchup. That estimate may or may not be the same as or even remotely similar to his actual platoon splits, even over his entire career. Those actual splits will surely help us in this estimate, but the was is often quite different than the is.
Let me digress a little and invoke the ole’ coin flipping analogy in order to explain how sample size and spread of talent come into play when it comes to estimating a true anything for a player – in this case platoon splits.
Note: If you want you can skip the “coins” section and go right to the “platoon” section.
Let’s say that we have a bunch of fair coins that we stole from our kid’s piggy bank. We know of course that each of them has a 50/50 chance of coming up head or tails in one flip – sort of like a pitcher with exactly even true platoon splits. If we flip a bunch of them 100 times, we know we’re going to get all kinds of results – 42% heads, 61% tails, etc. For the math inclined, if we flip enough coins the distribution of results will be a normal curve, with the mean and median at 50% and the standard deviation equal to the binomial standard deviation of 100 flips, which is 5%.
Based on the actual results of 100 flips of any of the coins, what would you estimate the true heads/tails percentage of that coin? If one coin came up 65/35 in favor of heads, what is your estimate for future flips? 50% of course. 90/10? 50%. What if we flipped a coin 1000 or even 5000 times and it came up 55% heads and 45% tails? Still 50%. If you don’t believe or understand that, stop reading and go back to whatever you were doing. You won’t understand the rest of this article. Sorry to be so blunt.
That’s like looking at a bunch of pitchers platoon stats and no matter what they are and over how many TBF, you conclude that the pitcher really has an even split and what you observed is just noise. Why is that? With the coins it is because we know beforehand that all the coins are fair (other than that one trick coin that your kid keeps for special occasions). We can say that there is no “spread in talent” among the coins and therefore regardless of the result of a number of flips and regardless of how many flips, we regress the result 100% of the way toward the mean of all the coins, 50%, in order to estimate the true percentage of any one coin.
But, there is a spread of talent among pitcher and batter platoon splits. At least we think there is. There is no reason why it has to be so. Even if it is true, we certainly can’t know off the top of our head how much of a spread there is. As it turns out, that is really important in terms of estimating true pitcher and batter splits. Let’s get back to the coins to see why that is. Let’s say that we don’t have 100% fair coins. Our sly kid put in his piggy bank a bunch of trick coins, but not really, really tricky. Most are still 50/50, but some are 48/52, 52/48, a few less are 45/55, and 1 or 2 are 40/60 and 60/40. We can say that there is now a spread of “true coin talent” but the spread is small. Most of the coins are still right around 50/50 and a few are more biased than that. If your kid were smart enough to put in a normal distribution of “coin talent,” even one with a small spread, the further away from 50/50, the fewer coins there are. Maybe half the coins are still fair coins, 20% are 48/52 or 52/48, and a very, very small percentage are 60/40 or 40/60. Now what happens if we flip a bunch of these coins?
If we flip them 100 times, we are still going to be all over the place, whether we happen to flip a true 50/50 coin or a true 48/52 coin. It will be hard to guess what kind of a true coin we flipped from the result of 100 flips. A 50/50 coin is almost as likely to come up 55 heads and 45 tails as a coin that is truly a 52/48 coin in favor of heads. That is intuitive, right?
This next part is really important. It’s called Bayesian inference, but you don’t need to worry about what it’s called or even how it technically works. It is true that if you flipped a coin and got 60/40 heads that that coin was much more likely to be a true 60/40 coin than it is to be a 50/50 coin. That should be obvious too. But here’s the catch. There are many, many more 50/50 coins in your kid’s piggy bank than there are 60/40. Your kid was smart enough to put in a normal distribution of trick coins.
So even though it seems like if you flipped a coin 100 times and got 60/40 heads, it is more likely you have a true 60/40 coin than a true 50/50 coin, it isn’t. It is much more likely that you have a 50/50 coin that got “heads lucky” than a true 60/40 coin that landed on the most likely result after 100 flips (60/40) because there are many more 50/50 coins in the bank than 60/40 coins – assuming a somewhat normal distribution with a small spread.
Here is the math: The chances of a 50/50 coin coming up exactly 60/40 is around .01. Chances of a true 60/40 coin coming up 60/40 is 8 times that amount, or .08. But, if there are 8 times as many 50/50 coins in your piggy bank as 60/40 coins, then the chances of your 60/40 coin being a fair coin or a 60/40 biased coin is only 50/50. If there 800 times more 50/50 coins than 60/40 coins in your bank, as there is likely to be if the spread of coin talent is small, then it is 100 times more likely that you have a true 50/50 coin than a true 60/40 coin even though the coin came up 60 heads in 100 flips.
