## Matt Garza, Fifth No-Hitter of 2010July 26, 2010

Posted by tomflesher in Baseball.
Tags: , , , , , ,
1 comment so far

Tonight, Matt Garza pitched the fifth no-hitter of 2010. He joins Edwin Jackson, Roy Halladay, Dallas Braden, and Ubaldo Jimenez in the Year of the Pitcher club.

As I pointed out when Jackson hit his no-hitter, no-hit games are probably Poisson distributed. Let’s update the chart.

The Poisson distribution has probability density function

$f(n; \lambda)=\frac{\lambda^n e^{-\lambda}}{n!}$

Maintaining our prior rate of 2.45 no-hitters per season, that means $\lambda = 2.45$. Our function is then

$f(n; \lambda = 2.5)=\frac{2.45^n (.0864)}{n!}$

The probabilities remain the same:

 n p cumulative 0 0.0863 0.0863 1 0.2114 0.2977 2 0.2590 0.5567 3 0.2115 0.7683 4 0.1296 0.8978 5 0.0635 0.9613 6 0.0259 0.9872 7 0.0091 0.9963 8 0.0028 0.9991 9 0.0008 0.9998 10 0.0002 1.0000

And though the expectation (E(49)) and cumulative expectation (C(49)) remain the same, the observed values shift slightly:

 E(49) Observed C(49) Total 4.23 5 4.23 5 10.36 11 14.59 16 12.69 8 27.28 24 10.36 17 37.65 41 6.35 1 43.99 42 3.11 5 47.10 47 1.27 1 48.37 48 0.44 0 48.82 48 0.14 1 48.95 49 0.04 0 48.99 49 0.01 0 49.00 49

The tailing observations (say, for 4+ no-hitters) don’t quite match the expected frequencies, but the cumulative values match quite nicely. There might be some unobserved variables that explain the weirdness in the upper tail. Still, cumulatively, we have 47 seasons with 5 or fewer no-hitters, which is almost exactly what’s expected. This is unusual, but not outside the realm of statistical expectation.

## Edwin Jackson, Fourth No-Hitter of 2010June 25, 2010

Posted by tomflesher in Baseball, Economics.
Tags: , , , , , , , , ,

Tonight, Edwin Jackson of the Arizona Diamondbacks pitched a no-hitter against the Tampa Bay Rays. That’s the fourth no-hitter of this year, following Ubaldo Jimenez and the perfect games by Dallas Braden and Roy Halladay.

Two questions come to mind immediately:

1. How likely is a season with 4 no-hitters?
2. Does this mean we’re on pace for a lot more?

The second question is pretty easy to dispense with. Taking a look at the list of all no-hitters (which interestingly enough includes several losses), it’s hard to predict a pattern. No-hitters aren’t uniformly distributed over time, so saying that we’ve had 4 no-hitters in x games doesn’t tell us anything meaningful about a pace.

The first is a bit more interesting. I’m interested in the frequency of no-hitters, so I’m going to take a look at the list of frequencies here and take a page from Martin over at BayesBall in using the Poisson distribution to figure out whether this is something we can expect.

The Poisson distribution takes the form

$f(n; \lambda)=\frac{\lambda^n e^{-\lambda}}{n!}$

where $\lambda$ is the expected number of occurrences and we want to know how likely it would be to have $n$ occurrences based on that.

Using Martin’s numbers – 201506 opportunities for no-hitters and an average of 4112 games per season from 1961 to 2009 – I looked at the number of no-hitters since 1961 (120) and determined that an average season should return about 2.44876 no-hitters. That means

$\lambda = 2.44876$

and

$f(n; \lambda = 2.44876)=\frac{2.44876^n (.0864)}{n!}$

Above is the distribution. p is the probability of exactly n no-hitters being thrown in a single season of 4112 games; cdf is the cumulative probability, or the probability of n or fewer no-hitters; p49 is the predicted number of seasons out of 49 (1961-2009) that we would expect to have n no-hitters; obs is the observed number of seasons with n no-hitters; cp49 is the predicted number of seasons with n or fewer no-hitters; and cobs is the observed number of seasons with n or fewer no-hitters.

It’s clear that 4 or even 5 no-hitters is a perfectly reasonable number to expect.

 2.44876

## NL Cy Young: Heating up earlyMay 31, 2010

Posted by tomflesher in Baseball.
Tags: , , , , , ,

There’s considerable debate, following Roy Halladay‘s perfect game, as to whether he or Ubaldo Jimenez should be considered the top contender for the National League’s Cy Young Award. Of course, it’s way too early to make those sorts of decisions, but let’s take a look at some of the data quickly.

Jimenez is sitting at 3.7 Wins Above Replacement and 38 Runs Above Replacement in 10 starts:

Year Age Tm Lg IP GS R Rrep Rdef aLI RAR WAR Salary
2010 26 COL NL 71.1 10 7 45 0 1.0 38 3.7 $1,250,000 5 Seasons 577.2 93 241 362 0 1.0 121 12.2$2,392,000
Provided by Baseball-Reference.com: View Original Table
Generated 5/31/2010.

Halladay has considerably less, with 22 RAR and 2.4 WAR:

Year Age Tm Lg IP GS R Rrep Rdef aLI RAR WAR Salary
2010 33 PHI NL 86.0 11 23 45 3 1.0 22 2.4 $15,750,000 13 Seasons 2132.2 298 893 1407 19 1.0 514 49.8$88,991,666
Provided by Baseball-Reference.com: View Original Table
Generated 5/31/2010.

Of course, 10 or 11 starts is far too small a sample to draw conclusions from this early in the season. Halladay has a perfect game; Jimenez has a no-hitter. Still, there’s no reason to believe that a perfect game, in and of itself, is enough to get Doc a Cy Young Award. After all, Mark Buehrle didn’t win the Cy last year, and Dallas Braden isn’t even in contention.

If both players keep pitching at or near this level, Halladay becomes a realistic contender, because at that point his marginal contribution may make the difference between whether the Phillies make the playoffs or not. As it stands right now, the NL East is entirely too volatile to make that decision.

(Incidentally, I love Baseball-Reference.com’s new stat sharing and player link tools!)

## Roy Halladay's Perfect GameMay 30, 2010

Posted by tomflesher in Baseball.
Tags: , , , , ,

Just what the Doctor ordered.

Andy at Baseball-Reference.com has an interesting blog entry about Doc’s perfect game. Roy Halladay was 0-3 in the game with two strikeouts, threw 115 pitches to 27 batters, and had a 98 Game Score.

Compared to Dallas Braden, Doc was much, much more likely to achieve this. Halladay’s opposing OBP is a miniscule $.297$ career, $.258$ this year, with his complementary probabilities of getting a batter out at $.703$ and $.742$. Using his career numbers, his probability of getting 27 consecutive batters out would be $.703^{27}$, or $.0000738$, which is approximately $7/100000$.

Interestingly, the last 3 perfect games have all had Florida teams as the victim.

## Quickie: Dallas Braden's Perfect GameMay 11, 2010

Posted by tomflesher in Baseball.
Tags: , , , , , , ,