 Felix Hernandex, Seattle Mariners Pitcher and the last pitcher to throw a perfect game (as of 2014)

The perfect game: the greatest feat a baseball pitcher can accomplish. Over a quarter of a million MLB games have been played and yet only 23 perfect games have ever been thrown. As a statistics aficionado, I would like to model the probability that a great pitcher throws a perfect game in a single season.

Throwing a perfect game means allowing no walks or hits over the entire game. Conveniently, there is a baseball statistic that works almost perfectly. Walks and Hits per Inning Pitched, a.k.a. WHIP, can be simply multiplied by 9 to obtain the average number of walks and hits allowed by a pitcher per game. The average number of baserunners a pitcher allows can be modeled by the Poisson Distribution with $$\lambda = 9$$ multiplied by the pitcher’s WHIP. The Poisson Distribution says that with an average number of occurrences, $$\lambda$$, the probability of x successes (in this case walks & hits) in the interval is equal to

\begin{align} \frac{\lambda^x * e^{-y}}{x!} \end{align}

We need to find the probability that the pitcher allows no baserunners.

Earlier I said I would find the probability that a great pitcher throws a perfect game in a single season. Well, what exactly is a “great pitcher?” In 2014, Clayton Kershaw had an incredible year with a WHIP of only 0.86. That is the sort of thing you see from a great pitcher once in their career. We’re talking Bob Gibson or Sandy Koufax during their best season kind of stuff. For this calculation, let’s say our “great pitcher” has a WHIP of 0.85. Taking that WHIP of $$0.85 * 9$$ gives us a $$\lambda$$ of $$7.65$$. Accordingly, the probability of this “great pitcher” allowing zero baserunners in any given game comes out to $$(7.65)^0 * e^-7.65/0! = 0.000476044$$. Wow! So the probability of a great pitcher allowing no batters to reach through 9 innings pitched is only about 0.05%. Clayton Kershaw, Los Angeles Dodgers Pitcher, Winner of the 2014 Cy Young Award and 2014 NL MVP

Okay, well what’s the probability that this “great pitcher” throws a perfect game over the span of a season then? To model that probability, we use the Binomial Distribution with a probability of success, p, equal to $$0.000476044$$. What about the number of trials? Well, in 2014, ten pitchers tied for most games started with 34, so let’s choose $$n=34$$. We want to find the probability that one or more perfect games are thrown which is equal to 1 minus the compliment (no perfect games thrown). Hence the probability of 1 or more perfect games being thrown is $$1-34!/(0!*34!) * 0.000476044^0 * (1- 0.000476044)^{34}$$. If you throw all that into your calculator, you should get $$0.01606$$.

Therefore, the probability that a great pitcher throws a perfect game in his best season is only around 1.6% – a fine accomplishment indeed.

Just to further prove the point that perfect games are a statistical wonder, let’s calculate the probability of this same great pitcher allowing just one base runner. The probability of allowing one base runner through nine innings is $$(7.65)^1 * e^-7.65/1! = 0.00364174$$. That looks pretty small (it’s still a pretty awesome feat). Small is relative though. $$0.00364174 / 0.000476044 = 7.65$$. In other words, it’s almost 8 times as likely for this great pitcher to throw a complete game allowing just one baserunner as it is for him to throw a perfect game! The probability allowing just one baserunner during at least one game of the season is 11.67%.

Time for the disclaimer and a fair warning for you fellow stats lovers; those probabilities are not necessarily accurate. In order to make use of these statistical models, we make some assumptions that aren’t necessarily valid. Basically, we assume the data set is “nice.” We assume it’s random, independent, and normally distributed. That may not be true. For example, if a pitcher walks two straight batters in a crucial situation, he may get a little flustered. That could then lead to him leaving a pitch out over the plate or walking another batter. In other words, baserunners are probably not independent of each other. Either way, this calculation probably provides a somewhat fair approximation and it certainly emphasizes the true difficulty in throwing a perfect game.