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Author Topic:   Discontinuing research about ID
Dubreuil
Member (Idle past 3041 days)
Posts: 84
Joined: 04-02-2015


Message 181 of 393 (756340)
04-18-2015 3:42 PM


RAZD writes:
So P.WeC- and P.WeC+ are not appearances -- does the person (P.WeC in this case) need to be present for +/- to occur?
No. For example:
P.Da: "I feel so sick"
P.Ri: "Does this medicine help?"
P.Da: "Yes, it helps a lot. I feel better now"
P.Ri: "I will go to P.WeC. This medicine will also heal him."
is: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+
(This row of appearances would break the pattern)
RAZD writes:
Can you have P.WeC without it being * P.WeC, P.WeC- or P.WeC+ ? I'm just trying to understand your marking system.
There are only *P.WeC, P.WeC- or P.WeC+. P.WeC without *,+,- is not defined in the notation and only describes the person.
RAZD writes:
M1, M2, M3, M4, M5, M6, M7, M10, M11, M12, M13, and M14 -- M8 and M9 seem to be missing from the "pattern" table 4
Yes, M8 and M9 are missing. There are only 12 M's.
RAZD writes:
Would I be correct in thinking that anything not covered by this list is ignored in the "pattern" derivation and applications?
Yes.
RAZD writes:
The question you did not answer was why\how should I conclude that these three hypothetical episodes should be part of the same pattern:
A data source of only three hypothetical episodes is to small to create a non-arbitrary pattern. Therefore it is not possible for only three episodes.
RAZD writes:
and how do I correct D and E so they fit the pattern?
Episode D:
E2: Not *P.WeC, but *P.Ya
E7: Not *P.Wo, but *P.Da
E8: Not *P.Da, but M6
E9: Not *P.BW, but *P.Ya
E12: Not M2, but *P.BW
Episode E:
E15: Not *P.BeC, but M4
RAZD writes:
I plan to, but as an initial comment you say that the calculation is based on the observed incidents rather than on an accounting of the possibilities, yes?
You can also approximate a probability by accounting possibilities of single appearances. For E1 there are 25 occurrences that fit with the pattern (P.Al, P.BW, P.Da, P.LF, P.Pi, P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, P.Al-, P.BW+, P.Tr+, P.WeC-, P.BeC, P.Ri, P.Ya, M4, P.BW-, P.Da) and 26 occurrences that break the pattern (*P.En, M3, M10, M11, M12, M14, P.Al+, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Pi-, P.Ri+, P.Ri-, P.Tr-, P.WeC+, P.Wo+, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya-). Altogether the probability would probably be about 0.5 that random data fits with the pattern. The test on random data resulted in a probability of 0.625. But at (00:00) the probability is 0.95. That is the big difference that can't be explained by chance.
RAZD writes:
If I throw a di 10 times and only get numbers between 1 and 3, can I calculate with confidence the probability of what the next throw will be? Would you calculate that probability based on the number of 1's the number of 2's and the number of 3's in those 10 throws to predict the next toss? Certainly I can take the results of those 10 throws and put them through standard probability calculations while ignoring possibilities that did not occur during the data gathering phase, yes?
If we assume that we don't know what shape the di was -- Tetrahedron (four faces), Cube or hexahedron (six faces), Octahedron (eight faces), Dodecahedron (twelve faces), Icosahedron (twenty faces) -- then the only evidence we have for the possibilities is what is observed during the data gathering phase, yes?
Let's say we know the die is a Tetrahedron (four faces) or an Octahedron (eight faces) and if we throw it 10 times we only get numbers between 1 and 4. The probability that the di is an Octahedron can then be calculated with the probability mass function:
The resource cannot be found.
Probability of success: 0.5
An unbiased Octahedron (eight faces) will only show numbers between 1 and 4 every second time (1-4, 5-8).
Successes: 10
