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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.
Originally posted by tsurfer2000h
reply to post by AnthraAndromda
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.
Originally posted by tsurfer2000h
reply to post by AnthraAndromda
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.
Well I know this wasn't a reply to me,but I did get another opinion and he said you are wrong..
I asked Phage.
(not really I just couldn't resist)
Seriously though it looks as though you have a need to be right and those who don't agree with your conclusions are wrong, why is that?
Oh? Can you show us all just where I was wrong?
For this to occur there must be two events , representing each "path".
www.abovetopsecret.com...
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
Originally posted by Phage
reply to post by AnthraAndromda
For this to occur there must be two events , representing each "path".
Yes, in order for it to occur there must be two tracks otherwise there is no possibility for the intersection of two tracks. Without 2 tracks it is null. But you didn't seem to notice that I changed my calculations to include the probability of that first track being produced. So, if you want to include images with less than two tracks you can.
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
No. It is the probability that the path will occur over a range of elevation values. The length of the track is a function of the elevation of the path. With an elevation of 90º the track will be one pixel long. With an elevation of 2º the track will be 80 pixels long. This is why you need to specify a track length (or range of lengths). The length of the track affects the probabilities.
And this is the probability that a path will occur at a specific elevation angle. Not rhe probability that an intersecting path. I know you think you have, however, the length of "track" isn't important.
There are two component to the path, the vertical and the horizontal. To calculate the probabilities of a path creating a track only the vertical component is considered. The azimuth comes into play only if those two tracks exist. Without those two tracks there can be no intersection.
" Probability of the first track (32-55 pixels in length) existing: 0.008" How is this possible since the probability of the path that creats the track is 0.0000078?
In order for the "extended" tracks to intersect (as you did in your example) the only requirement is that the azimuths are different (otherwise the tracks are parallel). This means there are 359 degrees available for that intersection to occur. 359/360 = 0.99 (or 179/180).
" Probability of that track having an intersecting azimuth: 0.000064 x 0.99 ..." Just where did the "0.99" come from?
To raise it to a power indicates "tries".
No. Please note that item #2 calculates the probability of two tracks existing.
The roblem here is that ou are still attempting to apply a single event model to a multi-event system.
Yes. Which means there is a 0.02 probability that it will happen. In order to calculate the probability that an event will occur over multiple tries it is necessary to consider the probability it will not occur. We then "not" that result to produce the probability that it will. wow.joystiq.com...
So there is a 0.02 probability that your event that won't happen won't happen?
Originally posted by Phage
reply to post by AnthraAndromda
We've been all through this.
And this is the probability that a path will occur at a specific elevation angle. Not rhe probability that an intersecting path. I know you think you have, however, the length of "track" isn't important.
No. It is the probability that the path will occur over a range of elevation values. The length of the track is a function of the elevation of the path. With an elevation of 90º the track will be one pixel long. With an elevation of 2º the track will be 80 pixels long. This is why you need to specify a track length (or range of lengths). The length of the track affects the probabilities.
There are two component to the path, the vertical and the horizontal. To calculate the probabilities of a path creating a track only the vertical component is considered. The azimuth comes into play only if those two tracks exist. Without those two tracks there can be no intersection.
" Probability of the first track (32-55 pixels in length) existing: 0.008" How is this possible since the probability of the path that creats the track is 0.0000078?
No. Please note that item #2 calculates the probability of two tracks existing.
Yes. Which means there is a 0.02 probability that it will happen. In order to calculate the probability that an event will occur over multiple tries it is necessary to consider the probability it will not occur. We then "not" that result to produce the probability that it will. wow.joystiq.com...
So there is a 0.02 probability that your event that won't happen won't happen?
We are talking about the probability of a track being produced at all. We must start with that. The only factor which determines if a track is produced is elevation. Your "original probability" includes azimuth and elevation. We are only concerned with elevation. The probability of one track being produced is 0.008.Therefore the probability of two tracks being produced is 0.008^2 = 0.000064.
If you insist on your 0.008, then we have to multiply it by the original probability
No.
It appears that you may be thinking that the detector is at the center
But according to you it is the intersecting tracks which are of interest. From the OP:
Non-intersecting tracks are permitted
It is far more probable that tracks will be intersecting than not intersecting because there is only one azimuth which will not produce an intersect, the azimuth which is identical.
That is; in all three of the second set of images there are two cosmic rays with intersecting vectors. This makes the whole proposition much less tenable.
You are doing the same thing (using incorrect values). You are attempting to calculate the single probability for two events occurring. I guess you don't realize that in doing so you are, in effect, calculating the probability of a single event. That event being the occurrance of two intersecting tracks.
And here you revert to the single event model.
It has nothing to do with game theory but it has everything to do with probabilities just as rolling dice has everything to do with probabilities. Would an explanation from a programming point of view help? wpquestions.com...
While this equation may apply in "gaming theory", it does not apply here.
Originally posted by Phage
reply to post by AnthraAndromda
We are talking about the probability of a track being produced at all. We must start with that. The only factor which determines if a track is produced is elevation. Your "original probability" includes azimuth and elevation. We are only concerned with elevation. The probability of one track being produced is 0.008.Therefore the probability of two tracks being produced is 0.008^2 = 0.000064.
