So did Sinclair ZX80
on
Scanjet Music
·
· Score: 2, Interesting
In the 80's, the nerd thing to do was to write assembly programs for the Sinclair. IIRC, you would convert the opcodes (which we all knew by heart) to ASCII, write a REM line with that, and run it (which I don't recall how).
I'd write loops inside loops, with changing and interdependent step sizes, and it would generate sounds on a FM radio sitting on the computer top (I KNOW my Z80 clock was 3.57MHz, way below FM; it was most probably due to harmonics interference or the radio IF).
I could get beats and interesting disco-like effects, and make alien phony calls. Then computers started shipping with speakers and sound processors and spoiled all the fun.
Agreed, and I guess not only soil cushioned the impact. If the canisters broke loose from the bulk of the section of spacecraft they were in, then they might have hit the ground at a lower speed (there's a velocity point where the G pull downwards equals the aerodynamic drag - "terminal velocity", so it will depend on mass and aerodynamics of the canister).
If they hit the ground inside a big chunk of the spacecract, then the deformation of this chunk also absorved part of the impact.
2,295 Gs of impact... I wonder where they got this number from. A rough estimative based on estimated numbers and estimated facts to produce a catchy figure, or did they actually have accelerometers on the canisters and had planned to crash-test them on the original mission?
Besides, the forces acting upon a tiny, low mass body when decelerating on impact are proportional to the mass (F=m*a), so if you drop a PET bottle with an ant inside from a 10-story building it will most probably survive the fall, and that doesn't make it Atom Ant. If you do the same and suvive getting inside a PET bottle you won't survive the impact, both because you will attain a higher terminal velocity, and because even if you hit the ground with the same speed (i.e. same decceleration), the forces acting on your body parts will be much, much higher.
IANARS (I am not a rocket scientist), but I suppose that debris entering the atmosphere will only burn if they have enough size/initial speed.
Space dust falling on Earth should get in unharmed (if being bombarded by the Sun's UV and all sorts of radiation before getting under the atmosphere blanket can be called unharmed).
I don't need a W800i. I don't need a W800i. It's from the evil Sony and uses their own memory card standard. I don't need a W800i bundled with a 512MB proprietary MemoryStick Duo memory card. I don't need a W800i bundled with a 512MB memory card and a 2MPx camera. I don't need it.
Specially if it comes bundled with Microsoft Visual Basic for Supercomputing - Science Edition, allowing any computer-illiterate scientist to easily put together simulations without needing to care about multiple-processor optimizations. It will have ready-to-use MSAtomicNuclei.ocx, MSAtmosphere.dll, etc.
You only need to go: Create a form. Drag the AtomView object to the form. Dim MyAtom(1E100) For i = 1 to 1E100
My Atom(i) = MSAtom.new(x,y) Next Atom(0).Nucleus.Split
Who wants to learn C, and all those threads things??
Once this program is trained enough, join it to a noise generator and a "natural selection" algorythm (typing intended), and you'll have an automated hit composer!!
It will eventually compose the "perfect hit", and kill art as we know it.
We're all concentrating on the electronic, switching power supply stuff.
What about the big power guzzlers in the house: Refrigerators and Air conditioners? Those AC motors suck power directly from 220/110 VAC, and isn't AC better for these cheap induction motors?
Although I disagree with "Efficiency. DC loses more energy per foot / mile than AC.", that could be well replaced by:
Efficiency: Low voltages (generally what your home devices use as DC) loses more energy per foot/mile than higher voltages.
This because if your home appliance needs 50W to work, sending 110VAC to it will require small currents through the power cable (~0.5A). Then it's converted to, say, 12VDC and travels a few inches (~4A). (using ~ not to engage in pedantic discussions).
The more current you send over a wire, the more power you dissipate on the wire (or the thicker wire you need), and/or the more voltage drop you have.
Why not sent 110V DC then? It's more expensive to convert DC to DC (from 110 to 12). For AC/DC you can use a transformer then a rectifier (or a switching power supply). For DC, you always need switching.
Wouldn't it be possible to send a specially crafted audio stream to VoIP programs such as Skype to explore eventual vulnerabilities on the audio codec routines?
I know it sounds far-fetched, but you know, jpegs were once safe too. Skype had its vulnerabilities (even on Linux), but were there any on the audio codec?
