Per BBREF:
John Kruk: 170 pounds
David Eckstein: 175 pounds
Hat tip Dan Szymborski
Wednesday, June 2, 2010
Monday, May 10, 2010
Thursday, April 22, 2010
Boston sends its best wishes to Heidi Watney for a full and fast recovery
The first comments on the Boston Globe's story about Heidi Watney's concussion-induced absence from NESN:
-We removed nuzzy's comment.
-We removed corkey's comment.
-We removed mrscoota's comment.
-We removed thommy's comment.
But we can imagine.
At least Mike Lowell's bat awaits her return, eagerly.
Update: The Globe, apparently discovering that its readers behave as if from Boston, has disabled the ability to comment on the Watney piece.
-We removed nuzzy's comment.
-We removed corkey's comment.
-We removed mrscoota's comment.
-We removed thommy's comment.
But we can imagine.
At least Mike Lowell's bat awaits her return, eagerly.
Update: The Globe, apparently discovering that its readers behave as if from Boston, has disabled the ability to comment on the Watney piece.
Sunday, December 13, 2009
Dexter finale: just so I can say I called it
I'm betting the central element of Dexter's fourth season finale will be Trinity capturing Harrison and Harrison being reborn in a pool of blood, much like Dexter was. Dexter will kill Trinity and save Harrison's life, but in future seasons he'll have to raise a son with dark urges -- forcing Dexter to decide how far he accepts and rejects Harry's upbringing of him.
Come on, that's a good ending even if the prediction proves wrong.
Come on, that's a good ending even if the prediction proves wrong.
Saturday, October 31, 2009
Tuesday, October 27, 2009
Comment on MGL on Bunts
I planned on posting his on The Book Blog, but sadly it doesn't seem to want to take it. It is a comment on this, from a slightly more mathematical perspective. I'm not sure how much reader('s) of this basically dormant blog will enjoy it, but it seems like a reasonable place to put it. What follows is the comment.
This is very good, although I am not sure I agree with some of the details. I've been grading a lot of freshman calculus exams lately, so apologies in advance for this little model, which is obviously heavily inspired by MGL's discussion.
We suppose both managers behave optimally. Let t ranging from 0 to 1 be the possible positions the infielders can play, so t=0 means that the infielders are playing as far in as possible, and t=1 means that they are playing as deep as possible (or as deep as realistically possible, since I guess they could play as deep as the fences!) We assume for simplicity that there is only one free parameter in terms of how "deep" the defense can play, while of course there are many in reality.
For a given hitter, in a given game state, with a given defense, we have the functions b(t) and h(t), where b(t) is the expected value of the bunt with the defense playing position t and h(t) the expected value of hitting away. We assume that b(t) is increasing, h(t) is decreasing, h(0) > b(0) (so hitting is preferred to bunting with the defense playing in) and b(1) > h(1). At the Nash equilibrium, the defense will play some position t_0 between zero and 1 and the offense will bunt some fraction 0 < c < 1.
t_0 is easy to find, as MGL indicates. Namely, our assumptions about h(t) and b(t) guarantee that there is a unique point x between 0 and 1 where h(x) = b(x). MGL's argument in the link shows that if the defense is playing any position t' not equal to x, it cannot be in a Nash equilibrium (the offense would have to bunt or hit 100% of the time, and the defense could then improve by changing their position.) Of course, if the hitter is a good bunter then b(t) is bigger, so x is closer to 0, while if he is a poor bunter or a good hitter x is closer to 1, all of which obviously makes intuitive sense.
Now, what should the batter do? He bunts some fraction c of the time, and no matter what of c he picks the total value of his PA is c*b(x)+(1-c)*h(x)=(c+1-c)*b(x)=b(x) since b(x)=h(x). However, there is still a constraint: for the optimal value c, the function f(t) = c*b(t) + (1-c)*h(t) (which is the value of the PA as a function of t if the batter bunts with probability c) must have a local maximum at t=x. Otherwise, by either moving in or out, the defense can decrease the value of the PA, which contradicts the Nash equilibrium assumption.
So we must have c*b(t) + (1-c)*h(t) with a local max at t=x. From freshman calculus, we know that a necessary condition for a local max is that the derivative vanishes at that point. So
0 = f'(x) = c*b'(x)+(1-c)*h'(x)
Which gives
c = - h'(x)/(b'(x)-h'(x))
Our assumptions about b(t) and h(t) guarantee that this is always strictly between 0 and 1, which is obviously good.
Note, though, and this is the key point, that the optimal ratio is independent of how good a bunter the batter is! All that the matter is the LOCAL behavior of the two functions at the indifference point.
For instance, suppose ARod is a bad bunter and a good hitter, so the defense plays him very far back, say at .9. But at .9, it may well be the case that b'(.9) >> h'(.9), or in other words, that the marginal improvement in ARod's bunting value when the defense steps back is much bigger than the corresponding marginal decrease in his hitting ability. In this case, ARod should be bunting most of the time. Similarly, you can be a very good bunter and poor hitter, but still be advised to swing away when the defense is playing you optimally. So I don't think I agree with MGL's recommendations about how much different players should bunt
This explains why the offensive manager's job is much harder the the defensive manager's. The defense only has to know when the two functions are roughly equal, which is something that can be discovered implicitly through trial and error. It is very hard, however, to get a sense of the derivatives of these functions through trial and error. So it makes a lot of sense that offensive manager's make more bunting mistakes than defensive ones.
