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Of Cosmic Rays, UFO, and SOHO

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posted on Aug, 12 2012 @ 08:06 PM
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reply to post by drakus
 


Just instaled. Nice! Thank you!
For those who haven't looked. This is from Harvard. Several flavors; PC, MAC, Linux, the source is also available.




posted on Aug, 13 2012 @ 01:11 AM
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reply to post by Phage
 





So that kind of beggs the question; "How much did you process that image before you posted your thumbnail?

I did nothing to the thumbnail image except convert the FITS file to a TIF so it could be uploaded to ATS.


New tols can be such a wonderful thing, and they can be a bitch.

You did nothing?
Here is your thumbnail:


Here is the original (converted to PNG)

sharpp.nrl.navy.mil...

Notice the differences?

The "cosmic Ray" streak is in there but requires rather extensive edge enhancement and contrast adjustment as well. This streak is barely above background, and is not even visibleuntil serius modification are done.

So ,,, you did not post "raw" data, you posted highly processed data.



edit on 13-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 13 2012 @ 03:18 AM
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reply to post by AnthraAndromda
 


I think we have a problem. In the image you referenced above, and provided a link to sees a bit odd.
I don't see anything odd. The wide dynamic range makes it necessary to adjust the levels. You might find this program more useful than Gimp. It's better suited to the wide dynamic range.
www.spacetelescope.org...



So ,,, you did not post "raw" data, you posted highly processed data.
My apologies, when you said thumbnail I thought this is what you meant:

I would consider the images I posted in this thread to be detail views rather than thumbnails. That's the source of my confusion.
The clickable thumbnail (first posted in your other thread) was not processed other than converting to a TIF. The details I posted had the levels adjusted (as well as being cropped). You can call that "highly processed" if you wish but I don't. Do you think I was trying to conceal something? I posted the link to the raw data. I expected you to look at it.


Dynamic range of what? The hardware out there will handle 256 levels of gray, that's it.
The dynamic range of each pixel.


if it's what I'm thinking of, is an automatic process performed by the CCD to eliminate any substrate noise it may have.
You're thinking wrong. As I said, background subtraction is processing which is performed on the FITS data. It is not performed by the CCD.


I also doubt that there will be enough "added" data to make a difference. So, in the end we will only have spent a lot of time producing what we already have.
There is no "added" data in the source but could very well be data loss in the processed images. As I said, I didn't know what you were going to do so I had no way of knowing what effect it would have on your plans. All I did was provide some information which you obviously did not have.


Oh, but I do understand the nature of that processing. It brings up the impossibly weak "signals" so that we can work with them.
Apparently you don't. You don't know what process of background subtraction is or involves.
edit on 8/13/2012 by Phage because: (no reason given)



posted on Aug, 13 2012 @ 03:19 AM
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reply to post by AnthraAndromda
 


You will notice that my calculation is different from yours; cosmic rays must interact with the sensing portion of the device to be recorded, thus those entering from the back aren't "seen" as they do not reach the sensing areas
Cosmic rays can easily penetrate the entire thickness of the CCD. They produce a track in the photoactive layer no matter which side they originate from.


60 microns!?! Seriously? Soho isn't that old. A modern device has an active layer more on the order of 0.2 to 0.3 nanometers, much less than you 60 microns (60,000nm).
SOHO was launched in 1995.
Only 0.2 to 0.3 nanometers? That's 2,000 times less than the wavelength of visible light. That's remarkable. Can you provide some documentation for that? I'd like to find out more about how that is done and how such a thin layer would react to light. Can you explain how you came up with an angle of incidence of "4 or 5 degrees"? It seem that with a pixel width of 21 microns and a depth of 0.3nm we would only hit one pixel with such an angle, not create a track across multiple pixels.


I suspect your numbers are reasonably accurate, given the huge error.
That would seem to be an oxymoron. I suspect I'm missing your point.


However, your estimation of the probability is wholly inaccurate.
Not according to experimental evidence (as provided). I read it to show the probability of a track of 32 to 55 pixels in length as being 0.008. That matches my figures.