It’s like the AIDS test contradiction. If you are a healthy, heterosexual, non-drug user, and you take an AIDS test which has a 1% false positive rate and you test positive, you are extremely unlikely to have AIDS. There are very few people with AIDS in your population so it is much more likely that you do not have AIDS and got a false positive (1 in 100) than you did have AIDS in the first place (maybe 1 in 100,000) and tested positive. Out of a million people in your demographic, if they all got tested, 10 will have AIDS and test positive (assuming a 0% false negative rate) and 999,990 will not have AIDS, but 10,000 of them (1 in 100) will have a false positive. So the odds you have AIDS is 10,000 to 10 or 1000 to 1 against.
In the coin example where the spread of coin talent is small and most coins are still at or near 50/50, pretty much no matter what we get when flipping a coin 100 times, we are going to conclude that there is a good chance that our coin is still around 50/50 because most of the coins are around 50/50 in true coin talent. However, there is some chance that the coin is biased, if we get an unusual result.
Now, it is awkward and not particularly useful to conclude something like, “There is a 60% chance that our coin is a true 50/50 coin, 20% it is a 55/45 coin, etc.” So what we usually do is combine all those probabilities and come up with a single number called a weighted mean.
If one coin comes up 60/40, our weighted mean estimate of its “true talent” may be 52%. If we come up with 55/45, it might be 51%. 30/70 might be 46%. Etc. That weighed mean is what we refer to as “an estimate of true talent” and is the crucial factor in making decisions based on what we think the talent of the coins/players are likely to be in the present and in the future.
Now what if the spread of coin talent were still small, as in the above example, but we flipped the coins 500 times each? Say we came up with 60/40 again in 500 flips. The chances of that happening with a 60/40 coin is 24,000 times more likely than if the coin were 50/50! So now we are much more certain that we have a true 60/40 coin even if we don’t have that many of them in our bank. In fact, if the standard deviation of our spread in coin talent were 3%, we would be about ½ certain that our coin was a true 50/50 coin and half certain it was a true 60/40 coin, and our weighted mean would be 55%.
There is a much easier way to do it. We have to do some math gyrations which I won’t go into that will enable us to figure out how much to regress our observed flip percentage to the mean flip percentage of all the coins, 50%. For 100 flips it was a large regression such that with a 60/40 result we might estimate a true flip talent of 52%, assuming a spread of coin talent of 3%. For 500 flips, we would regress less towards 50% to give us around 55% as our estimate of coin talent. Regressing toward a mean rather than doing the long-hand Bayesian inferences using all the possible true talent states assumes a normal distribution or close to one.
The point is that the sample size of the observed measurement is determines how much we regress the observed amount towards the mean. The larger the sample, the less we regress. One season observed splits and we regress a lot. Career observed splits that are 5 times that amount, like our 500 versus 100 flips, we regress less.
But sample size of the observed results is not the only thing that determines how much to regress. Remember if all our coins were fair and there were no spread in talent, we would regress 100% no matter how many flips we did with each coin.
So what if there were a large spread in talent in the piggy bank? Maybe a SD of 10 percent so that almost all of our coins were anywhere from 20/80 to 80/20 (in a normal distribution the rule of thumb is that almost of the values fall within 3 SD of the mean in either direction)? Now what if we flipped a coin 100 times and came up with 60 heads. Now there are lots more coins at true 60/40 and even some coins at 70/30 and 80/20. The chances that we have a truly biased coin when we get an unusual result is much greater than if the spread in coin talent were smaller, even in 100 flips.
So now we have the second rule. The first rule was that the number of trials is important in determining how much credence to give to an unusual result, i.e., how much to regress that result towards the mean, assuming that there is some spread in true talent. If there is no spread, then no matter how many trials our result is based on, and no matter how unusual our result, we still regress 100% toward the mean.
All trials whether they be coins or human behavior have random results around a mean that we can usually model as long as the mean is not 0 or 1. That is an important concept, BTW. Put it in your “things I should know” book. No one can control or influence that random distribution. A human being might change his mean from time to time but he cannot change or influence the randomness around that mean. There will always be randomness, and I mean true randomness, around that mean regardless of what we are measuring, as long as the mean is between 0 and 1, and there is more than 1 trial (in one trial you either succeed or fail of course). There is nothing that anyone can do to influence that fluctuation around the mean. Nothing.
