As others have remarked, many stores either have or have had them in. However, AFAIK, they are all the "standard" configurations with the 120GB hard drive -- to get the 160 or the 200, I believe you have to order online.
OK, 2) and 3) are good points I hadn't considered. 1) of course I'm aware of, and 4) and 5) are not a big deal to me.
2) I would not have guessed. Apple now claims 5 hours for the 15, and 5.5 for the 17, but I'm assuming that's only when, among other things, the screen is cranked down to minimum. If it's at full brightness, my expcation would have been that the 17 would run out before the 15, even with the bigger battery.
Re 3)... question, what's the RPM of the hard drive in your 15?
Could you elaborate on preferring the 17 to the 15? I've been using a 17 for the past 3.5 years, but I'm about ready to switch back. I have 23" displays in my two work environments, and I carry my PowerBook around daily. Still, I am a bit concerned I'll miss the space when I'm not at one of my displays. OTOH, the new 15 has the same resolution as my 17.
Sure. Apple's always had a good environmental image. But they got a bit of a black eye from the Greenpeace report, bogus or not, and it's no surprise Jobs would take an opportunity like this to talk about how green Apple is. I mean, there is just no news here.
It was news when Greenpeace announced their list some time back (and Apple disputed their ranking). Is anybody remotely surprised Apple wants to up their environmental profile?
Well, a fair enough criticism, but it's going to be hard to find someone with an informed opinion who hasn't participated and had either a good or a bad experience. In my case, although my team did take 3rd one year, I had my share of issues with judging. The second year at nationals we bombed, and before the first year we bombed at regionals. In each case it was easy to find many causes to blame on the judges. In fact the first year I sent a nasty (and pretty stupid) letter to the contest organizer. But when you step back and look at these cases objectively, 99 times out of 100 it's the team's fault. So, I think I know where the GP was coming from.
I completely disagree with the claim that "the problems they ask are much prefabricated problems - if you know their solution, you're in". This sentiment can only reflect a lack of much experience with the contests. As to what constitutes "real" programming skills, well, that is more a matter of opinion. But the ACM contests certainly test something a lot more interesting than typing skills and memorization. And I don't even understand the bit about "sponsorship promoting ACM products and courses". ACM courses? Huh?
I guess that depends on what you mean by hacking. In my experience, often it's raw coding speed that carries the day, and having lots of tricks at your disposal helps enormously. But that will only get you so far without an ability to suss out the clever, concise solutions quickly. In any case, I agree with your "sour grapes" assessment of the parent.
- Bob Hearn, member, Rice 1986 (3rd place) & 1987 ACM programming teams
Wikipedia was already updated even before the Slashdot submission. The really interesting bit is this change history of that page. The current text replaces this:
"It is rumored that Harvey Mudd will attempt to steal the cannon again this year in celebration of the 20th anniversary of their original theft. Be warned. Thankfully Dabney house has planned ahead, and relocated the cannon to a safehouse in Compton."
So, was this "relocation" the actual MIT appropriation?
As of a year or so ago, no kidding, they're building the Overwhelmingly Large Telescope (official name).
So what name does this one get?
The Staggeringly Large Telescope? Not as big as "overwhelming".
Oh, but it shouldn't be as big as overwhelming - the Overwhelmingly Large Telescope, after all, would have a diameter of 100m(!!!), and more surface area than all previous professional telescope mirrors put together. At $1.2 trillion, a bargain, compared to $750 million for the 30m Canadian telescope. Hell, the secondary mirror alone for the OLT would be almost as big as this Canadian telescope.
"It's just not clear to me why a probability measure should satisfy this countably additivity condition"
Well... let's see. It seems to me that once you start talking in terms of a pdf, you are already granting countable additivity, really by definition of what the pdf means. P( [0, infinity) ) = sum(i=0, infinity) P( [i, i+1) ), because each probability is an integral of the pdf over the given range, and integration is linear. This doesn't use countable additivity; rather, it satisfies the condition of countable additivity - thus allowing one to reason using results in probability theory that depend on the axiom of countable additivity.
So (after doing my homework) I don't see how the concept of a pdf that doesn't satisfy countable additivity could be meaningful.
Oh, about your answer to the problem of doing better than never switching - very clever! However, that's not the answer I had in mind. With your solution you should only expect to do better if the number you see in the envelope is in a certain range. But there's also a strategy in which you should expect to do better no matter what number you see!
