As a life-long Cincinnati Reds fan, I was very pleased that Homer Bailey accomplished the feat of pitching a no-hitter last night. He was also responsible for the last one, on Sep. 28 of 2012 against the Pirates (See list here). Much has been made of the fact that this is the second no-hitter of his career, and even he said:
Apparently it is much less likely to be a fluke, or simply good luck, to throw two no-hits games, which could be the case with just one. In fact, only 30 pitchers have more than one to their name. To record a no-hitter, the pitcher must record 27 outs without conceding a hit. Opposing players can still reach base via a Walk or Error.
Notice how Homer only gives up a single walk, which actually helped preserve the no-hitter.
In this case, since batters are hitting .241 against Bailey, the chance that each at-bat (which excludes walks) ends with an out (or error) is p= 1-0.241 = 0.759. Since the game had only one walk and no errors, let’s simplify the situation to make it a series of Bernoulli trials, in which success entails retiring a batter, which occurs 75.9% of the time, and failure is the concession of a hit. How likely is it that 27 outs will be recorded without a single hit? The probably is just (0.759)^27 = 0.00058415479, or about 1 in 1700.
Since there are 2430 major league games every season, and both pitchers have a chance to record a no-no, it is expected that at least one will be thrown every year. This, although an amazing accomplishment, there is still the sense that “luck” played a large role. This is certainly true, since a ball hit in just a slightly different manner, or a great defensive play, are all that separates a no-hitter from a more prosaic game.
So how likely is a no hitter? Not very. According to Bill James, only one pitcher in history was “expected” to throw more than one in his career, given his rate of allowing hits and total starts: Nolan Ryan.