The second rule is that the spread of talent also matters in terms of how much to regress the actual results toward the mean. The more the spread, the less we regress the results for a given sample size. What is more important? That’s not really a specific enough question, but a good answer is that if the spread is small, no matter how many trials the results are based on, within reason, we regress a lot. If the spread is large, it doesn’t take a whole lot of trials, again, within reason, in order to trust the results more and not regress them a lot towards the mean.
Let’s get back to platoon splits, now that you know almost everything about sample size, spread of talent, regression to mean, and watermelons. We know that how much to trust and regress results depends on their sample size and on the spread of true talent in the population with respect to that metric, be it coin flipping or platoon splits. Keep in mind that when we say trust the results, that it is not a binary thing, as in, “With this sample and this spread of talent, I believe the results – the 60/40 coin flips or the 50 point reverse splits, and with this sample and spread, I don’t believe them.” That’s not the way it works. You never believe the results. Ever. Unless you have enough time on your hands to wait for an infinite number of results and the underlying talent never changes.
What we mean by trust is literally how much to regress the results toward a mean. If we don’t trust the stats much, we regress a lot. If we trust them a lot, we regress a little. But. We. Always. Regress. It is possible to come up with a scenario where you might regress almost 100% or 0%, but in practice most regressions are in the 20% to 80% range, depending on sample size and spread of talent. That is just a very rough rule of thumb.
We generally know the sample size of the results we are looking at. With Siegrest (I almost forgot what started this whole thing) his career TBF is 604 TBF, but that’s not his sample size for platoon splits because platoon splits are based on the difference between facing lefties and righties. The real sample size for platoon splits is the harmonic mean of TBF versus lefties and righties. If you don’t know what that means don’t worry about it. A shortcut is to use the lesser of the two which is almost always TBF versus lefties, or in Siegrest’s case, 231. That’s not a lot, obviously, but we have two possible things going for Maddon, who played his cards like Siegrest was a true reverse split lefty pitcher. One, maybe the spread of platoon skill among lefty pitchers is large (it’s not), and two, he has a really odd observed split of 47 points in reverse. That’s like flipping a coin 100 times and getting 70 heads and 30 tails or 65/35. It is an unusual result. The question is, again, not binary – whether we believe that -47 point split or not. It is how much to regress it toward the mean of +29 – the average left-handed platoon split for MLB pitchers.
While the unusual nature of the observed result is not a factor in how much regressing to do, it does obviously come into play, in terms of our final estimate of true talent. Remember that the sample size and spread of talent in the underlying population, in this case, all lefty pitchers, maybe all lefty relievers if we want to get even more specific, is the only thing that determines how much we trust the observed results, i.e., how much we regress them toward the mean. If we regress -47 points 50% toward the mean of +29 points, we get quite a different answer than if we regress, say, an observed -10 split 50% towards the mean. In the former case, we get a true talent estimate of -9 points and in the latter we get +10. That’s a big difference. Are we “trusting” the -47 more than the -10 because it is so big? You can call it whatever you want, but the regression is the same assuming the sample size and spread of talent is the same.
The “regression”, by the way, if you haven’t figured it out yet, is simply the amount, in percent, we move the observed toward the mean. -47 points is 76 points “away” from the mean of +29 (the average platoon split for a LHP). 50% regression means to move it half way, or 38 points. If you move -47 points 38 points toward +29 points, you get -9 points, our estimate of Siegrest’s true platoon split if the correct regression is 50% given his 231 sample size and the spread of platoon talent among LH MLB pitchers. I’ll spoil the punch line. It is not even close to 50%. It’s a lot more.
How do we determine the spread of talent in a population, like platoon talent? That is actually easy but it requires some mathematical knowledge and understanding. Most of you will just have to trust me on this. There are two basic methods which are really the same thing and yield the same answer. One, we can take a sample of players, say 100 players who all had around the same number of opportunities (sample size), say, 300. That might be all full-time starting pitchers in one season and the 300 is the number of LHB faced. Or it might be all pitchers over several seasons who faced around 300 LHB. It doesn’t matter. Nor do the number of opportunities. They don’t even have to be the same for all pitchers. It is just easier to explain that way. Now we compute the variance in that group – stats 101. Then we compare that variance with the variance expected by chance – still stats 101.