Trials: 10
Result: 0.00097656 = 1:10^3
Therefore the next throw will also show a number between 1 and 4 with a residual uncertainty of 1:10^3. More throws are necessary to increase the certainty. If we throw the di 20 times and we only get numbers between 1 and 4, then the probability is 0.00000095 = 1:10^6. The found pattern was tested 47 times and the beginning was very different to the random data source (15/24,45/47).
RAZD writes:
Essentially you are assuming that the three observed results are the only possibilities. Nor do we know if the di was weighted so we are assuming that the results are not biased.
For the probability calculation was assumed that the two observed results are the only possibilities, the pattern fits or it doesn't fit.
Cat Sci writes:
It is not the odds of getting the pattern, it is the odds of its existence.
Yeah, that's what I thought.
Your whole argument is flawed: You don't figure out why and how a pattern exists by calculating the odds of it existing.
You are right about this. The probability calculation is solely about the patterns existence. But there is also a residual uncertainty of 1:10^3 about a triune God.
Cat Sci writes:
Your pattern could just be a natural result of the TV series making process, you have not eliminated that possibility and that the chance of it happening is very low doesn't either.
For the origin I will point back to something I already said. From [Msg=162]:
quote:
1. Do you agree there is an coincidental contribution?
2. Do you agree that a coincidental contribution will change the row of appearances?
3. Do you agree that a change in the row of appearances will cause the pattern to not fit sometimes?
4. Do you agree that if the pattern doesn't fit that often, then the pattern will have only a low residual uncertainty like 1:10^2?
If all this questions are answered with Yes, then the involvement of chance precludes a pattern with a residual uncertainty of 1:10^7 because: 1.->2.->3.->4.
I point back to that, because until now no one has answered these four question with for example "Yes, Yes, Yes, No" or "Yes, No, No, No".
Cat Sci writes:
You also haven't refuted this counter argument:
I did. From [Msg=120]:
quote:
RAZD writes:
If you think you can remove elements and still maintain your pattern then they are not essential to the pattern and should NOT be included in the profile/s
You could be right about that. For example it should be easily possible to remove "green", "big/wide/a lot", "lack of knowledge", "do nothing", "very old", "standby", "science", "stone", "death" and "4" without affecting anything. Only to remove "holiday" and "starships" would change something. There are often similarities like "a science starship" or similarities between "very old" and "death" therefore I added them there. The removable elements appear also at the events of P.Al, but they don't cause a next event, therefore it is unimportant if they are a part of P.Al or not. To observe the behaviour of these elements I preliminary added them to P.Al.
"holiday" and "starships" was added to P.Al to make it fit with season 1, 3 and 4. It should not affect the distinctness of the pattern for season 5 and 6 therefore. Even if so, it would not significantly reduce the certainty about the pattern. 5.3 sigma is a very high certainty.
Cat Sci writes:
TV shows follow rules and they are going to have patterns. Quantizing events in the shows and then looking for patterns in the notations is going to make more complicated patterns that are going to have lower odds of occurring.
There can be actually patterns like this with a residual uncertainty of 1:10 or 1:10^2. For example there are at least three persons discussing with each other in the first 30 seconds. This could probably happen in 9 out of 10 times. The found pattern has a residual uncertainty of 1:10^7, not 1:10. And the found pattern is more complex and includes complex pattern on its own.