The probability of a cosmic ray having the "correct" path remains the same 1:255,102.04
No, it doesn't. There are a range of values for elevation which will produce a track, not a just a single radian. This increases the probability that a track will be produced beyond 1/360 (or 1/180, if that's what you used). You have not produced the odds for the "correct" path, you have produced the odds for matching path, a path of the same length and azimuth.


So yes we have two "tracks" each with the sae probability, which hasn't changed from it's original 1:255,102.04 (your wee song and dance was mostly deleted
Only if the first track is there can we then calculate the probabilty that the second one will meet the required conditions. How can a single track be "special"? It's like calculating the odds of rolling snakeyes with one die. It's a null value. An again, the odds you calculated are for a matching path, not an intersecting path.


Actually "vectors" and "azimuths" are both incorrect, we shall call them "paths". And no we are not working on a single plane.
Vector is incorrect, azimuth is not incorrect because the azimuth is the direction of the path. Yes we are working on a single plane. The vertical plane only determines the length of the track. You made no mention of the importance of the length of the track therefore we are only dealing with the azimuth of the track. The azimuth is the direction of the path in the horizontal plane so it is the correct term but path will also work. In order to intersect the only requirement is that the azimuths of the two paths are different.


You like the term "flus" which is really just anther way of quantizing the probability.
Not really, it's actually a measure of the level of radiation. But flux levels will affect the odds of any particular track occuring. The more events, the more likely the event is to occur. The more times you roll the dice the better the odds of getting snake eyes.


This value was found by "counting" cosmic ray strikes on 86 SOHO images
Interesting. I assume that you were counting paths rather than strikes. What was the minimum path length you used? This would relate to the question of the definition of "special". But if you used the browse images you would have missed (at the minimum) any strikes behind the occulter disk and outside the optical path. On the C2 images that amounts to about 10% of the image.


You are aware that it is not possible to have probabilities greater than 1, right?
Yes. I am aware of that and that was a blunder on my part. The problem came from the way I applied the number of hits to the odds of an intersecting track occurring. The proper formula would be 1-(1-p)^n, where p is the probability and n is the number of "tries". This revises the probability for an intersecting track (assuming that there is a track for it to intersect) to 0.938. Close to certainty but not quite because; a) there must be at least two tracks of the required length (between 32 and 55 pixels) and b) the pair must not have the same azimuth.

edit on 8/13/2012 by Phage because: (no reason given)



posted on Aug, 13 2012 @ 10:31 AM
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reply to post by Phage
 


Well, then perhaps I could send a bit of an apolgy your way. My "software engineer" thinking can get in the way at times. In my world, anything that isn't original "bitmap" is a "thumbnail". This is likely due to the fact that to build one of these "things", I call the ".tThumbnail( ... )" method in the graphics library. Hence; if it's not original is a "thumbnail". I didn't even "think" that there might be some confusion here.

Anyway, I shall attept to be more clear.

Also, thank you for the App link. I have DS9 fm Harvard; nice software, reasonabley well thought out, runs well, qick, intuitive. A bit short on documentation, not much by way of "tool tips" either. But, still a god aplication. I'm still "playing" with it.

I've already installed from your link and will get to know it througout the day.

Are there, to the best of your knowledge, any threads here (on ATS about this software I know I can do a search, but, I'm being lazy, well sort of anyway. I don't have much real wrk this morning and I want to play with my new toys
ANyway, if there are no threads (and I'll do my own seach soon anyway), perhaps there should be. Whie either of these ay not solve the "cosmic ray" question, either can go a very long way to deterining"if" soething might be considered "solid".or an object.

By the way, I've found soething that "may" be an actual object, and not a planet, or Terrestrial "hardware". Although, I don't think either of the programs will determine "what" it is that I have found. I'll post ore when I have a better handle on it.

