THE BOOK--Playing The Percentages In Baseball

Let’s say that we have two pitchers’ stats for a single season and they are exactly equal, so that we are all in agreement that they had exactly the same season, quality-wise.

And let’s say that we are comfortable adjusting each pitcher’s stats for the quality of their opponents and that we are all in agreement that that will be the tie-breaker.

Pitcher A has faced a collective offense that had an OPS of .750 against all teams during the season. 

Pitcher B faced a collective offense that had an OPS of .760 overall during the season.

We agree that we are not going to look at exactly what batters each pitcher faced.  Too much work.  We are only looking at the teams that each pitcher faced and using each of those team’s offensive OPS for the season to determine “strength of schedule” for each pitcher. 

So far, it appears that pitcher B faced tougher opponents, so he is crowned as the “best” pitcher between the two, for the season.

But…

What if we find out (either using regressions, individual player projections, or G-d comes down and tells us - it doesn’t matter how) that the true talent of pitcher A’s opponents was actually better, offensively, than that of pitcher B. In other words, the fact that pitcher A’s opponents had a .750 OPS this season and pitcher B’s, a .760 OPS, was just a fluke, as could easily be the case of course, as even an entire team OPS can vary quite a bit (from its “true talent” level) by chance in only one season, since, even though a team has 6,000 some odd PA in a year, those 6,000 PA are made up of 15 or so individuals with 100-600 PA each (or so).  That makes a big difference.

Anyway, so which pitcher, including a “strength of schedule (opponent)” adjustment, was the better pitcher during the season?

A somewhat (but not really) related question?  What are the chances that the team with the best record in baseball is the “best” team in baseball (true talent, going forward, etc.)?  Assuming that everyone’s true talent stays roughly the same throughout the entire season (no major injuries, trades, etc.).  Best record in one league (AL or NL)?

What are the chances that the team with the best record in baseball is the “best” team in baseball (true talent, going forward, etc.)?  Assuming that everyone’s true talent stays roughly the same throughout the entire season (no major injuries, trades, etc.).  Best record in one league (AL or NL)?

I’ll have the answer using a season sim, a typical spread of talent among teams, and a typical schedule.

I’d like to see some guesses though.  I guess I’ll also throw in how often the pennant winners are the best teams in each league and how often the WS winner is the best team in baseball.

I seem to remember having a thread on this.  I even created some sort of sim.  If I remember correctly, the true-talent best team would have something like a 10% or 15% chance of winning the World Series (if we presume that the talent level of teams is spread out normally, with 1 SD = .060 wins per game).  Let me see if I can find that thread…

I can’t seem to find the thread at the moment.  I did find this one:

http://www.insidethebook.com/ee/index.php/site/comments/will_a_rod_win_a_world_series/

which seems to to be a precursor to a longer thread on the subject.

Guesses:
Chances team with best record is the best team in baseball: 40%
Chances the team with the best record is the World Series winner: 19%

I feel like my first number is pretty high, but it just seems like certain teams (like the Yankees and Mariners in the late-90’s) are just SO much better than everyone else, that there’s a pretty much 100% certainty that they were “truly” the best team at the time.

This might be the thread you were looking for:

http://www.insidethebook.com/ee/index.php/site/comments/true_talent_levels_for_sports_leagues/

Mike, obviously if one or even two teams are well above the others in true talent, it changes everything, but for a typical season with the talent being some what normally distributed, your numbers are too high.

I have only run this year’s teams using my pre-season estimates of team talent.  The SD of talent for those estimates is .0412.  .06 seems a little high, but it could be right. .06 would imply of course that there was greater than a 50% chance that at least 1 team per season had a true wp of .620 or better (and at least one who was .380 or worse).  That seems very high.  I think that is a rare team. In my pre-season estimates, I had the Red Sox at .562 and the Mets at .561 - the best teams in each league.

Of course, we can estimate pretty closely the SD of true talent over the last X years by looking at the variance of records and subtracting the expected variance by chance in 162 games, which is probably around .0018 wins (using the binomial variance plus a little more for different “p’s” every game).

Anyway, if I change the team true talent wp to a distribution with a SD of .06 rather than .04 wins per game, it changes the numbers of course.  Changes them a lot actually.  Especially if one team in each league stands well above the others.

(BTW, does anyone know how to generate a random normal distribution, given a certain SD and number of elements, such as a 30 team league with a certain SD of true talent wp?)

So, I guess for now, I’ll base the question on these true talents, which were my estimates going into this season:

ari .541
atl .536
chn .533
cin .485
col .490
flo .469
hou .434
lan .542
mil .525
nyn .561
phi .535
pit .461
sdn .530
sln .495
sfn .419
was .447
ala .514
bal .442
bos .562
cha .480
cle .541
det .543
kca .446
min .467
nya .557
oak .499
sea .480
tba .524
tex .450
tor .503

BTW, as far as the other question - what to use to adjust for “strength of schedule” or “opponent” (true talent or actual performance against everyone during that time period), I don’t know the answer off the top of my head.  I think it is true talent, but I am not sure.