Replies to this message:
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NoNukes
Inactive Member


(1)
Message 182 of 393 (756343)
04-18-2015 4:00 PM
Reply to: Message 175 by Dubreuil
04-18-2015 10:04 AM


Earth shattering ideas not fit for the public..
I wanted to follow up on something Cat Sci posted. I fully expect that you will ignore this message, but ..
Cat Sci writes:
That you can identify these patterns and calculate the odds of them existing, says nothing about how and why they exist.
In fact, a more complicated pattern identifying process coupled with convoluted the rules are for labeling items in the pattern has two effects:
1) the use of complicated pattern rules increases the probability of identifying a pattern. You can actually increase the number of alternatives for identifying a particular pattern until you insure that the pattern is matched.
2) increasesing the complexity of the identified pattern thereby resulting in an increasingly low probability in the calculations of the type you describe.
In short the more BS your process is, the more likely that your calculated 'probabilities' are going to conceal the BS. And some of your rules are doozies.
The individual items of the pattern are not independent. Some items cannot happen unless other items have happened. And none of the items in a TV show are random. Before I even pick up the paper, my first question would be whether or not the calculated probabilities are meaningful and correct based on these considerations.
Here is another view.
If in fact, there is something of value in the paper, it is not just of interest to mathematicians. The ideas here are potentially world transforming. In short they suggest the existence of a logical and inescapable argument that Jesus is Lord and that special creation is real. And as the author of the paper, you claim to have no time to translate this earth shattering, world transforming idea into English for mass consumption.
Just how reasonable is that anyway?

Je Suis Charlie
Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

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RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 183 of 393 (756352)
04-18-2015 5:01 PM
Reply to: Message 181 by Dubreuil
04-18-2015 3:42 PM


clarification of elements
RAZD writes:
So P.WeC- and P.WeC+ are not appearances -- does the person (P.WeC in this case) need to be present for +/- to occur?
No. For example:
P.Da: "I feel so sick"
P.Ri: "Does this medicine help?"
P.Da: "Yes, it helps a lot. I feel better now"
P.Ri: "I will go to P.WeC. This medicine will also heal him."
is: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+
So we have *P.Da, (observed) P.Da-, (adversly affected) *P.Ri, (observed) *P.Da, (observed) P.Da+, (positively affected) *P.Ri, (observed) *P.WeC, (observed) P.WeC+ (positively affected)
ie no, P.WeC- and P.WeC+ are not appearances, but yes the person (P.WeC in this case) need to be present\observed for +/- to occur, is this correct?
Thus we could define elements as
  1. (*P.(A), not affected) = element 1
  2. *P.(A) becomes positively affected: P.(A)+ = element 2
  3. *P.(A) becomes negatively affected: P.(A)- = element 3
  4. (*P.(B), not affected) = element 4
  5. *P.(B) becomes positively affected: P.(A)+ = element 5
  6. *P.(B) becomes negatively affected: P.(A)- = element 6
    etc
... and 13 people would generate 39 different elements.
Because of the way you define appearance\observation as including being named, it seems that it would not be possible to affect a person without identifying the person being affected and counting that as an appearance\observation, yes?
Yes, M8 and M9 are missing. There are only 12 M's.
Plus the 39 people elements would make a total of 51 elements.
RAZD writes:
Would I be correct in thinking that anything not covered by this list is ignored in the "pattern" derivation and applications?
Yes.
So we could add #52: Something Else not included in the other 51, to cover this otherwise ignored element.
RAZD writes:
and how do I correct D and E so they fit the pattern?
Episode D:
E2: Not *P.WeC, but *P.Ya
E7: Not *P.Wo, but *P.Da
E8: Not *P.Da, but M6
E9: Not *P.BW, but *P.Ya
E12: Not M2, but *P.BW
Episode E:
E15: Not *P.BeC, but M4
Thank you, that would then give 5 hypothetical episodes:
Events Episode A Episode B Episode C Episode D Episode E
Event #1 P.Al P.BW P.Da P.LF P.Tr
Event #2 P.BeC M5 P.LF P.Ya P.Ya
Event #3 P.En P.Pi P.Ri P.Tr P.Wo
Event #4 P.Wo P.Al M4 M10
Event #5 P.Da P.En P.Wo P.Ya
Event #6 M1 P.Ri P.Al M6
Event #7 P.BW P.Tr P.WeC P.Da
Event #8 M4 P.Wo P.BW M6
Event #9 P.Ri P.BeC P.En P.Ya M2
Event #10 P.WSA P.Pi P.Da
Event #11 P.WeC P.LF M7
Event #12 P.Tr P.Ya P.BeC P.BW M10
Event #13 M5 M6 M1 M7 P.LF
Event #14 M2 M3 M13 P.BeC P.Pi
Event #15 M14 M4 M12 P.En M4
But I was looking to make 5 "episodes" that
  1. don't repeat any elements in other "episodes" for the same event
    (D and E now both have P.Ya in event #2 and B and E now both have M4 in event #15
    and
  2. don't repeat any elements within that episode
    (D has P.Ya repeated 3 times and M6 twice)
Would you be able to change that so each hypothetical episode matches your "pattern" and is unique?
RAZD writes:
The question you did not answer was why\how should I conclude that these three hypothetical episodes should be part of the same pattern:
A data source of only three hypothetical episodes is to small to create a non-arbitrary pattern. Therefore it is not possible for only three episodes.
I'm not sure you are understanding the question: I'm trying to determine how you developed your groups so that I could reproduce your work. The reason that I wanted 5 hypothetical episodes with no repeated element across the event levels or within each episode is so you can describe, and I can understand, how they get grouped during analysis.
Enjoy
Edited by RAZD, : "presenobserved" to present\observed
Edited by RAZD, : m6 in red