I hate these "must-have-a-firewall-passage" kind of programs, and I've so far managed to keep them out of my network, but now I'm trying hard to convince my boss not to install Skype on a CAD user's PC "because a customer wants to talk with him"!! CAN'T HE USE A PHONE??!
I agree, but I look like an overloaded miner's donkey with a phone, camera, mp3 player, and PDA all hanging on my belt and pockets. Not to mention the portable game, and a notebook (the right tool for email processing).
I don't expect my 1.3Mpix phone to match my 35mm (film) SLR, but it's better than nothing if I need to capture something unexpected and I'm not carrying the real camera. And the two extra lenses. And the tripod. And the silica baggies. And the lens cleaning brush....
Sure the mp3/pda/game/camera/email cell phone may be bigger than a plain phone, but it's smaller than my old Nokia 2160. And if I need small, I just pull the SIM from the V800 and slap it on a T610 or C65. (no, wait, these also have cameras. A T68 then.)
You're right. "mails" is really a shame (since "mail" is already a collective of stuff carried by the postal service). Thanks. I got lost in translation, but I think I'd rather have written "...processing his mail" instead (and I believe that's correct English this time).
Somehow, saying Darwin was troubled processing letters sound to me like he had alphabetization problems...
On "Selected Letters on Evolution and Origin of Species", it is interesting that some of the letters really have a conversation "sequence", considering the long "latency" time between each packet. This was specially true during Darwin's trip, but also when he was at home.
Something like we will experience when exchanging emails with colonies on other planets or solar systems: You write, and your grandson gets the answer.
When a quick response was expected, they'd send a messenger and ask that recipient answered by return mail (and the messenger would wait for the answer to be written).
Also, something as easy as sending an article you wrote for a friend to review (attach/send today) would require that someone hand-copied your writings or that you send the only original and wait for it to come back with the review. You didn't keep a copy on your "sent items".
In the book, Darwin's son says his father was troubled by the chore of processing mails, and spent a lot of time just doing that.
(please note that I'm not willing to engage a flame war or something like that - I'm really intrigued by this crazy thing, and I confess you got me hanging by a thread)
Maybe problem here is that there are two different comparisons being mixed up. It seems like we both agree in the results, but are arguing about different comparisons.
1st comparison: The odds of winning between: "having the option of switching in the end" vs "not having the option of switching in the end".
2nd comparison: Having the option of switching in the end, the odds of winning between: "switch" vs "not switch".
On the first comparison, we both agree that having the option of switching increases the odds. This is because once you changed the rules to have the switch option, your choices fall from 100 to 2. As soon as the rule is changed, it doesn't matter how many doors you have initially, in the end there will be 2. Opting for this game rule will increase my odds from 1/100 to 1/2 (I understand that for you, it increases from 1/100 to 99/100)
I think that your (and many others) logic goes that when you first pick a door, it has 1/100 probability of being wrong, then in the end it retains this probability so you are faced with the option of keeping this 1/100 win chance or switching to the 99/100 win chance door.
Please consider that I do start with a 1/100 chance with my first pick, but after you open the first door, I have 1/99, then 1/98, then 1/97, etc. This because you are eliminating more and more "wrong" doors. When you get to the 98th door, my chance of being right is 1/2 and so is my chance of being wrong.
I don't agree with your simulation because to calculate the odds of an outcome you need to count all possible outcomes, not just some of them (again, I'm intrigued here, and trying to find the flaw that makes the simulation seem to disagree with "my" theory):
> First go. (Prize is behind door 84). You pick door 1. I open doors 2-83, 85-100. Switcher picks remaining door (84) and gets the prize. Non-switcher gets a goat.
A general formulation could be: First go. (prize is behind door N). I pick door M (M may be = to N). You open all doors but M and N, and in case M=N you open all doors except D and N.
If M!=N (and there's 99/100 chance of that), there is only one possible combination of doors to be opened by you, so the only outcome is that I end up having to choose between M and N, and switching to N will be the winning option. For each of the 99 possible M!=N, there is one outcome, all saying switch is the winning choice.
Now, when M=N (1/100 of chance), you have 99 options of D when picking the doors you will open, so there are 99 possible pairs in the end, and switching will always give me the goat. For each of the 99 possible D selections, there is one outcome, all saying switch is the bad choice.
So the number of possible outcomes is not 100 as I was considering, but 396. 198 of which I win by switching, 198 of which I win by not switching. Thus, 50% (man I which I remembered enough math to put this in terms of combination/permutation stuff...).