Anyway, Thanks for the software link.

edit on 13-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 13 2012 @ 12:12 PM
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reply to post by Phage
 



The proper formula would be 1-(1-p)^n, where p is the probability and n is the number of "tries".


And this is the formula for what exactly?

By what mechanism is the probability given such great help?

I'm trying to understand "why" the probability would increase from such a small value to almost 1.

edit on 13-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 13 2012 @ 12:37 PM
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reply to post by AnthraAndromda
 


And this is the formula for what exactly?
The formula for the probability that a single event of probability p will occur given multiple "tries". Simplest case scenario:

The probability of flipping a coin once and getting heads is 0.5:
1-(1-0.5)^1 = 0.5

The probability of getting heads at least once in two tosses:
1-(1-.5)^2 = 0.75

The probability of getting heads at least once in three tosses:
1-(1-.5)^3 = 0.87

The probability of getting heads at least once in four tosses:
1-(1-.5)^4 = 0.94

The probability of getting heads at least once in six tosses:
1-(1-.5)^6 = 0.98

And so on. The more tries, the greater the chances of getting at least one occurence but the probability will never reach 1.0


 

In the case of cosmic ray tracks we have a probability 0.0079 for an intersecting azimuth. We have 350 "tries" in a given frame. The probability of getting at least one intersecting azimuth:
1-(1-0.0079)^350 = 0.938



posted on Aug, 13 2012 @ 08:22 PM
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Originally posted by Phage
reply to post by AnthraAndromda
 


And this is the formula for what exactly?
The formula for the probability that a single event of probability p will occur given multiple "tries". In the case of cosmic ray tracks we have a probability 0.0079 for an intersecting azimuth. We have 350 "tries" in a given frame. The probability of getting at least one intersecting azimuth:
1-(1-0.0079)^350 = 0.938


Except we are working with two events; each path is a separate, independant event. Thus the joint probability becomes; P(A and B) = P(A) * P(B). Giving us the results I've already posted.

The probability you gave is correct for a single event model, but does not apply to a multiple event model. Your probability is the "probability" that another event will occur, remember your system has only one event. When there is more than 1 event, the probability of each individual event ust be accounted for. (BTW; when I say "event" you might want to think "event type" i.e coin flip, cosmic ray on specific path, etc)


edit on 13-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 14 2012 @ 12:29 AM
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reply to post by AnthraAndromda
 

I understand the point. I look at it this way; If there is no "first" track at all there can be no relationship between two tracks. It is a null event. So calculating the odds doesn't really make sense in my point of view. But, if you like, we can include the odds for the first track occurring. It does reduce the odds but not to an extreme degree.


Thus the joint probability becomes; P(A and B) = P(A) * P(B). Giving us the results I've already posted.
The formula is correct but the variables you used are not. Again, you are calculating the probability for an exact duplication of the first track, this is not a requirement for your definition of an event. Your probability for the first track is also too restrictive since there is a range of values for the elevation of the vector.

If we do include that null case in which the first track does not exist this is how it is calculated:

1) Probability of the first track (32-55 pixels in length) existing: 0.008
2) Probability of a second track of the same length range existing: 0.008^2 = 0.000064
3) Probability of that track having an intersecting azimuth: 0.000064 x 0.99 = 0.000063
4) Probability of that pair of tracks appearing in any frame: 1-(1-0.000063)^350 = 0.0219


I think I understand part of the reason for your incredulity. The trouble is that you have not rigidly defined the requirements of a "special" event. As I have pointed out these are the probabilities for two tracks of lengths between 32 and 55 pixels with extended intersecting azimuths appearing anywhere on a frame. By stating an angle of "4 or 5 degrees" you implied tracks of this length as stipulated by a nominal thickness of the photoactive layer. The only other stipulation you stated was the intersecting azimuths. This is a very broad definition so we end up with seemingly high probabilities. By further refining these criteria the probability may (or may not) change significantly. So, I used your three "unusual" events to apply some constraints to the definition of "special".

