Your example assumes you called a certain known endpoint (a person, or an automated telephone answering system) and interacted directly with it.
BitTorrent downloads from, and uploads to, unknown endpoints that happen to have or want the file, respectively.
On the one hand, you authorise your BitTorrent client to communicate with these hosts on your behalf, and your goal is the same (to get and give the file); this may constitute a form of interaction.
On the other hand, you have no control over which hosts your BitTorrent client contacts. These people may be people you know or strangers; people in the same or another jurisdiction. The link may be difficult to establish.
>> According to the slides, the Ceylon Project aims to create a programming language and SDK for business computing, designed with an eye to the successes and failures of the Java. It is built to run on the JVM
"Nintendo CEO Satoru Iwata went on the offensive today against his smartphone counterparts, arguing that the model pursued by individuals like Peter Vesterbacka is 'dying.' In a panel discussion at the Game Developers Conference in San Francisco, Iwata said that innovation wasn't coming from independent game coders, but from large and established companies like his own. Iwata also pointed to the major concern over the price model for smartphone games. Compared to games on established consoles, which hover around fifty dollars, mobile titles like Angry Birds run for 99 cents and make their developers little money due to the policies of online app stores. At these price points, "there's no motivation [for] high-value video games," Iwata said. Still, the executive did admit that the business model for console games had yet to be completely figured out."
Mandate from CRTC disallows operation. Decision from cabinet to overturn the mandate allows operation. Ruling from court to reject the decision disallows operation.
There's a mistaken assumption here: Google doesn't save $6.5 million as long as Youtube is still encoding H.264 video. This is a purely political move on Google's part.
That is right, but if Google and Mozilla get their way, Google will have saved themselves $6.5 million a year by not needing to encode H.264 anymore for YouTube, and everyone else for not needing to put in support anymore. I seem to have tripped up in my explanations again.
The other side of that coin is that, if Google decided to include H.264 support in Chrome, other browser makers, without or with little money, could have been unable to pay the royalties needed to be paid to patent owners to support it in their browsers. Does Google saving itself $6.5 million annually pale in comparison to Google saving itself and everyone else up to $6.5 million annually?
The Web does not need H.264 to function, so if everyone drops support for it preemptively, then even the small browser makers (I'm thinking of Flock here, which is based on Chromium but has a user base of its own) won't need to pay royalties to anyone.
This is like the GIF patents by Unisys* driving adoption of the PNG format, except that the patents are being assessed now because we know of a few formats that are** suitable for the job but unencumbered already. For instance, WebM can, like H.264, be implemented by anyone, but in addition to that, WebM intentionally doesn't require that royalties be paid. It can become a standard, like PNG originally wasn't a standard but then became an ISO/IEC standard 8 years later.
________________ * = As well as other factors such as the lack of true-color support, alpha and ICC in a lossless format, but that's off-topic. ** = May not be for now, due to lack of hardware acceleration, but could become in the future. Some people don't really care about power usage or CPU usage, so it's already suitable for them.
That's because you can't really use HTML 5 to make an ad that is going to be served to users of IE below version 9, in all of which support for HTML 5 does not exist.
So what do you use to advertise IE 9? Either Flash, or Java, or HTML 4 + JavaScript, or some other solution.
Java is bulky. HTML 4 + JavaScript is not that fast in IE 8 and earlier, so it's liable to freeze IE up and disrupt page navigation. Other solutions may mess up even further. You're left with Flash.
Slight inaccuracy, you can't DM a user on Twitter unless you're following that user and they also follow you back. You can still @reply to them, and they'll see it in their Mentions tab, which I don't know how many people check and how often.
You're entirely right about it being easy to search for keywords the company is interested in, so that it can know who is talking about it, though. Even more so with the new Streaming API...:)
500 KB Web fonts on a Slashdotted page... I guess one thing you can do now is replace that declaration with "font-face: Arial, Helvetica, sans-serif;" everywhere.
33,000 USD of hard drives, currently at about 1.5 TiB for 80 USD, is 633,600 GiB.
633,600 GiB can store 158,400,000 songs, at 4 MiB apiece.
The second trial of Jammie Thomas awarded the RIAA 1,920,000 USD for 24 songs, which comes out to 80,000 USD apiece.
For 158,400,000 songs, the RIAA would be awarded 12,672,000,000,000 USD (12 trillion short scale). That's only a bit less than the national US debt, which is 13,208,593,598,669 USD (13 trillion short scale) as of this comment!
It simply means that there is a 50% chance that Peter has a brother. Peter's sex is not given and it has 50% chance that it may be a girl.:)
It is given. "At least one of whom is a boy" means that the number of boys is guaranteed to be greater than 0. Therefore, one child is a boy, and I pinpointed him as being the given of the problem, and gave him a name to differentiate him from the other unspecified child. If the problem had stated "Exactly one of whom is a boy", then the probability of both children being boys is 0 (0%), because the number of boys is guaranteed to be greater than 0 and less than 2.
