Year of the No Hitter – What are the Odds?

Yesterday, Matt Garza threw the first no-hitter in Rays history, a gem in which he faced a minimum of 27 batters, the only running back eliminated on a double play. This game brings the total no-hitters thrown in 2010 to five, the most in a year since 1991, when there were seven no-hitters thrown. How rare is it to throw a no-hitter? There have been 268 no-hitters thrown in Major League history and there have been 393,396 games. Multiply the number of games by two to account for the fact that each game has two starters, then divide 268 by that sum (786,792) to get the probability that a starter will throw a no-hitter on any given day. We find this probability to be 0.000681247, or about 0.068%.

So far this season, 1,487 games have been played, which means that 2,974 (1,487 * 2) pitchers have started. To find the probability that a single no-hitter will be thrown this year, we must first find the expected number of no-hitters thrown this year. We can do this by multiplying the probability that a starter will throw a no-hitter by the number of games started.

0.000681247 * 2974 = 2.02603

To find the probability that a no-hitter will be thrown in this set of games, we use a mathematical formula called the Poisson Random Variable. Since a Poisson random variable has a mean y, the probability that the random variable is equal to x is given by the formula:
(yxe-yx) / x!

Excel (and Open Office) has a function (= FISH (x, y, True)) to find the probability that a variable with mean y assumes a value of x. Plugging our numbers into this equation, we find that there is an 86.866% chance that at least one no-hitter will be thrown this year.

To find the probability that 5 no-hitters will be thrown, we simply change the variable x from 1 to 5. If we plug in this new number, we find that the probability that 5 no-hitters will be thrown this year is 0.0377%; rare, but not out of the realm of possibility.