Does this make sense? Is there some other parameter that people think should be included in this ridiculously simple model?
This is very good, although I am not sure I agree with some of the details. I've been grading a lot of freshman calculus exams lately, so apologies in advance for this little model, which is obviously heavily inspired by MGL's discussion.
We suppose both managers behave optimally. Let t ranging from 0 to 1 be the possible positions the infielders can play, so t=0 means that the infielders are playing as far in as possible, and t=1 means that they are playing as deep as possible (or as deep as realistically possible, since I guess they could play as deep as the fences!) We assume for simplicity that there is only one free parameter in terms of how "deep" the defense can play, while of course there are many in reality.
For a given hitter, in a given game state, with a given defense, we have the functions b(t) and h(t), where b(t) is the expected value of the bunt with the defense playing position t and h(t) the expected value of hitting away. We assume that b(t) is increasing, h(t) is decreasing, h(0) > b(0) (so hitting is preferred to bunting with the defense playing in) and b(1) > h(1). At the Nash equilibrium, the defense will play some position t_0 between zero and 1 and the offense will bunt some fraction 0 < c < 1.
t_0 is easy to find, as MGL indicates. Namely, our assumptions about h(t) and b(t) guarantee that there is a unique point x between 0 and 1 where h(x) = b(x). MGL's argument in the link shows that if the defense is playing any position t' not equal to x, it cannot be in a Nash equilibrium (the offense would have to bunt or hit 100% of the time, and the defense could then improve by changing their position.) Of course, if the hitter is a good bunter then b(t) is bigger, so x is closer to 0, while if he is a poor bunter or a good hitter x is closer to 1, all of which obviously makes intuitive sense.
Now, what should the batter do? He bunts some fraction c of the time, and no matter what of c he picks the total value of his PA is c*b(x)+(1-c)*h(x)=(c+1-c)*b(x)=b(x) since b(x)=h(x). However, there is still a constraint: for the optimal value c, the function f(t) = c*b(t) + (1-c)*h(t) (which is the value of the PA as a function of t if the batter bunts with probability c) must have a local maximum at t=x. Otherwise, by either moving in or out, the defense can decrease the value of the PA, which contradicts the Nash equilibrium assumption.
So we must have c*b(t) + (1-c)*h(t) with a local max at t=x. From freshman calculus, we know that a necessary condition for a local max is that the derivative vanishes at that point. So
0 = f'(x) = c*b'(x)+(1-c)*h'(x)
Which gives
c = - h'(x)/(b'(x)-h'(x))
Our assumptions about b(t) and h(t) guarantee that this is always strictly between 0 and 1, which is obviously good.
Note, though, and this is the key point, that the optimal ratio is independent of how good a bunter the batter is! All that the matter is the LOCAL behavior of the two functions at the indifference point.
For instance, suppose ARod is a bad bunter and a good hitter, so the defense plays him very far back, say at .9. But at .9, it may well be the case that b'(.9) >> h'(.9), or in other words, that the marginal improvement in ARod's bunting value when the defense steps back is much bigger than the corresponding marginal decrease in his hitting ability. In this case, ARod should be bunting most of the time. Similarly, you can be a very good bunter and poor hitter, but still be advised to swing away when the defense is playing you optimally. So I don't think I agree with MGL's recommendations about how much different players should bunt
This explains why the offensive manager's job is much harder the the defensive manager's. The defense only has to know when the two functions are roughly equal, which is something that can be discovered implicitly through trial and error. It is very hard, however, to get a sense of the derivatives of these functions through trial and error. So it makes a lot of sense that offensive manager's make more bunting mistakes than defensive ones.
Does this make sense? Is there some other parameter that people think should be included in this ridiculously simple model?
Saturday, July 11, 2009
A perfect game
Jonathan Sanchez is getting gypped. Officially, he threw a no-hitter Friday. But because he allowed no walks and hit no batters, only third baseman Juan Uribe's fielding error separated Sanchez from a perfect game -- a far rarer and more prestigious feat than a no-no. Why should Sanchez be punished because his teammate choked? Not only was Sanchez's performance tantamount to a perfect game, but it was actually a bit better than perfect, because he had to record an extra out to make up for Uribe's screw up. The stat community should from now on determine perfect games independently of errors.
Wednesday, May 27, 2009
Bonzai
First Pearl Harbor. Now Dice-K.
How much bombing from Japan is this country going to take?
Truly, if the CIA had forced senior al Qaeda detainees to watch Daisuke Matsuzaka throw 100+ pitches in 5 innings, allowing 12 baserunners, serving up 690 wild pitches, all while wiggling his ass like an earthworm on a fish hook, Osama would have been toast on 9/12.
But to the detainees, it would have felt like forever.
How much bombing from Japan is this country going to take?
Truly, if the CIA had forced senior al Qaeda detainees to watch Daisuke Matsuzaka throw 100+ pitches in 5 innings, allowing 12 baserunners, serving up 690 wild pitches, all while wiggling his ass like an earthworm on a fish hook, Osama would have been toast on 9/12.
But to the detainees, it would have felt like forever.
Thursday, May 21, 2009
Travesty
You're a baseball announcer. How do you ruin back-to-back-to-back home runs? Right here. It comes naturally if you're John Asshat Sterling.
No amount of waterboarding could ever suffice to punish the use of the term "Swish-a-licious." But by golly we should try.
No amount of waterboarding could ever suffice to punish the use of the term "Swish-a-licious." But by golly we should try.
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