Assuming that all families have exactly 2 children with random sex distribution:
1. Is at least one of your children a boy? - Yes, it is. Then the possibility that your other child is a son is 1/3.
Because 75% of families have at least one boy and 25% have two boys.
You gave the right probability here, but for the wrong reason. "Is at least one of your children a boy?" answered in the affirmative means that you can now answer the question as if one boy was a given. The question now is, "Given that I have at least one son, what is the probability that I have 2 sons?"
Per this page, this can be written as P(2 sons | at least 1 son) = P(2 sons and at least 1 son) / P(at least 1 son) = (1/4) / (3/4) = 1/3.
And I just invalidated all of my other comments on this thread... Ouch!
Would you agree that "I rolled a die and it landed on 6. What's the probability that it landed on 6?" yields a probability of 1 (100%)? That's because it's a given of the problem. We don't even have to know if it's unweighted, or 6-sided; it could be 100-sided, and it still wouldn't change the fact that it landed on 6. If you don't agree with this, then you say "the probability is undefined, because I have insufficient information about your die".
Would you agree that, in "I tossed a coin twice, and at least one of these landed heads. What's the probability of both having landed heads?", a given of the problem is that there was 1 Heads, and therefore that the probability we're looking for is 0.5 (50%) for the unspecified coin? If you didn't agree with saying that a given of a problem has P = 1, then the universe of the problem is {(H,H), (H,T), (T,H), (T,T)}, each occurring with equal probability, and the answer is 0.25 (25%). Note that, if you don't accept givens to problems, (T,T) is not impossible, because you ignore "and at least one of these landed heads".
The problem posed in TFA amounts to "My wife gave birth twice, and at least one of the children was a boy. That boy was born on a Tuesday. What's the probability of both being boys?" Therefore the problem is exactly like the two coin problem, with 1 boy being a given, and the birth weekday being extra information that isn't used in the problem's question therefore doesn't affect the resulting probability. What's the probability now?
Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?
This is the question posed without the birth weekday specified. TFA actually tries to say that there are 4 outcomes for the pair of children, one of which is impossible, so they remove it. Since "boy, boy" is only one of the 3 outcomes, then the probability must be 1/3. Right?
Wrong.
The boy (let's call him Peter) being a boy is a given of the problem, so it has P = 1. The other child -- we don't care about it being born before or after Peter -- is independent, so the probability that it's a boy is 0.5*. The 4 outcomes are as follows:
Peter, Boy = 0.25* Peter, Girl = 0.25* Boy, Peter = 0.25* Girl, Peter = 0.25*
So, whichever way we slice this problem, the solution is 0.5*.
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.
Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g., What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.
I bet you would freak out if there was snow in California. Hell, the UK was like a bunch of pussies last winter because of about 15 cm of snow (half a foot), using up all of their grit, and Canada gets more than that every winter.
The COINS dataset, which is talked about in the summary, was recently published in CSV files (not proprietary), compressed with Zip (not proprietary) and distributed using HTTP (not proprietary) and BitTorrent (not proprietary) at the user's option.
I think that the clearing of private data in Firefox is a bit counter-productive, because deleting from SQLite databases merely marks the rows' storage space as being reclaimable within the file.
I once cleared private data for a day when my places.sqlite was around 70 MiB, then checked the file size and saw that it hadn't even changed by one byte. It wouldn't surprise me if the URLs were still in there -- all of them, intact, until you visit other pages to make Firefox overwrite the reclaimable pages in places.sqlite.
Even if Firefox truncated places.sqlite when the user clicked "delete everything", the URLs would still be readable on the underlying storage device. Firefox would have to shred(1) or zero out the file. I doubt that's going to happen.
Slashdot Mirror
Your example assumes you called a certain known endpoint (a person, or an automated telephone answering system) and interacted directly with it.
BitTorrent downloads from, and uploads to, unknown endpoints that happen to have or want the file, respectively.
On the one hand, you authorise your BitTorrent client to communicate with these hosts on your behalf, and your goal is the same (to get and give the file); this may constitute a form of interaction.
On the other hand, you have no control over which hosts your BitTorrent client contacts. These people may be people you know or strangers; people in the same or another jurisdiction. The link may be difficult to establish.
Clearly all the Slashdot commenters are busy patching their bosses' JBoss servers against this vulnerability.
From TFA:
>> According to the slides, the Ceylon Project aims to create a programming language and SDK for business computing, designed with an eye to the successes and failures of the Java. It is built to run on the JVM
>>> the Ceylon Project is built to run on the JVM
Whoops!