Let’s take BA, for example. If we have a bunch of players with 400 AB each, what is the variance in BA among the players expected by chance? Easy. Binomial theorem. .000625 in BA. What if we observe a variance of twice that, or .00125? Where is the extra variance coming from? A tiny bit is coming from the different contexts that the player plays in, home/road, park, weather, opposing pitchers, etc. A tiny bit comes from his own day-to-day changes in true talent. We’ll ignore that. They really are small. We can of course estimate that too and throw it into the equation. Anyway, that extra variance, the .000625, is coming from the spread of talent. The square root of that is .025 or 25 points of BA, which would be one SD of talent in this example. I just made up the numbers, but that is probably close to accurate.
Now that we know the spread in talent for BA, which we get from this formula – observed variance = random variance + talent variance – we can now calculate the exact regression amount for any sample of observed batting average or whatever metric we are looking at. It’s the ratio of random variance to total variance. Remember we need only 2 things and 2 things only to be able to estimate true talent with respect to any metric, like platoon splits: spread of talent and sample size of the observed results. That gives us the regression amount. From that we merely move the observed result toward the mean by that amount, like I did above with Siegrest’s -47 points and the mean of +29 for a league-average LHP.
The second way, which is actually more handy, is to run a regression of player results from one time period to another. We normally do year-to-year but it can be odd days to even, odd PA to even PA, etc. Or an intra-class correlation (ICC) which is essentially the same thing but it correlates every PA (or whatever the opportunity is) to every other PA within a sample. When we do that, we either use the same sample size for every player, like we did in the first method, or we can use different sample sizes and then take the harmonic mean of all of them as our average sample size.
This second method yields a more intuitive and immediately useful answer, even though they both end up with the same result. This actually gives you the exact amount to regress for that sample size (the average of the group in your regression). In our BA example, if the average sample size of all the players were 500 and we got a year-to-year (or whatever time period) correlation of .4, that would mean that for BA, the correct amount of regression for a sample size of 500 is 60% (1 minus the correlation coefficient or “r”). So if a player bats .300 in 500 AB and the league average is .250 and we know nothing else about him, we estimate his true BA to be (.300 – .250) * .4 + .250 or .270. We move his observed BA 60% towards the mean of .250. We can easily with a little more math calculate the amount of regression for any sample size.
Using method #1 tells us precisely what the spread in talent is. Method 2 tells us that implicitly by looking at the correlation coefficient and the sample size. With either method, we get the amount to regress for any given sample size.
Let’s look at some year-to-year correlations for a 500 “opportunity” (PA, BA, etc.) sample for some common metrics. Since we are using the same sample size for each, the correlation tells us the relative spreads in talent for each of these metrics. The higher the correlation for any given sample, the higher the spread in talent (there are other factors that slightly affect the correlation other than spread of talent for any given sample size but we can safely ignore them).
Pitcher ERA: .240
BABIP for pitchers (DIPS): .155
BABIP for batters: .450
Now let’s look at platoon splits:
This is for an average of 200 TBF versus a LHP, so the sample size is smaller than the ones above.
Platoon wOBA differential for pitchers (200 BF v. LHB): .135
Platoon wOBA differential for batters (200 BF v. LHP): .180
Those numbers are telling us that, like DIPS, the spread of talent among batters and pitchers with respect to platoon splits is very small. You all know now that this, along with sample size, tells us how much to regress an observed split like Siegrest’s -47 points. Yes, a reverse split of 47 points is a lot, but that has nothing to do with how much to regress it in order to estimate Siegrist’s true platoon split. The fact that -47 points is very far from the average left-handed pitcher’s +29 points means that it will take a lot of regression to moved it into the plus zone, but the -47 points in and of itself does not mean that we “trust it more.” If the regression were 99% then whether the observed were -47 or +10, we would arrive at nearly the same answer. Don’t confuse the regression with the observed result. One has nothing to do with the other. And don’t think in terms of “trusting” the observed result or not. Regress the result and that’s your answer. If you arrive at answer X it makes no difference whether your starting point, the observed result, was B, or C. None whatsoever. That is a very important point. I don’t know how many times I have heard, “But he had a 47 point reverse split in his entire career!” You can’t possibly be saying that you estimate his real split to be +10 or +12 or whatever it is.” Yes, that’s exactly what I’m saying. A +10 estimated split is exactly the same whether the observed split were -47 or +5. The estimate using the regression amount is the only thing that counts.