About the problem implying a pdf, note that the word "random" does not occur in the problem. It is simply stated that I (who put the checks in the envelopes) choose a number, and you don't know what it is. To me that's an unspecified pdf.
About proving there's no uniform distribution over the reals... hmm. I've often seen it stated, with the justification being that it would be impossible to normalize. It seems self-evident - the density must be finite over a given range. But then it could not be uniform if it were normalized. Is that a proof? If not, I'm not sure why not. I'm not sure how countable additivity might enter into it. I would be interested to learn how, if I'm wrong.
Actually, most definitions I see of uniform distribution are spefically over an interval [a,b] - so then by definition one could not have a uniform distribution over the reals. More properly, I guess, what the above argument shows is that if you have a distribution over the reals, it cannot assign equal probability to all equal intervals.
Note that the paradoxical distributions related to the coin-flipping problem are *not* uniform.
If, after 4 days, the 4 marked monks he sees are still alive, he knows he must be marked as well - otherwise they would all have deduced they were the marked ones already.
On day N + 1 the state isn't the same as on day N. It's proof by induction. We have to assume that all the monks are perfect logicians, but that's given.
The theorem they each deduce is that if there are N marked monks, then after N days the marked ones will know who they are, and will kill themselves.
The base case is for N = 1, which I think is clear.
The inductive step is, suppose it is day N. Then the marked monks all see N - 1 marked monks. (The unmarked ones see N.) By induction, if there were only N - 1 marked, they would have committed suicide after day N - 1. But they didn't. Therefore, they all now know they are marked as well, and will then commit suicide, before day N + 1.
Which I think is pretty much what I said before... but hopefully I made it a bit more clear. 2 is just a special case of N.
If by "problems" you mean "problems with the obvious answers", then yes, you have hit a few of them. But if you mean "problems with the paradox", in the sense of it being a flawed puzzle, then I disagree. It is an excellent puzzle as stated.
The issue of utility is not really relevant for the basic two-envelope paradox. That only arises when you have infinite expectation, which you do not since no particular probability distribution function was specified in the problem.
You mention the lack of a pdf as a probem - well, yes, that's the crux. The apparent paradox is resolved when you realize that you are implicity assuming a uniform distribution over the reals to choose the values, and no such distribution exists.
You're right that applying concepts from the St. Petersberg Paradox (coin-flipping until it lands tails) can lead to a two-envelope paradox variant which does yield paradoxical expectations. But, as you point out, then the expectations are infinite, and it is meaningless to compare two different strategies each yielding infinite expectation.
However - you still haven't answered my question about the existence of a strategy which does better than not switching. I claim that such a strategy exists, independent of the pdf. You can restrict it to pdf's with finite expectation, if you wish.
Also, here's where it really gets interesting. It turns out that both the arguments above about whether to switch are wrong! You can do better than to always keep the one you drew, but not by always switching. Since you know this puzzle already, I'll bet you're very surprised by this, and likely do not believe it. You can email me if you want me to prove it, though.
First, suppose that there is just one marked monk. Then the next day, when they all go to the balcony, he will see that nobody else has a mark. Therefore, since he knows that somebody is marked, he must be marked, and thus he should commit suicide.
Now suppose there are two. Each of them sees just one other marked monk. Now, the next day, they *still* see one other marked monk. But, if there were only one, he would already have commited suicide. Therefore, each then knows that he must be marked as well, so both should then commit suicide.
In general, if N are marked, they will each see N - 1 other marked monks. If, after N - 1 days, they have not committed suicide, then they all know there must be N, and they should then commit suicide.
Slashdot Mirror
As others have remarked, many stores either have or have had them in. However, AFAIK, they are all the "standard" configurations with the 120GB hard drive -- to get the 160 or the 200, I believe you have to order online.
Does anyone know different?
OK, that I can understand; the dark side can be hard to resist. But sure are you that Yoda instead it was not?
"After being encouraged to viral market Serenity, the studio has started legal action against fans"
I hate what the Internet has done to basic language skills.
OK, 2) and 3) are good points I hadn't considered. 1) of course I'm aware of, and 4) and 5) are not a big deal to me.
2) I would not have guessed. Apple now claims 5 hours for the 15, and 5.5 for the 17, but I'm assuming that's only when, among other things, the screen is cranked down to minimum. If it's at full brightness, my expcation would have been that the 17 would run out before the 15, even with the bigger battery.