But according to you it is the intersecting tracks which are of interest. From the OP:
It is far more probable that tracks will be intersecting than not intersecting because there is only one azimuth which will not produce an intersect, the azimuth which is identical.
You are doing the same thing (using incorrect values). You are attempting to calculate the single probability for two events occurring. I guess you don't realize that in doing so you are, in effect, calculating the probability of a single event. That event being the occurrance of two intersecting tracks.
It has nothing to do with game theory but it has everything to do with probabilities just as rolling dice has everything to do with probabilities. Would an explanation from a programming point of view help? wpquestions.com...
It is possible to draw many traces in that area where no traces are parallel, nor intersecting.
No. That is the probability of each path. The probability of all paths is 1.0.
And, you are failing to understand that the probability for all paths is the same (0.0000078)
In effect you are. You are using the probability of producing one particular path. That's what the "1" in 1/360 represents, one degree out of 360. Just like the "1" in 1/6 represents one particular number on the die. The odds of rolling doubles is (1/1) x (1/6) = 1/6. The first probability is 1.0 because no matter what you do, you will get a number. The second probability is 0.167 because you have 1 chance out of 6 to get a number which matchs that first number. If you want to roll snake eyes (or any other specified double, the odds are (1/6) x (1/6) = 1/36. The point is, for doubles all you have to do is roll the first die. The combined odds do not depend on what that number is.
I am not "trying to roll snake-eyes", I'm simply producing another path.
That would make sense. However sometimes a part of the compression routine is employed to delete the area of the disk from the data for bandwidth considerations. It only seems to be there in older images though. But it's easy to determine when it's in place because the pixels have a value of 0. BTW, if you're interested, in the FITS files the image data begins at 21C0. Each pixel has two bytes, making each pixel row 2048 bytes long.
Would you agree that the occulting disk will block all light (photons)? If so, we may have an area of the image within which we can determing cosmic ray flux.
Originally posted by Phage
reply to post by AnthraAndromda
No. That is the probability of one path. The probability of all paths is 1.0.
And, you are failing to understand that the probability for all paths is the same (0.0000078)
In effect you are. You are using the probability of producing one particular path. That's what the "1" in 1/360 represents, one degree out of 360. Just like the "1" in 1/6 represents one particular number on the die. The odds of rolling doubles is (1/1) x (1/6) = 1/6. The first probability is 1.0 because no matter what you do, you will get a number. The second probability is 0.167 because you have 1 chance out of 6 to get a number which matchs that first number. If you want to roll snake eyes (or any other specified double, the odds are (1/6) x (1/6) = 1/36. The point is, for doubles all you have to do is roll the first die. The combined odds do not depend on what that number is.
The probability for any path is 0.0000078. Although the probability for "all" paths is most certainly not 1.
No. That is not what I said. I said the odds of rolling doubles is 1:6. Doubles is any pair. Snake eyes is a particular pair. A particular pair. The odds of rolling the first ace is 1:6, the odds of rolling the second ace is 1:6. The odds of rolling both aces is 1:36. The odds of rolling a non-specific pair is 1:6 because the odds of rolling any number with the first die are 6/6. This is what you seem to have a problem grasping.
No, again you have missed the mark. the probability for snake-eyes is 1:36, as you have shown. Unfortunately only a few words eariler you said it was 1:6 You are contradicting yourself.
No. You don't understand the problem. I have been searching for a way to get you to understand the mistake you are making. You are calculating the odds for "snake eyes", not the odds for "doubles". You should be calculating the odds for "doubles". As a starting point.
You have failed to construct an accurate model of the "problem", and now seem to be "searching" for something that will overturn my argument. Problem with that is, you won't find anything.
What you "think" I am calculating is not the probability for "snake-eyes", nor is it "doubles". I am calculating the probabilities of a multi-event model.
The first element has an equal probability.
Did I somewhere imply that any given vector had a different probability than any other? I don't believe I did, but in the off chance I was interpreted that way; all vectors have equal probability.
Originally posted by Phage
reply to post by AnthraAndromda
By the way: the probability for rolling doubles is still 1:36 whether you predefine the "doubles" you want or not. The probability is still 1:36 The first element has an equal probability.
That's what I said. There is only one definition for doubles; a pair of any two numbers. And the odds of rolling doubles is 1/1 * 1/6 = 1/6. Snake eyes is a special case of doubles with reduced odds because you do not have equal probabilities of getting an ace with the first die.
And there is an equal probability that a cosmic ray will follow any path. Remember? You said it yourself:
Did I somewhere imply that any given vector had a different probability than any other? I don't believe I did, but in the off chance I was interpreted that way; all vectors have equal probability.
www.abovetopsecret.com...edit on 8/15/2012 by Phage because: (no reason given)
So which is it? 1:6 or 1:36?
www.thinkingbettor.com...
For some reason, a lot of people have trouble grasping that concept. The chances of rolling doubles with a single toss of a pair of dice is 1 in 6. People want to believe it’s 1 in 36, but that’s only if you specify which pair of doubles must be thrown.
No one will accuse you of having an underactive imagination. I neither said nor implied any such thing.
And yes yu did state that some paths have greater probability. (it might have been closer to an implicatin, but, it seemed rather "overt" to me)
You are quite simply, wrong.
Your concepts of this, the logic you are using, quite simply, don't work.