I am assuming you have only applied a "pixel resize" which would maintain the original pixel count. Please inform me if this is incorrect. This is why links to your source data would be useful, you did not provide dates and times for the images you used.. Using your "thumbnails" I see these criteria:
1) The tracks should be between 80 and 150 pixels in length. This constraint is quite important. The probabilities increase quite a bit with decreased lengths.
2) The angle of intercept should be between 43º and 118º. What is important here is the span of the range, obviously. The wider the range of acceptable angles the higher the probabilities of an event.
3) The nearest endpoints should be no greater than 130 pixels apart. Of course, the closer the endpoints have to be, the lower the probabilities for an event.

With these constraints we find:
1) Probability of one track of specified length occurring: 0.006 (angle of incidence = 1º to 2º)
2) Probability of a second track of such length occurring: 0.000036
3) Probability of two tracks of that length having an intercept angle within the specified range: 0.0000151
4) Probability of those two tracks occurring with at least one endpoint of each within the specified proximity: 0.0000002
5) Probability of those two tracks appearing in any frame: 0.00007.

With about 100 frames per day that means we would see a special event an average of every 143 days. But a very slight change in flux levels has a great effect on this figure. If, instead of 4/cm2/sec, we use 5 it drops to 114 days. With 6 we see 95 and 8 gives us 71 days. The effect of the flux level is dramatic enough so even a slight increase in Solar cosmic rays is highly significant. A Solar event of too low intensity to qualify as a proton storm would drastically change the probabilities on any given day. This would be another reason the dates and times of the images you used might be useful.

edit on 8/14/2012 by Phage because: (no reason given)



posted on Aug, 14 2012 @ 11:21 AM
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Originally posted by Phage
reply to post by AnthraAndromda
 

I understand the point. I look at it this way; If there is no "first" track at all there can be no relationship between two tracks. It is a null event. So calculating the odds doesn't really make sense in my point of view. But, if you like, we can include the odds for the first track occurring. It does reduce the odds but not to an extreme degree.


I'm guessing this pretty much details your understanding of probability




Thus the joint probability becomes; P(A and B) = P(A) * P(B). Giving us the results I've already posted.


The formula is correct but the variables you used are not. Again, you are calculating the probability for an exact duplication of the first track, this is not a requirement for your definition of an event. Your probability for the first track is also too restrictive since there is a range of values for the elevation of the vector.


No, Phage. That "calculation" is for any event. You are misinterpreting my "definition" of an event.
Event: Any Cosmic Ray.



If we do include that null case in which the first track does not exist this is how it is calculated:

1) Probability of the first track (32-55 pixels in length) existing: 0.008
2) Probability of a second track of the same length range existing: 0.008^2 = 0.000064
3) Probability of that track having an intersecting azimuth: 0.000064 x 0.99 = 0.000063
4) Probability of that pair of tracks appearing in any frame: 1-(1-0.000063)^350 = 0.0219


Firstly; there is no "null event"! The logic that if the first event never occurred, is faulty. If the first event does not occur, there are no events..



I think I understand part of the reason for your incredulity. The trouble is that you have not rigidly defined the requirements of a "special" event. As I have pointed out these are the probabilities for two tracks of lengths between 32 and 55 pixels ...


You are attempting to apply your definition of "special event" prematurely. At this point in the discussion, we have not reach a point where any of that even matters. Yet you insist on applying it; incorrectly.

And this makes all the rest of your comments irrelevant.

All we are talking about is the existence of two cosmic ray in the same event period with any specific path.

Anyway, I have already shown the formula for the calculation of multiple events during an observation period. There is no amount of misunderstand,creative mathematics, nor, indeed, anything else that can change the mathematical facts here.