There; it makes sense to me (I'm glad I could figure this out, it was more like a gut feeling when I started...); doesn't it make sense to you? It looks like the flaw is that when counting the possible outcomes one tend to see the case where M=N as having only one possible outcome, when there are many.
(if this thread is closed, we can go on; email me at the_name_of_this_site at rf dot com point br)
To lower the requirement of back-and-forth posting, I'll make my choices in advance, then you can run your simulation as many times as you want (of course, if you run less than 50, results may be non-conclusive).
Since the only comparison needed is between the "Switch" and "not Switch" scenario, here are my choices for each case:
For the "not switch scenario", I'll take door 1. You will agree that I have a 1/100 chance of being right at this moment, before you open the 98 doors. Correct?
For the "switch scenario", I'll take door not-1 (meaning I picked 1, then switched to whatever door was left after you opened the 98 remaining ones). At this moment, before you open the 98 doors, you will also agree that the probability of each door is 1/100, so not-1 gives me 1/100 too.
In any scenario, I don't need to see you opening the 98 doors, because I ALREADY know they won't have the prize, so there is NO NEW information when you open them. Besides, I've decided beforehand if I'm switching or not, so there is no point in seeing you theatrically opening the virtual doors ta-daa!!!
So it's 1/100 for the "not switch" scenario, and 1/100 for the "switch" scenario. No difference between the two scenarios.
Want me to decide "switch"/"not switch" only AFTER you open the 98 doors? Ok. I pick door 1.
Again, I don't need to see you opening the 98 doors, because I ALREADY know they won't have the prize. All that matters is that you will in the end present me two doors, "1" and "not-1". Each one has 1/2 probablility of having the prize. The decision to switch now will give me 50% of chance against 50% of chance if I don't switch. No difference again.
What is your point again? That if I see you opening the 98 doors I will gain knowledge? I'll know that one of the two remaining doors have the prize? Well, I have magical powers so I already know that.
> looking at a picture of the light that left NGC891 24 million years ago. right? wrong?
Wrong, due to the accelerating expansion of the universe. If the light of something out there 24 million light-years away hits us now, it doesn't mean that this light left the origin 24 million years before.
There are no "less likely" outcomes when it gets to the last decision (between the two remaining choices).
No matter what happens, and no matter how you play your 100-door, 1000-door, whatever simulation, the last choice will be between 2 doors. Whatever happened in the past does not change the odds of the prize to be in one or on the other.
The only difference would be if you had GAINED some true information about the location of the prize - which didn't happen, since you knew from the start that you would end up with two doors to choose from, one having the prize, other not.
Considering past facts as an influence to the choice here is akin to think that because you rolled a dice and it gave you "6", on the next roll the odds of it giving "6" again have magically diminished because the odds of two consecutive "6" are smaller than "6" and "anything".
The odds of two consecutive rolls result in the same number is smaller than resulting in different numbers, but for each individual roll, the odds are always 1/6. The past events won't affect the physics making the "6" face heavier.
The same way, unless you gained knowlegde from the past events (which you didn't, since the final outcome was fixed: you having to pick between the two last doors), whatever happened before your last choice won't magically change your odds of picking the right/wrong door - unless that is a Schrodinger's goat which is in the car-goat state until you actually see it, and is affected by 'spooky action at a distance' "condensing the probablilities" (whatever that is) only on the door you aren't going to select as the other doors are opened!
> the 99% probability of the car being in one of the doors you haven't chosen "condenses" into the door that he didn't open
No, the 99% probability of the car being in one of the TWO DOORS HE DIDN'T OPEN "condenses" into the TWO DOORS HE DIDN'T OPEN, so the same probablility of winning is on any of the closed doors, hence, no advantage or disadvantage in switching.
The problem with the reasoning is that some people think in the outcome as the explanation on Wikipedia:
1 - you pick goat A, Monty shows goat B, switch = WIN 2 - you pick goat B, Monty shows goat A, switch = WIN 3 - you pick the car, Monty shows any of the goats, switch = LOOSE
Which means 2/3 of WINs if you switch, but the correct balance is:
1 - you pick goat A, Monty shows goat B, switch = WIN 2 - you pick goat B, Monty shows goat A, switch = WIN 3 - you pick the car, Monty shows goat A, switch = LOOSE 4 - you pick the car, Monty shows goat B, switch = LOOSE
There are 4, not 3 final outcomes, 2 of which are wins, 2 are loose.