we are limited in our ability to understand
by our ability to understand
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Dubreuil
Member (Idle past 3041 days)
Posts: 84
Joined: 04-02-2015


Message 184 of 393 (756357)
04-18-2015 6:54 PM


RAZD writes:
Thus we could define elements as
  1. (*P.(A), not affected) = element 1
  2. *P.(A) becomes positively affected: P.(A)+ = element 2
  3. *P.(A) becomes negatively affected: P.(A)- = element 3
  4. (*P.(B), not affected) = element 4
  5. *P.(B) becomes positively affected: P.(A)+ = element 5
  6. *P.(B) becomes negatively affected: P.(A)- = element 6
    etc
It would be:
  1. P.(A) appears: *P.(A) = element 1
  2. P.(A) becomes positively affected: P.(A)+ = element 2
  3. P.(A) becomes negatively affected: P.(A)- = element 3
  4. P.(B) appears: *P.(B) = element 4
  5. P.(B) becomes positively affected: P.(A)+ = element 5
  6. P.(B) becomes negatively affected: P.(A)- = element 6
    etc
RAZD writes:
ie no, P.WeC- and P.WeC+ are not appearances, but yes the person (P.WeC in this case) need to be presenobserved for +/- to occur, is this correct?
RAZD writes:
Because of the way you define appearance\observation as including being named, it seems that it would not be possible to affect a person without identifying the person being affected and counting that as an appearance\observation, yes?
Yes, a person must be identified first. But there can also be references with "he" or "she" that can create +/- without an appearance, if it is clear who "he" or "she" is.
RAZD writes:
Would you be able to change that so each hypothetical episode matches your "pattern" and is unique?
This is actually not possible. Episode A-D shall have a E3->E4 transition. That are 4 episodes, but there are only 3 possible transitions: P.Al-, M4, M10. *P.Wo and *P.Al doesn't trigger E4 at E3 in the episodes A and B. There are other transitions like this:
E1->E2: 6 possible transitions
E2->E3: 13 possible transitions
E3->E4: 3 possible transitions
E5->E6: 6 possible transitions
E10->E11: 11 possible transitions
It is also nearly impossible that there is no element within a episode that repeats itself. In this example:
*P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+
appeared *P.Da and *P.Ri already repeatedly. *P.Da and *P.Ri would triggered new events repeatedly. An episode without repetitions would be rare to find. If you want to group elements yourself, then simultaneous appearances would be important. If there is a simultaneous appearance of P.BeC, P.Da, P.LF, P.Ri and P.WeC, then you will need a part for your pattern that contains all this appearances together. But I don't think all this is necessary. The pattern was created for the first 76 episodes (table 5 on page 8) and tested on a random data source and further 47 episodes. The test showed, that the pattern fits at (00:00), but not for random data. The answer to your question could be table 5 on page 8. It is shown there from where the elements were added to E1-E15.
NoNukes writes:
the use of complicated pattern rules increases the probability of identifying a pattern. You can actually increase the number of alternatives for identifying a particular pattern until you insure that the pattern is matched.
The pattern was created for the first three seasons of the data source and has no predictive power for them.
NoNukes writes:
The individual items of the pattern are not independent. Some items cannot happen unless other items have happened.
For example? There is no appearance that can't happen until an other appearance has happened, for example.
NoNukes writes:
Before I even pick up the paper, my first question would be whether or not the calculated probabilities are meaningful and correct based on these considerations.
To understand a paper about a theory you mostly have to read the paper. You would have to read it to understand it. But you don't want to read it, before you have found out that the content is correct. That's not how to acquire knowledge.
NoNukes writes:
If in fact, there is something of value in the paper, it is not just of interest to mathematicians. The ideas here are potentially world transforming. In short they suggest the existence of a logical and inescapable argument that Jesus is Lord and that special creation is real. And as the author of the paper, you claim to have no time to translate this earth shattering, world transforming idea into English for mass consumption.
I already spend some time to explain it here.
For your other opinions: Add some reasons, references and/or examples and not your opinion solely. Otherwise there can't be a discussion.