What about the certainty of the result? The certainty of the estimate depends mostly on the sample size of the observed results. If we never saw a player hit before and we estimate that he is a .250 hitter we are surely less certain than if we have a hitter who has hit .250 over 5000 AB. But does that change the estimate? No. The certainty due to the sample size was already included in the estimate. The higher the certainty the less we regressed the observed results. So once we have the estimate we don’t revise that again because of the uncertainty. We already included that in the estimate!
And what about the practical importance of the certainty in terms of using that estimate to make decisions? Does it matter whether we are 100% or 90% sure that Siegrest is a +10 true platoon split pitcher? Or whether we are only 20% sure – he might actually have a higher platoon split or a lower one? Remember the +10 is a weighted mean which means that it is in the middle of our error bars. The answer to that is, “No, no and no!” Every decision that a manager makes on the field is or should be based on weighted mean estimates of various player talents. The certainty or distribution rarely should come into play. Basically the noise in the result of a sample of 1 is so large that it doesn’t matter at all what the uncertainty level of your estimates are.
So what do we estimate Siegrest’s true platoon split, given a 47 point reverse split in 231 TBF versus LHB. Using no weighting for more recent results, we regress his observed splits 1 minus 230/1255, or .82 (82%) towards the league average for lefty pitchers, which is around 29 points for a LHP. 82% of 76 points is 62 points. So we regress his -47 points 62 points in the plus direction which gives us an estimate of +15 points in true platoon split. That is half the split of an average LHP, but it is plus nonetheless.
That means that a left-handed hitter like Coghlan will hit better than he normally does against a left-handed pitcher. However, Coghlan has a larger than average estimated split, so that cancels out Siegrest’s smaller than average split to some extent. That also means that Soler or another righty will not hit as well against Siegrest as he would against a LH pitcher with average splits. And since Soler himself has a slightly smaller platoon split than the average RHB, his edge against Siegrest is small.
We also have another method for better estimating true platoon splits for pitchers which can be used to enhance the method we use using sample results, sample size, and means. It is very valuable. We have a pretty good idea as to what causes one pitcher to have a smaller or greater platoon split than another. It’s not like pitchers deliberately throw better or harder to one side or the other or that RH or LH batters scare or distract them. Pitcher platoon splits mostly come from two things: One is arm angle. If you’ve ever played or watched baseball that should be obvious to you. The more a pitcher comes from the side, the tougher he is on same-side batters and the larger his platoon split. That is probably the number one factor in these splits. It is almost impossible for a side-armer not to have large splits.
What about Siegrest? His arm angle is estimated by Jared Cross of Steamer, using pitch f/x data, at 48 degrees. That is about a ¾ arm angle. That strongly suggests that he does not have true reverse splits and it certainly enables us to be more confident that he is plus in the platoon split department.
The other thing that informs us very well about likely splits is pitch repertoire. Each pitch has its own platoon profile. For example, pitches with the largest splits are sliders and sinkers and those with the lowest or even reverse are the curve (this surprises most people), splitter, and change.
In fact, Jared (Steamer) has come up with a very good regression formula which estimates platoon split from pitch repertoire and arm angle only. This formula can be used by itself for estimating true platoon splits. Or it can be used to establish the mean towards which the actual splits should be regressed. If you use the latter method the regression percentage is much higher than if you don’t. It’s like adding a lot more 50/50 coins to that piggy bank.
If we plug Siegrest’s 2015 numbers into that regression equation, we get an estimated platoon from arm angle and pitch repertoire of 14 points, which is less than the average lefty even with the 48 degree arm angle. That is mostly because he uses around 18% change ups this year. Prior to this season, when he didn’t use the change up that often, we would probably have estimated a much higher true split.
So now rather than regressing towards just an average lefty with a 29 point platoon split, we can regress his -47 points to a more accurate mean of 14 points. But, the more you isolate your population mean, the more you have to regress for any given sample size, because you are reducing the spread of talent in that more specific population. So rather than 82%, we have to regress something line 92%. That brings -47 to +9 points.
So now we are down to a left-handed pitcher with an even smaller platoon split. That probably makes Maddon’s decision somewhat of a toss-up.
His big mistake in that same game was not pinch-hitting for Lester and Ross in the 6th. That was indefensible in my opinion. Maybe he didn’t want to piss off Lester, his teammates, and possibly the fan base.Who knows?