If there are 100, 1000, 10000 doors, in the end you just know the car must be in one of the remaining 2. There is no probablity advantage in switching or not. It's 50%/50% now. It was 1/3 on your fist pick, ant it's 1/2 after Monty eliminates one door, so your chances ALREADY grew from 1/3 to 1/2 - but switching or not won't make them 2/3.
Is the data riding the low-voltage lines (110/220V), or the primary distribution lines (15/13.8kV)? (note: voltages may vary depending on your country)
IIRC, there was an article years ago on Wired saying that data over power lines could work better (only?) in some countries due to the architecture of the power distribution grid.
If it relied in big central transformers, then wide areas covered by a low-voltage grid it would be easier to implement (the continuous "media" would cover a large area).
If it had high-voltage lines all over the place, then a myriad of small-power (~10kVA) transformers to serve low-voltage for every neighborhood/block/building, then it was a problem, because every transformer would have to have a "data bypass" to interconnect the "subnets". (this is the Brazil scenario).
The Wired article was about a guy who claimed to have solved the bypass issue, but had failed to show anything more than pretty numbers.
Was that problem solved, or Australia has the "good case" power grid?
A friend just came back from Japan, where his cousin was paying groceries et all with his cellphone, which had a "sweep-type" fingerprint scanner (and videophone, and fast internet, etc).
I also heard years ago that somewhere in Scandinavia you could pay some soda vending machines just by calling the phone number on its front with your cell phone.
It is interesting to see phone companies grabbing part of the credit card market.
Maybe it'll converge to using your phone/phone account as an ID, driver's license, bank account, credit card, and even to call people!
Instead of money, you'll be paid in talktime credits...
Slashdot Mirror
In the 80's, the nerd thing to do was to write assembly programs for the Sinclair. IIRC, you would convert the opcodes (which we all knew by heart) to ASCII, write a REM line with that, and run it (which I don't recall how).
I'd write loops inside loops, with changing and interdependent step sizes, and it would generate sounds on a FM radio sitting on the computer top (I KNOW my Z80 clock was 3.57MHz, way below FM; it was most probably due to harmonics interference or the radio IF).
I could get beats and interesting disco-like effects, and make alien phony calls. Then computers started shipping with speakers and sound processors and spoiled all the fun.
Agreed, and I guess not only soil cushioned the impact. If the canisters broke loose from the bulk of the section of spacecraft they were in, then they might have hit the ground at a lower speed (there's a velocity point where the G pull downwards equals the aerodynamic drag - "terminal velocity", so it will depend on mass and aerodynamics of the canister).
If they hit the ground inside a big chunk of the spacecract, then the deformation of this chunk also absorved part of the impact.
2,295 Gs of impact... I wonder where they got this number from. A rough estimative based on estimated numbers and estimated facts to produce a catchy figure, or did they actually have accelerometers on the canisters and had planned to crash-test them on the original mission?
Besides, the forces acting upon a tiny, low mass body when decelerating on impact are proportional to the mass (F=m*a), so if you drop a PET bottle with an ant inside from a 10-story building it will most probably survive the fall, and that doesn't make it Atom Ant. If you do the same and suvive getting inside a PET bottle you won't survive the impact, both because you will attain a higher terminal velocity, and because even if you hit the ground with the same speed (i.e. same decceleration), the forces acting on your body parts will be much, much higher.
IANARS (I am not a rocket scientist), but I suppose that debris entering the atmosphere will only burn if they have enough size/initial speed.
Space dust falling on Earth should get in unharmed (if being bombarded by the Sun's UV and all sorts of radiation before getting under the atmosphere blanket can be called unharmed).
> You want to run an interpreted language on a supercomputer? Why not hitch a pony up to a Porsche?
Sorry, you're right. Why would anyone run poorly designed, bloated, unsuited software that could make even the fastest hardware seem slow?
I don't need a W800i. I don't need a W800i. It's from the evil Sony and uses their own memory card standard. I don't need a W800i bundled with a 512MB proprietary MemoryStick Duo memory card. I don't need a W800i bundled with a 512MB memory card and a 2MPx camera. I don't need it.
DANG IT'S NOT WORKING!!! HELP!
s/My Atom|^Atom/MyAtom/g
Well, it's a computer-illiterate scientist's program...
Specially if it comes bundled with Microsoft Visual Basic for Supercomputing - Science Edition, allowing any computer-illiterate scientist to easily put together simulations without needing to care about multiple-processor optimizations. It will have ready-to-use MSAtomicNuclei.ocx, MSAtmosphere.dll, etc.