1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
2. Probability is calculated by specific equation: P = P(a) * P(b) * ... p(n).
3. The probability of two cosmic rays intersecting is: 0.0000000038950081 or 1 in 256,738,875.5879
4. The probability of three such separate "events" (images from 3 observations) is
5.9091511031674153381441e-26 or 1 chance in 16,922,904,534,696,723,631,056,456.459405 (you will notice that this number has grown; the reason for this is to accommodate your 360 X 360 model. You insisted on more "possible" event sources, so you got it: 1:16 Septillion. But hey, it could still happen


I think this well establishes my original contention, and that some of these things are not cosmic rays.
despite what any of us may want, the math dictates reality in this case. You would do well to accept reality.

edit on 14-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 14 2012 @ 11:39 AM
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reply to post by AnthraAndromda
 


1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
So we should only see a cosmic ray track once in 160 days of images?


3. The probability of two cosmic rays intersecting is: 0.0000000038950081 or 1 in 256,738,875.5879
That translates to one in about 7,000 years of images. Yet intersecting tracks are easy to find.


So what do you do rather than reanalyzing your approach? You conclude that they must not be cosmic rays. The problem is your math is wrong and your science is none.


You are attempting to apply your definition of "special event" prematurely. At this point in the discussion, we have not reach a point where any of that even matters. Yet you insist on applying it; incorrectly.

I had to do so since you haven't. I used the three images you provided (still no dates and times) to try to deduce what is is you call "special".


edit on 8/14/2012 by Phage because: (no reason given)



posted on Aug, 14 2012 @ 12:13 PM
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reply to post by Phage
 




So what do you do rather than reanalyzing your approach? You conclude that they must not be cosmic rays. The problem is your math is wrong and your science is none.

You are attempting to apply your definition of "special event" prematurely. At this point in the discussion, we have not reach a point where any of that even matters. Yet you insist on applying it; incorrectly.


I had to do so since you haven't. I used the three images you provided (still no dates and times) to try to deduce what is is you call "special".


You think I haven't been doing this all along? "Reanalyzing" the problem, trying to find other factors, other method. Sir, you know nothing of engineering!

No, Phage, my Math is correct, and my science is as well. All you have done throughout this discussion is misapply mathematics, apply logic that isn't quite (logic that is), and attempt to shift the whole misunderstand onto me. That won't work!

To paraphrase Issac Newton; "Sir, I have studied these matters, you have not" These matters being; Mathematics , physics, Engineering, and perhaps more importantly; Data analysis.

All that you have done here is make the issue increasingly less tenable for you, and make the probabilities work better for me.

Unless you can bring something "real" to the table, I think we are done here.

The probability of some of these SOHO image "formations" being "cosmic Rays" is so small that the idea isn't worthy of thought or consideration. What these things actually are is still a mystery, as there is not enough information to determine that, but, we can say with a good level of confidence; they are not cosmic Rays

You have lost this one, please do so with grace.



posted on Aug, 14 2012 @ 12:19 PM
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reply to post by AnthraAndromda
 


1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
So we should only see a cosmic ray track once in 160 days of images?



No, Phage, my Math is correct, and my science is as well.

Sure it is. I think you should report your findings to Joseph Gurman.
I'm sure he would be fascinated to discover that what he has be thinking are cosmic rays can't be because your math proves it.

edit on 8/14/2012 by Phage because: (no reason given)



posted on Aug, 14 2012 @ 12:38 PM
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Originally posted by Phage
reply to post by AnthraAndromda
 


1. The probability of a event path is still: 0.00006241 or 1 in 16,023.073225444640282006088767826.
So we should only see a cosmic ray track once in 160 days of images?


Again, your math and logic has broken down.

In as much as there are 129,600 possible "paths", I should think we would see many cosmic rays. However, that says nothing about their "path".

In view of the mechanics, logic, science, and Mathematics; My computations stand.




No, Phage, my Math is correct, and my science is as well.

Sure it is. I think you should report your findings to Joseph Gurman.
I'm sure he would be fascinated to discover that what he has be thinking are cosmic rays can't be because your math proves it.


