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Monday, May 23, 2011
Bunting with Granderson
Larry offers many of the considerations, quotes The Book, and it takes a while to get there, but he also brought up game theory:
We should also consider that it’s not always good strategy to have strong hitters avoid the bunt, particularly when the strong hitter is fast and a good bunter. According to “The Book”, strong hitters make better bunters because the defense expects weak hitters to bunt and strong hitters to swing away. When a good hitter is at the plate, on average a bunt attempt can produce nearly as high a run expectancy as swinging away, providing that the defense is not expecting a bunt. I do NOT think the Mets were expecting Granderson to bunt yesterday!
I think he does a good job overall in forcing the reader to think, to accept that there is a debate, and it’s not open and shut.
I would like to see the following: where do the fielders field when Curtis Granderson is up to bat compared to say Paul Konerko or Billy Butler. Game theory would dictate the fielders to play in a little bit more with Granderson than with the other two guys, because of the possibility of the bunt (and that Granderson is also faster too… a better test case than Granderson would be a hitter that is the same speed as Konerko/Butler, but bunts more often than those two guys).
You can look at an extreme situation, like the shift. The fielders are practically telling the batter that if he bunts down the third base line, he gets a free base (it’s the equivalent of an intentional walk… the intentional hit). Given that kind of setup, it’s surprising (disappointing) how little the shifted hitters bunt. So, if they bunt too little on a massive shift, I can see how teams are not going to play in more on good hitters who have shown bunt on occasion.
***
How successful of a hitter does he need to be if shifted with bases empty? If he’s a good hitter, say +.05 runs per PA, and the run value of an out is about -.18 runs and a single/walk is +.27 runs, then we solve for this:
+.05 = .27p - .18(1-p)
Solving for p gives us
p = (.05 + .18) / (.27 + .18) = 51%
So, pretty much, the batter needs to bunt successfully 50% of the time, in order for the shift to be a breakeven point for him. Of course, it might be a hard sell to get Ortiz to practice his bunting.
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