You only need to go: Create a form. Drag the AtomView object to the form.
Dim MyAtom(1E100)
For i = 1 to 1E100
My Atom(i) = MSAtom.new(x,y)
Next
Atom(0).Nucleus.Split
Who wants to learn C, and all those threads things??
Once this program is trained enough, join it to a noise generator and a "natural selection" algorythm (typing intended), and you'll have an automated hit composer!!
It will eventually compose the "perfect hit", and kill art as we know it.
We're all concentrating on the electronic, switching power supply stuff.
What about the big power guzzlers in the house: Refrigerators and Air conditioners? Those AC motors suck power directly from 220/110 VAC, and isn't AC better for these cheap induction motors?
Although I disagree with "Efficiency. DC loses more energy per foot / mile than AC.", that could be well replaced by:
Efficiency: Low voltages (generally what your home devices use as DC) loses more energy per foot/mile than higher voltages.
This because if your home appliance needs 50W to work, sending 110VAC to it will require small currents through the power cable (~0.5A). Then it's converted to, say, 12VDC and travels a few inches (~4A). (using ~ not to engage in pedantic discussions).
The more current you send over a wire, the more power you dissipate on the wire (or the thicker wire you need), and/or the more voltage drop you have.
Why not sent 110V DC then? It's more expensive to convert DC to DC (from 110 to 12). For AC/DC you can use a transformer then a rectifier (or a switching power supply). For DC, you always need switching.
I think there is at least one reason not to distribute DC inside the house: The same reason car battery contacts get yucky after some time.
AC prevents that galvanic(?) effect to occur on the house outlets.
Wouldn't it be possible to send a specially crafted audio stream to VoIP programs such as Skype to explore eventual vulnerabilities on the audio codec routines?
I know it sounds far-fetched, but you know, jpegs were once safe too. Skype had its vulnerabilities (even on Linux), but were there any on the audio codec?
I hate these "must-have-a-firewall-passage" kind of programs, and I've so far managed to keep them out of my network, but now I'm trying hard to convince my boss not to install Skype on a CAD user's PC "because a customer wants to talk with him"!! CAN'T HE USE A PHONE??!
I agree, but I look like an overloaded miner's donkey with a phone, camera, mp3 player, and PDA all hanging on my belt and pockets. Not to mention the portable game, and a notebook (the right tool for email processing).
...
I don't expect my 1.3Mpix phone to match my 35mm (film) SLR, but it's better than nothing if I need to capture something unexpected and I'm not carrying the real camera. And the two extra lenses. And the tripod. And the silica baggies. And the lens cleaning brush.
Sure the mp3/pda/game/camera/email cell phone may be bigger than a plain phone, but it's smaller than my old Nokia 2160. And if I need small, I just pull the SIM from the V800 and slap it on a T610 or C65. (no, wait, these also have cameras. A T68 then.)
You're right. "mails" is really a shame (since "mail" is already a collective of stuff carried by the postal service). Thanks. I got lost in translation, but I think I'd rather have written "...processing his mail" instead (and I believe that's correct English this time).
Somehow, saying Darwin was troubled processing letters sound to me like he had alphabetization problems...
On "Selected Letters on Evolution and Origin of Species", it is interesting that some of the letters really have a conversation "sequence", considering the long "latency" time between each packet. This was specially true during Darwin's trip, but also when he was at home.
Something like we will experience when exchanging emails with colonies on other planets or solar systems: You write, and your grandson gets the answer.
When a quick response was expected, they'd send a messenger and ask that recipient answered by return mail (and the messenger would wait for the answer to be written).
Also, something as easy as sending an article you wrote for a friend to review (attach/send today) would require that someone hand-copied your writings or that you send the only original and wait for it to come back with the review. You didn't keep a copy on your "sent items".
In the book, Darwin's son says his father was troubled by the chore of processing mails, and spent a lot of time just doing that.
Those were the times.
(please note that I'm not willing to engage a flame war or something like that - I'm really intrigued by this crazy thing, and I confess you got me hanging by a thread)
Maybe problem here is that there are two different comparisons being mixed up. It seems like we both agree in the results, but are arguing about different comparisons.
1st comparison: The odds of winning between: "having the option of switching in the end" vs "not having the option of switching in the end".
2nd comparison: Having the option of switching in the end, the odds of winning between: "switch" vs "not switch".