Perhaps you should ask him! Though, I'll bet he will agree with me. (seriously, any small application for PC, absolutely free, and I'll give you all the rights too. IF your "man" proves me wrong)

Oh, and, my "math" doesn't "prove" anything; it only demonstrates, and illustrates probability.

edit on 14-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 14 2012 @ 12:44 PM
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Y'all have been eading this, I'm sure soe are not convinced either way. SO I will suggest this; Take this discussion to virtually any high school math teacher, or better to a local college and ask someone who is a bit better versed in this than either of us (although I should tell you ; that I have been using these methods and techniques (probability) professionally for 30+ years, hasn't failed me yet.).

But, y'all should get another opinion.



posted on Aug, 14 2012 @ 02:40 PM
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AnthraAndromda, I think Phage does not need a teacher to argue his case convincingly, also in this thread.

Instead, the problem is that your stuff got pwned beyond recovery already on page 2, but you are seemingly incapable of understanding that, or psychologically incapable of admitting it.

You shan't be allowed to conclude that your case is as credible as Phage's when Phage has demonstrated he is correct, and you are not.

Let it go and move on.



edit on 14-8-2012 by ScientificUAPer because: typo



posted on Aug, 14 2012 @ 03:02 PM
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Originally posted by ScientificUAPer
AnthraAndromda, I think Phage does not need a teacher to argue his case convincingly, also in this thread.

Instead, the problem is that your stuff got pwned beyond recovery already on page 2, but you are seemingly incapable of understanding that, or psychologically incapable of admitting it.

You shan't be allowed to conclude that your case is as credible as Phage's when Phage has demonstrated he is correct, and you are not.

Let it go and move on.



edit on 14-8-2012 by ScientificUAPer because: typo


Phage has not demonstrated his correctness. All he has demonstrated is a lack of expertise in probability, a failure to understand the mechanics, and physics. He has at no time demonstrated even a "flicker" of understanding.

You are showing the same level of diminished comprehension. Instead of argueing this, why don't you go and get another "expert" opinion.

Most of Phage's argument can only work in a single event model, and we are working with a multi=event model. So tell me; just how does this equate to being right?

The fact that Phage did not recognize this as a multi-event model in the beginning illustrates that Phage is not an engineer. The way he tried to twist probability (probabilities greater than 1, inappropriate calculation method, etc) shows that he is neither a mathematician, nor a physicist.

But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.


edit on 14-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 14 2012 @ 03:20 PM
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Originally posted by AnthraAndromda
..
But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.

Your stuff got busted, end of story. The fact that you claim to be an expert doesn't make it any better.


edit on 14-8-2012 by ScientificUAPer because: typo



posted on Aug, 14 2012 @ 03:41 PM
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Originally posted by ScientificUAPer

Originally posted by AnthraAndromda
..
But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.

Your stuff got busted, end of story. The fact that you claim to be an expert doesn't make it any better.


No sir it was not "busted".

I have shown, with proper mathematics, the reality of this. You may disagree if you like, however, be sure of yourself first. Get another opinion.

If you had actually read, and understood what was written, by both of us, you would quickly see that Phage was never on solid "probability" ground. Perhaps you might like to do your own research. You could start with an education in math and probability. As I have already stated; probability has been my friend for 35+ years, and has never failed.

ETA: I nearly forgot this. Do you remember a website called uslottery.com? It may have been before your time. It was a "gamging" site run by an Idaho Indian Tribe, it was also my site (in an engineering sense; "I owned it"). Unfortunately, I also had to kill it when the feds decided you can't do that over a domestic network back in 1998.

Do you know what area of mathematics is primarily used in the gaming industry?

edit on 14-8-2012 by AnthraAndromda because: (no reason given)



posted on Aug, 14 2012 @ 04:03 PM
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reply to post by AnthraAndromda
 





But, I'll tell you what; get someone who is at least as expert as I, and we will see, what we will see.


Don't take this the wrong way, but what are you an expert in?




Most of Phage's argument can only work in a single event model, and we are working with a multi=event model. So tell me; just how does this equate to being right?


And how does this equate to being wrong?

So what your saying is because he is using this as a single event model he is wrong,but have you actually looked at this as a single event because if not you cannot say he wrong.



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