Was it good small-ball strategy? Was it horribly giving away a valuable out? Let's figure it out!
The logic behind having Alex Presley bunt is that it moves the tying run into scoring position. And, per baseball conventional wisdom, "at home you play for the tie."
But what is the likelihood of the Pirates tying the game with Chase d'Arnaud at the plate, runners on second and third, and one out?
d'Arnaud is hitting .225. However, a single doesn't automatically tie the game, because Matt Diaz isn't exactly a speedster - in his career, he's scored from second on a single only 56% of the time. Furthermore, 7 of d'Arnaud's singles were infield singles, on which Diaz will not score. So the chance of d'Arnaud tying the game is the chance of him hitting a single to the outfield times 56%, plus the chance of d'Arnaud hitting a double or triple. Seeing as how d'Arnaud has never homered in the majors, and homered only 18 times in 1567 minor league PAs, I'm disregarding that possibility.
d'Arnaud has 15 singles, 4 doubles, and 1 triple in 94 PAs. 7 were infield singles and one a bunt single, so 50% of d'Arnaud's non-bunt singles were infield singles on which Diaz will not score, and 50% were outfield singles on which he will score 56% of the time.
15/94 * .5 * .56 = 4.47% chance of d'Arnaud singling and Diaz scoring.
5/94 = 5.32% chance of d'Arnaud doubling or tripling (on which we assume Diaz scores).
So after a successful sacrifice bunt, there's about a 9.8% chance of d'Arnaud tying the game. Is a 10% chance worth giving up an out by taking the bat out of your hottest hitter's hands? You be the judge.
If Presley doesn't bunt, there are a range of possible outcomes:
- He could homer or triple, or could hit a double on which Diaz scores from first (which historically he's done 44% of the time), tying the game or putting the Pirates into the lead. With one home run, three triples, and four doubles in 88 ML PAs, the probability is (1/88) + (3/88) + (4/88) * .44 = 6.55% chance of Presley tying the game or putting the Pirates into the lead.
- He could double and Diaz not score, putting runners on second and third with no one out and the Pirates down 2-1: (4/88) * (1-.44) = 2.55%.
- He could single or walk, loading the bases with no one out. (I'm assuming that McKenry doesn't score from second on a single.). With 19 singles and 8 walks, the probability is 27/88 = 30.7%.
- He could ground into a double play. He's never done so in the majors, but doing some calculations to estimate DP opportunities in his minor league career, I estimated that he's got maybe a 5.9% chance of doing so.
- He could make a "productive out", putting runners at second and third with one out; I calculated that probability at about 32.6%.
- He could make an unproductive out, leaving runners at first and second with one out. I calcuated that probability at about 21.7%.
So basically, by not having Presley bunt, you have about a 40% chance of an outcome better than a successful sacrifice bunt. You have about a 33% chance of an outcome the same as a successful sac bunt, and about a 22% chance of an outcome the same as an unsuccessful sac bunt. And you have about a 6% chance of a worse outcome than an unsuccessful sac bunt.
To me, the upside outweighs the downside. Again, you be the judge.