On the first comparison, we both agree that having the option of switching increases the odds. This is because once you changed the rules to have the switch option, your choices fall from 100 to 2. As soon as the rule is changed, it doesn't matter how many doors you have initially, in the end there will be 2. Opting for this game rule will increase my odds from 1/100 to 1/2 (I understand that for you, it increases from 1/100 to 99/100)
I think that your (and many others) logic goes that when you first pick a door, it has 1/100 probability of being wrong, then in the end it retains this probability so you are faced with the option of keeping this 1/100 win chance or switching to the 99/100 win chance door.
Please consider that I do start with a 1/100 chance with my first pick, but after you open the first door, I have 1/99, then 1/98, then 1/97, etc. This because you are eliminating more and more "wrong" doors. When you get to the 98th door, my chance of being right is 1/2 and so is my chance of being wrong.
I don't agree with your simulation because to calculate the odds of an outcome you need to count all possible outcomes, not just some of them (again, I'm intrigued here, and trying to find the flaw that makes the simulation seem to disagree with "my" theory):
> First go. (Prize is behind door 84). You pick door 1. I open doors 2-83, 85-100. Switcher picks remaining door (84) and gets the prize. Non-switcher gets a goat.
A general formulation could be:
First go. (prize is behind door N). I pick door M (M may be = to N). You open all doors but M and N, and in case M=N you open all doors except D and N.
If M!=N (and there's 99/100 chance of that), there is only one possible combination of doors to be opened by you, so the only outcome is that I end up having to choose between M and N, and switching to N will be the winning option. For each of the 99 possible M!=N, there is one outcome, all saying switch is the winning choice.
Now, when M=N (1/100 of chance), you have 99 options of D when picking the doors you will open, so there are 99 possible pairs in the end, and switching will always give me the goat. For each of the 99 possible D selections, there is one outcome, all saying switch is the bad choice.
So the number of possible outcomes is not 100 as I was considering, but 396. 198 of which I win by switching, 198 of which I win by not switching. Thus, 50% (man I which I remembered enough math to put this in terms of combination/permutation stuff...).
There; it makes sense to me (I'm glad I could figure this out, it was more like a gut feeling when I started...); doesn't it make sense to you? It looks like the flaw is that when counting the possible outcomes one tend to see the case where M=N as having only one possible outcome, when there are many.
(if this thread is closed, we can go on; email me at the_name_of_this_site at rf dot com point br)
I'll take your "simulation".
And here's how it is broken:
To lower the requirement of back-and-forth posting, I'll make my choices in advance, then you can run your simulation as many times as you want (of course, if you run less than 50, results may be non-conclusive).
Since the only comparison needed is between the "Switch" and "not Switch" scenario, here are my choices for each case:
For the "not switch scenario", I'll take door 1. You will agree that I have a 1/100 chance of being right at this moment, before you open the 98 doors. Correct?
For the "switch scenario", I'll take door not-1 (meaning I picked 1, then switched to whatever door was left after you opened the 98 remaining ones). At this moment, before you open the 98 doors, you will also agree that the probability of each door is 1/100, so not-1 gives me 1/100 too.
In any scenario, I don't need to see you opening the 98 doors, because I ALREADY know they won't have the prize, so there is NO NEW information when you open them. Besides, I've decided beforehand if I'm switching or not, so there is no point in seeing you theatrically opening the virtual doors ta-daa!!!
So it's 1/100 for the "not switch" scenario, and 1/100 for the "switch" scenario. No difference between the two scenarios.
Want me to decide "switch"/"not switch" only AFTER you open the 98 doors? Ok. I pick door 1.
Again, I don't need to see you opening the 98 doors, because I ALREADY know they won't have the prize. All that matters is that you will in the end present me two doors, "1" and "not-1". Each one has 1/2 probablility of having the prize. The decision to switch now will give me 50% of chance against 50% of chance if I don't switch. No difference again.
What is your point again? That if I see you opening the 98 doors I will gain knowledge? I'll know that one of the two remaining doors have the prize? Well, I have magical powers so I already know that.
> looking at a picture of the light that left NGC891 24 million years ago. right? wrong?
C A-C523-1213-852383414B7F0147&pageNumber=5&catID=2)
Wrong, due to the accelerating expansion of the universe. If the light of something out there 24 million light-years away hits us now, it doesn't mean that this light left the origin 24 million years before.
It would be true if the universe was static.