"Nintendo CEO Satoru Iwata went on the offensive today against his smartphone counterparts, arguing that the model pursued by individuals like Peter Vesterbacka is 'dying.' In a panel discussion at the Game Developers Conference in San Francisco, Iwata said that innovation wasn't coming from independent game coders, but from large and established companies like his own. Iwata also pointed to the major concern over the price model for smartphone games. Compared to games on established consoles, which hover around fifty dollars, mobile titles like Angry Birds run for 99 cents and make their developers little money due to the policies of online app stores. At these price points, "there's no motivation [for] high-value video games," Iwata said. Still, the executive did admit that the business model for console games had yet to be completely figured out."
Okay, not exactly, but Iwata-san did say something against smartphones at the Game Developers Conference in San Francisco, a mere 13 days ago.
Mandate from CRTC disallows operation.
Decision from cabinet to overturn the mandate allows operation.
Ruling from court to reject the decision disallows operation.
Unscrambled as requested.
There's a mistaken assumption here: Google doesn't save $6.5 million as long as Youtube is still encoding H.264 video. This is a purely political move on Google's part.
That is right, but if Google and Mozilla get their way, Google will have saved themselves $6.5 million a year by not needing to encode H.264 anymore for YouTube, and everyone else for not needing to put in support anymore. I seem to have tripped up in my explanations again.
The other side of that coin is that, if Google decided to include H.264 support in Chrome, other browser makers, without or with little money, could have been unable to pay the royalties needed to be paid to patent owners to support it in their browsers. Does Google saving itself $6.5 million annually pale in comparison to Google saving itself and everyone else up to $6.5 million annually?
The Web does not need H.264 to function, so if everyone drops support for it preemptively, then even the small browser makers (I'm thinking of Flock here, which is based on Chromium but has a user base of its own) won't need to pay royalties to anyone.
This is like the GIF patents by Unisys* driving adoption of the PNG format, except that the patents are being assessed now because we know of a few formats that are** suitable for the job but unencumbered already. For instance, WebM can, like H.264, be implemented by anyone, but in addition to that, WebM intentionally doesn't require that royalties be paid. It can become a standard, like PNG originally wasn't a standard but then became an ISO/IEC standard 8 years later.
________________
* = As well as other factors such as the lack of true-color support, alpha and ICC in a lossless format, but that's off-topic.
** = May not be for now, due to lack of hardware acceleration, but could become in the future. Some people don't really care about power usage or CPU usage, so it's already suitable for them.
Disable your ad blocker. Ding, instant shitload of ads.
That's because you can't really use HTML 5 to make an ad that is going to be served to users of IE below version 9, in all of which support for HTML 5 does not exist.
So what do you use to advertise IE 9? Either Flash, or Java, or HTML 4 + JavaScript, or some other solution.
Java is bulky. HTML 4 + JavaScript is not that fast in IE 8 and earlier, so it's liable to freeze IE up and disrupt page navigation. Other solutions may mess up even further. You're left with Flash.
And I thought Google TiSP was just a joke...
I believe you meant http://www.jerseys-2010.xn--mgba3a4fra/
Iran, Islamic Republic of. ccTLDs: xn--mgba3a4f16a, xn--mgba3a4fra.
The Unicode whitelist on Slashdot is preventing us from having the Farsi reading, so see here.
Slight inaccuracy, you can't DM a user on Twitter unless you're following that user and they also follow you back. You can still @reply to them, and they'll see it in their Mentions tab, which I don't know how many people check and how often.
You're entirely right about it being easy to search for keywords the company is interested in, so that it can know who is talking about it, though. Even more so with the new Streaming API... :)
500 KB Web fonts on a Slashdotted page... I guess one thing you can do now is replace that declaration with "font-face: Arial, Helvetica, sans-serif;" everywhere.
Just for future reference,
33,000 USD of hard drives, currently at about 1.5 TiB for 80 USD, is 633,600 GiB.
633,600 GiB can store 158,400,000 songs, at 4 MiB apiece.
The second trial of Jammie Thomas awarded the RIAA 1,920,000 USD for 24 songs, which comes out to 80,000 USD apiece.
For 158,400,000 songs, the RIAA would be awarded 12,672,000,000,000 USD (12 trillion short scale). That's only a bit less than the national US debt, which is 13,208,593,598,669 USD (13 trillion short scale) as of this comment!
It simply means that there is a 50% chance that Peter has a brother. Peter's sex is not given and it has 50% chance that it may be a girl. :)
It is given. "At least one of whom is a boy" means that the number of boys is guaranteed to be greater than 0. Therefore, one child is a boy, and I pinpointed him as being the given of the problem, and gave him a name to differentiate him from the other unspecified child. If the problem had stated "Exactly one of whom is a boy", then the probability of both children being boys is 0 (0%), because the number of boys is guaranteed to be greater than 0 and less than 2.