MGL, I’ve done what you are suggesting in the past, and posted in the blog.  It was something like since 1961 (or whenever the 162-game schedule started), in seasons where there were 162 games scheduled to play (i.e., exclude 1981, 1994, 1995, I think those are it), the standard deviation for actual win% was something like 1 SD = .072.

Since the binomial is .039, that leaves us with a true of .060.

***

Your team spread shows only .042.  We should expect something low, since your estimate of true talent must be closer to the mean for every player.  That is, if Pujols has a true talent of god-certified .450 wOBA, the best you can do is estimate it as say .430.  And for rookies, you are in a bigger predicament, being forced to make an estimate closer to the mean than you’d like otherwise.

As an obvious example, say that you only have 20 PA for every single hitter (and that’s all you have), and you are asked to estimate the true OBP of each hitter.  You will undoubtedbly get almost all the hitters to have a true OBP of .330 to .350, with 1 SD = .003 or some ridiculously low number.

I left a long comment that didn’t get through. Maybe it is in the spam filter?

Ahhhh, a “typical” season.  I assumed you meant a “random” season.

Tango, good point about my estimates being “contracted.”

If the observed SD is .072, that is a variance of .005184.  The binomial random variance is .039 if everyone had the same wp per game.  Given that they don’t, the random SD is a little higher.  Let’s call it .041.  That is a variance of .0017.  Subtract .0017 from .005184 and take the square root, and you get .059.

With a SD of talent of .042, the best team in each league wins the most games around 29% of the time.  The best team in baseball wins the most games around 18% of the time. Pennant 20% and WS 11%.

If the SD of talent is around .065 (probably higher than what it is), then the numbers are:

best team in each league wins most games: 50%
best team in BB wins most games: 35%
best team in BB wins WS: 15%

dcj: there is nothing in the spam filter. 

Unfortunately, sometimes it happens that a post gets eaten up at some point.  My only suggestion is to write in notepad before posting, so that you will have a saved copy until you see it here.

Luckily, I did save a copy of the post. Let’s see if it gets through this time.

--

This is a complicated question. Suppose the Rays end up winning the pennant. MGL estimated in April that they had a true talent of .524. Now that they have played a full season at a .605 level (based on component stats), maybe the best guess for the actual true talent is .545. Really we should have a probability distribution: Tampa Bay’s true talent is likely around .545, but there’s a 15% chance it is above .570, and a 15% chance it is below .520. Numbers for illustration purposes only.

Say that we end up modeling Tampa Bay’s true talent with a normal distribution, mean of .545, SD of .020. We can do the same thing for all the other AL teams. Technically, the distributions aren’t independent—certainly not if we insist that the league average be exactly .500—but I imagine it’s close enough.

Example: there are 5 teams.

True talent of team 1 is normally distributed, mean .550, SD .020.
Team 2: mean .530, SD .020.
Team 3: mean .520, SD .020.
Team 4: mean .470, SD .020.
Team 5: mean .430, SD .020.

Team 2 wins the pennant. Assume independence of the true talent variables.

We want to compare team 2 against the best of teams 1, 3, 4, and 5. Let F_1(x) be the cumulative distribution function for team 1. That means, F_1(x) is the probability that team 1’s talent is less than x. The formula is

F_1(x) = Phi((x - .550)/.020)

where Phi is the cumulative distribution function for the standard normal variable with mean 0 and SD 1. It is the function NORMSDIST in Excel, so we have

F_1(x) = NORMSDIST((x - .550)/.020) or
F_1(x) = NORMDIST(x, .550, .020, TRUE)

Similarly define the functions F_2 through F_5. The cumulative distribution function for the maximum of teams 1, 3, 4, and 5 is

G(x) = F_1(x) * F_3(x) * F_4(x) * F_5(x)

The max of teams 1, 3, 4, and 5 is less than x exactly when all four teams are less than x. So, the probability that the max is less than x is the product of the chances that each team is less than x (assuming independence). This is the reasoning behind the formula for G(x) above.

We’re interested in the chance that team 2 is better than the max of teams 1, 3, 4, and 5. Let f_2(x) be the probability density function for team 2, i.e. the standard normal curve with mean .530 and SD .020. In Excel,

f_2(x) = NORMDIST(x, .530, .020, FALSE)

Now the chance that team 2 is the best team is the integral from x=0 to 1 of f_2(x) * G(x) dx. Doing a numerical approximation, I get a 21% chance. The probabilities for all five teams are

Team 1: 69%
Team 2: 21%
Team 3: 10%
Teams 4 and 5: 0%

I think I see why my method is so much more complicated than MGL’s #11. If we want to know what is the frequency that the best team in baseball wins the pennant/WS, you all are right that the only important thing is the SD of true talent among teams. So, there is no need to go through the long process in my comment.

If the question is for a specific year, I believe the calculation I laid out is necessary. Take the 2006 Cardinals. What is the probability that they were the best team in baseball? Sure, they won only 83 games in a weak division, but they were expected to do much better beforehand. The probability is higher than you would guess just looking at the regular season record, and lower than you would guess using the preseason projections. It ends up being a difficult question.