There was a VERY interesting article on Scientific American about common misconceptions about the big bang some months ago
(stripped-down web version http://www.sciam.com/article.cfm?articleID=0009F0
explaining things like that.
There are no "less likely" outcomes when it gets to the last decision (between the two remaining choices).
No matter what happens, and no matter how you play your 100-door, 1000-door, whatever simulation, the last choice will be between 2 doors. Whatever happened in the past does not change the odds of the prize to be in one or on the other.
The only difference would be if you had GAINED some true information about the location of the prize - which didn't happen, since you knew from the start that you would end up with two doors to choose from, one having the prize, other not.
Considering past facts as an influence to the choice here is akin to think that because you rolled a dice and it gave you "6", on the next roll the odds of it giving "6" again have magically diminished because the odds of two consecutive "6" are smaller than "6" and "anything".
The odds of two consecutive rolls result in the same number is smaller than resulting in different numbers, but for each individual roll, the odds are always 1/6. The past events won't affect the physics making the "6" face heavier.
The same way, unless you gained knowlegde from the past events (which you didn't, since the final outcome was fixed: you having to pick between the two last doors), whatever happened before your last choice won't magically change your odds of picking the right/wrong door - unless that is a Schrodinger's goat which is in the car-goat state until you actually see it, and is affected by 'spooky action at a distance' "condensing the probablilities" (whatever that is) only on the door you aren't going to select as the other doors are opened!
I beg to differ:
> the 99% probability of the car being in one of the doors you haven't chosen "condenses" into the door that he didn't open
No, the 99% probability of the car being in one of the TWO DOORS HE DIDN'T OPEN "condenses" into the TWO DOORS HE DIDN'T OPEN, so the same probablility of winning is on any of the closed doors, hence, no advantage or disadvantage in switching.
The problem with the reasoning is that some people think in the outcome as the explanation on Wikipedia:
1 - you pick goat A, Monty shows goat B, switch = WIN
2 - you pick goat B, Monty shows goat A, switch = WIN
3 - you pick the car, Monty shows any of the goats, switch = LOOSE
Which means 2/3 of WINs if you switch, but the correct balance is:
1 - you pick goat A, Monty shows goat B, switch = WIN
2 - you pick goat B, Monty shows goat A, switch = WIN
3 - you pick the car, Monty shows goat A, switch = LOOSE
4 - you pick the car, Monty shows goat B, switch = LOOSE
There are 4, not 3 final outcomes, 2 of which are wins, 2 are loose.
If there are 100, 1000, 10000 doors, in the end you just know the car must be in one of the remaining 2. There is no probablity advantage in switching or not. It's 50%/50% now. It was 1/3 on your fist pick, ant it's 1/2 after Monty eliminates one door, so your chances ALREADY grew from 1/3 to 1/2 - but switching or not won't make them 2/3.
Thanks! I wish I could mod you + Informative...
AAARGH! "small-power (~100kVA)" !!!
Is the data riding the low-voltage lines (110/220V), or the primary distribution lines (15/13.8kV)? (note: voltages may vary depending on your country)
IIRC, there was an article years ago on Wired saying that data over power lines could work better (only?) in some countries due to the architecture of the power distribution grid.
If it relied in big central transformers, then wide areas covered by a low-voltage grid it would be easier to implement (the continuous "media" would cover a large area).
If it had high-voltage lines all over the place, then a myriad of small-power (~10kVA) transformers to serve low-voltage for every neighborhood/block/building, then it was a problem, because every transformer would have to have a "data bypass" to interconnect the "subnets". (this is the Brazil scenario).
The Wired article was about a guy who claimed to have solved the bypass issue, but had failed to show anything more than pretty numbers.
Was that problem solved, or Australia has the "good case" power grid?
Thanks! I'll change that right now.
That "established connections" rule came with scripts I've downloaded about 8 years ago (ipchains?), and I never gave it a second thought...
Out goes the ! --syn...
A friend just came back from Japan, where his cousin was paying groceries et all with his cellphone, which had a "sweep-type" fingerprint scanner (and videophone, and fast internet, etc).
I also heard years ago that somewhere in Scandinavia you could pay some soda vending machines just by calling the phone number on its front with your cell phone.
It is interesting to see phone companies grabbing part of the credit card market.
Maybe it'll converge to using your phone/phone account as an ID, driver's license, bank account, credit card, and even to call people!
Instead of money, you'll be paid in talktime credits...