Alex Cobb makes strikeout history twice in unique outing
Alex Cobb struck out all four batters he faced in the third inning Friday night, yet still allowed a run that inning. He also struck out 13 hitters in less than five innings. Both stats were the first of their kind.
First, check out the play-by-play for the third inning Cobb had against the Padres:
Notice anything? Cobb faced four batters, struck out all four out and also allowed a run. After combing through the gamelogs of every instance in baseball history where a pitcher struck out four batters in an inning (thanks Baseball-Reference.com, again), we can report that no player post-1916 accomplished such a feat. Inning-by-inning gamelogs aren't available prior to 1916, so there are five games in that period where it could have happened -- but, really, what are the odds?
There were several times where a pitcher struck out four in an inning and also allowed at least one run -- it's just that those pitchers faced more than four batters in that particular inning. There were several times where a pitcher struck out all four batters he faced in an inning -- it's just that none of those allowed a run. There were zero times prior to Friday night where both, striking out all four batters one faced and allowing a run, happened. So Cobb made history there, at least post-1916 history, as far as we know.
But that wasn't all.
Cobb lasted only 4 2/3 innings but struck out a whopping 13 hitters.
Thanks to Baseball-Reference.com's play index, we can pass along that no player in baseball history -- again, post-1916 -- has ever struck out at least 13 hitters in less than five innings of work. Zack Greinke struck out 13 in five innings last September, but that was five full innings.
So Cobb made history twice in one night.
What an amazing stat line.
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