Re 3)... question, what's the RPM of the hard drive in your 15?
Could you elaborate on preferring the 17 to the 15? I've been using a 17 for the past 3.5 years, but I'm about ready to switch back. I have 23" displays in my two work environments, and I carry my PowerBook around daily. Still, I am a bit concerned I'll miss the space when I'm not at one of my displays. OTOH, the new 15 has the same resolution as my 17.
Sure. Apple's always had a good environmental image. But they got a bit of a black eye from the Greenpeace report, bogus or not, and it's no surprise Jobs would take an opportunity like this to talk about how green Apple is. I mean, there is just no news here.
It was news when Greenpeace announced their list some time back (and Apple disputed their ranking). Is anybody remotely surprised Apple wants to up their environmental profile?
Well, a fair enough criticism, but it's going to be hard to find someone with an informed opinion who hasn't participated and had either a good or a bad experience. In my case, although my team did take 3rd one year, I had my share of issues with judging. The second year at nationals we bombed, and before the first year we bombed at regionals. In each case it was easy to find many causes to blame on the judges. In fact the first year I sent a nasty (and pretty stupid) letter to the contest organizer. But when you step back and look at these cases objectively, 99 times out of 100 it's the team's fault. So, I think I know where the GP was coming from.
I completely disagree with the claim that "the problems they ask are much prefabricated problems - if you know their solution, you're in". This sentiment can only reflect a lack of much experience with the contests. As to what constitutes "real" programming skills, well, that is more a matter of opinion. But the ACM contests certainly test something a lot more interesting than typing skills and memorization. And I don't even understand the bit about "sponsorship promoting ACM products and courses". ACM courses? Huh?
I guess that depends on what you mean by hacking. In my experience, often it's raw coding speed that carries the day, and having lots of tricks at your disposal helps enormously. But that will only get you so far without an ability to suss out the clever, concise solutions quickly. In any case, I agree with your "sour grapes" assessment of the parent.
- Bob Hearn, member, Rice 1986 (3rd place) & 1987 ACM programming teams
Sorry, perhaps I should have added a smily to my reply. I didn't mean to be sarcastic.
:-) )
I did proofread, but I still missed Padadena.
(P.S. - "courtesy"
Wikipedia was already updated even before the Slashdot submission. The really interesting bit is this change history of that page. The current text replaces this:
"It is rumored that Harvey Mudd will attempt to steal the cannon again this year in celebration of the 20th anniversary of their original theft. Be warned. Thankfully Dabney house has planned ahead, and relocated the cannon to a safehouse in Compton."
So, was this "relocation" the actual MIT appropriation?
So I missed a key, so sue me! I also capitalized "Appropriated." And I found www.howeandser.com after submitting.
http://www.swiss.ai.mit.edu/~gjs/gjs.html
A round trip to the moon takes light more like 3 seconds, actually.
As of a year or so ago, no kidding, they're building the Overwhelmingly Large Telescope (official name).
So what name does this one get?
The Staggeringly Large Telescope? Not as big as "overwhelming".
Oh, but it shouldn't be as big as overwhelming - the Overwhelmingly Large Telescope, after all, would have a diameter of 100m(!!!), and more surface area than all previous professional telescope mirrors put together. At $1.2 trillion, a bargain, compared to $750 million for the 30m Canadian telescope. Hell, the secondary mirror alone for the OLT would be almost as big as this Canadian telescope.
Yep, that's what I had in mind.
"It's just not clear to me why a probability measure should satisfy this countably additivity condition"
Well... let's see. It seems to me that once you start talking in terms of a pdf, you are already granting countable additivity, really by definition of what the pdf means. P( [0, infinity) ) = sum(i=0, infinity) P( [i, i+1) ), because each probability is an integral of the pdf over the given range, and integration is linear. This doesn't use countable additivity; rather, it satisfies the condition of countable additivity - thus allowing one to reason using results in probability theory that depend on the axiom of countable additivity.
So (after doing my homework) I don't see how the concept of a pdf that doesn't satisfy countable additivity could be meaningful.
Oh, about your answer to the problem of doing better than never switching - very clever! However, that's not the answer I had in mind. With your solution you should only expect to do better if the number you see in the envelope is in a certain range. But there's also a strategy in which you should expect to do better no matter what number you see!