Assuming that all families have exactly 2 children with random sex distribution:
1. Is at least one of your children a boy? - Yes, it is. Then the possibility that your other child is a son is 1/3.
Because 75% of families have at least one boy and 25% have two boys.
You gave the right probability here, but for the wrong reason. "Is at least one of your children a boy?" answered in the affirmative means that you can now answer the question as if one boy was a given. The question now is, "Given that I have at least one son, what is the probability that I have 2 sons?"
Per this page, this can be written as P(2 sons | at least 1 son) = P(2 sons and at least 1 son) / P(at least 1 son) = (1/4) / (3/4) = 1/3.
And I just invalidated all of my other comments on this thread... Ouch!
*takes a huge bite of humble pie*
And now you're saying that it's twice as likely to have a girl as it is to have a boy?
You say:
P(Peter, Boy) + P(Boy, Peter) = 1/3
P(Peter, Girl) + P(Girl, Peter) = 2/3
Let's try to reverse-engineer this problem.
Would you agree that "I rolled a die and it landed on 6. What's the probability that it landed on 6?" yields a probability of 1 (100%)? That's because it's a given of the problem. We don't even have to know if it's unweighted, or 6-sided; it could be 100-sided, and it still wouldn't change the fact that it landed on 6. If you don't agree with this, then you say "the probability is undefined, because I have insufficient information about your die".
Would you agree that, in "I tossed a coin twice, and at least one of these landed heads. What's the probability of both having landed heads?", a given of the problem is that there was 1 Heads, and therefore that the probability we're looking for is 0.5 (50%) for the unspecified coin? If you didn't agree with saying that a given of a problem has P = 1, then the universe of the problem is {(H,H), (H,T), (T,H), (T,T)}, each occurring with equal probability, and the answer is 0.25 (25%). Note that, if you don't accept givens to problems, (T,T) is not impossible, because you ignore "and at least one of these landed heads".
The problem posed in TFA amounts to "My wife gave birth twice, and at least one of the children was a boy. That boy was born on a Tuesday. What's the probability of both being boys?" Therefore the problem is exactly like the two coin problem, with 1 boy being a given, and the birth weekday being extra information that isn't used in the problem's question therefore doesn't affect the resulting probability. What's the probability now?
Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?
This is the question posed without the birth weekday specified. TFA actually tries to say that there are 4 outcomes for the pair of children, one of which is impossible, so they remove it. Since "boy, boy" is only one of the 3 outcomes, then the probability must be 1/3. Right?
Wrong.
The boy (let's call him Peter) being a boy is a given of the problem, so it has P = 1. The other child -- we don't care about it being born before or after Peter -- is independent, so the probability that it's a boy is 0.5*. The 4 outcomes are as follows:
Peter, Boy = 0.25*
Peter, Girl = 0.25*
Boy, Peter = 0.25*
Girl, Peter = 0.25*
So, whichever way we slice this problem, the solution is 0.5*.
P(Peter, Boy) + P(Boy, Peter) = 0.5*
1 * P(Other is Boy) = 0.5*
- - - - - -
* May slightly differ due to the male:female ratio at birth. It is assumed here to be 1:1.
I don't know about the chance of it landing heads up, but if you threw it up, it has a rather high chance of being damaged by your stomach acid...
The problem is stated thus:
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.
Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g., What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.
What if the second child was born on Tuesday but was not a twin of the Tuesday boy?
Say, the Tuesday 75 weeks later. It's still a Tuesday...
I bet you would freak out if there was snow in California. Hell, the UK was like a bunch of pussies last winter because of about 15 cm of snow (half a foot), using up all of their grit, and Canada gets more than that every winter.
It just depends on what you're used to.
(Disclaimer: I am Canadian.)
It will probably be increases of 10.8%, 8.5% and 6.8% compared to the previous proportion, not the current entire Internet.
So if violence grew by 10.8%, but the previous proportion was 4.4%, you jump to about 4.9%.
The COINS dataset, which is talked about in the summary, was recently published in CSV files (not proprietary), compressed with Zip (not proprietary) and distributed using HTTP (not proprietary) and BitTorrent (not proprietary) at the user's option.
I think that the clearing of private data in Firefox is a bit counter-productive, because deleting from SQLite databases merely marks the rows' storage space as being reclaimable within the file.
I once cleared private data for a day when my places.sqlite was around 70 MiB, then checked the file size and saw that it hadn't even changed by one byte. It wouldn't surprise me if the URLs were still in there -- all of them, intact, until you visit other pages to make Firefox overwrite the reclaimable pages in places.sqlite.
Even if Firefox truncated places.sqlite when the user clicked "delete everything", the URLs would still be readable on the underlying storage device. Firefox would have to shred(1) or zero out the file. I doubt that's going to happen.