About the problem implying a pdf, note that the word "random" does not occur in the problem. It is simply stated that I (who put the checks in the envelopes) choose a number, and you don't know what it is. To me that's an unspecified pdf.
About proving there's no uniform distribution over the reals... hmm. I've often seen it stated, with the justification being that it would be impossible to normalize. It seems self-evident - the density must be finite over a given range. But then it could not be uniform if it were normalized. Is that a proof? If not, I'm not sure why not. I'm not sure how countable additivity might enter into it. I would be interested to learn how, if I'm wrong.
Actually, most definitions I see of uniform distribution are spefically over an interval [a,b] - so then by definition one could not have a uniform distribution over the reals. More properly, I guess, what the above argument shows is that if you have a distribution over the reals, it cannot assign equal probability to all equal intervals.
Note that the paradoxical distributions related to the coin-flipping problem are *not* uniform.
Oh, X doesn't have to be an integer. So you can't gain any information by noticing whether the dollar amount is even or odd.
r eshold=1&commentsort=0&tid=228&mode=thread&cid=138 03894
Somebody else posted the same puzzle after me, but that's where the discussion seems to be:
http://ask.slashdot.org/comments.pl?sid=165444&th
If, after 4 days, the 4 marked monks he sees are still alive, he knows he must be marked as well - otherwise they would all have deduced they were the marked ones already.
On day N + 1 the state isn't the same as on day N. It's proof by induction. We have to assume that all the monks are perfect logicians, but that's given.
The theorem they each deduce is that if there are N marked monks, then after N days the marked ones will know who they are, and will kill themselves.
The base case is for N = 1, which I think is clear.
The inductive step is, suppose it is day N. Then the marked monks all see N - 1 marked monks. (The unmarked ones see N.) By induction, if there were only N - 1 marked, they would have committed suicide after day N - 1. But they didn't. Therefore, they all now know they are marked as well, and will then commit suicide, before day N + 1.
Which I think is pretty much what I said before... but hopefully I made it a bit more clear. 2 is just a special case of N.
If by "problems" you mean "problems with the obvious answers", then yes, you have hit a few of them. But if you mean "problems with the paradox", in the sense of it being a flawed puzzle, then I disagree. It is an excellent puzzle as stated.
The issue of utility is not really relevant for the basic two-envelope paradox. That only arises when you have infinite expectation, which you do not since no particular probability distribution function was specified in the problem.
You mention the lack of a pdf as a probem - well, yes, that's the crux. The apparent paradox is resolved when you realize that you are implicity assuming a uniform distribution over the reals to choose the values, and no such distribution exists.
You're right that applying concepts from the St. Petersberg Paradox (coin-flipping until it lands tails) can lead to a two-envelope paradox variant which does yield paradoxical expectations. But, as you point out, then the expectations are infinite, and it is meaningless to compare two different strategies each yielding infinite expectation.
However - you still haven't answered my question about the existence of a strategy which does better than not switching. I claim that such a strategy exists, independent of the pdf. You can restrict it to pdf's with finite expectation, if you wish.
I already posted this problem, here:
r eshold=1&commentsort=0&tid=228&tid=4&mode=thread&c id=13802487
:-( Too many other puzzles...
http://ask.slashdot.org/comments.pl?sid=165444&th
Nobody has replied yet.
However, traditionally you are allowed to open the envelope before deciding whether to switch. (See, e.g., http://www.maa.org/devlin/devlin_0708_04.html ).
Also, here's where it really gets interesting. It turns out that both the arguments above about whether to switch are wrong! You can do better than to always keep the one you drew, but not by always switching. Since you know this puzzle already, I'll bet you're very surprised by this, and likely do not believe it. You can email me if you want me to prove it, though.
This is an oldie (but a goodie).
First, suppose that there is just one marked monk. Then the next day, when they all go to the balcony, he will see that nobody else has a mark. Therefore, since he knows that somebody is marked, he must be marked, and thus he should commit suicide.
Now suppose there are two. Each of them sees just one other marked monk. Now, the next day, they *still* see one other marked monk. But, if there were only one, he would already have commited suicide. Therefore, each then knows that he must be marked as well, so both should then commit suicide.
In general, if N are marked, they will each see N - 1 other marked monks. If, after N - 1 days, they have not committed suicide, then they all know there must be N, and they should then commit suicide.