Replies to this message:
 Message 185 by NoNukes, posted 04-18-2015 10:30 PM Dubreuil has not replied
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NoNukes
Inactive Member


Message 185 of 393 (756363)
04-18-2015 10:30 PM
Reply to: Message 184 by Dubreuil
04-18-2015 6:54 PM


For example? There is no appearance that can't happen until an other appearance has happened, for example.
No, but a disappearance must be preceded by an appearance. And people in a TV story rarely appear without some purpose even if it's just getting killed by the alien threat. So the fact of an appearance increases the probability that some other things will occur like "becomes negatively affected" prior to the occurrence of the corresponding disappearance. Perhaps what I meant was not obvious, but why did you pick out a stupid straw man to dismiss in lieu of trying to understand.
To understand a paper about a theory you mostly have to read the paper.
I am questioning the paper's methods. I can do that by referring to descriptions of the methods. I've explained my criticisms. You cannot seem to work up much to dismiss them. Are you suggesting that my comments about probability in general are in error?
I already spend some time to explain it here.
Not really. I don't have the credentials that Dr. Adequate possesses, but I can do a credible job of working my way through a probability analysis. I'm trying to work up a reason to bother with your paper. A short convincing presentation that you've found 'good news for modern man' might help.
Edited by NoNukes, : No reason given.

Je Suis Charlie
Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)
If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass

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 Message 184 by Dubreuil, posted 04-18-2015 6:54 PM Dubreuil has not replied

  
Admin
Director
Posts: 12998
From: EvC Forum
Joined: 06-14-2002
Member Rating: 2.3


(4)
Message 186 of 393 (756372)
04-19-2015 7:08 AM
Reply to: Message 177 by Dubreuil
04-18-2015 12:20 PM


Moderator Astonishment
Dubreuil writes:
I'm not a theologist. Ask theological questions to a theologist. I don't know what God prefers to do, if he exists. And I can't give evidences for anything God would hypothetical prefer to do. But I agree that God must be a bizarre being if he exists. How stated before, he leaves innocent babies in Africa to die. This is not a forum about religion, it's about science (EvC Forum ⇒ Science Forums ⇒ Intelligent Design). And I see no reason to engage in this absolutely bizarre religious discussions which completely lack evidences.
I have to say that this is the most unexpected statement I could imagine from someone who wrote a paper that includes the phrase "triune God" in the title, and that includes a section called "Testing the pattern for a triune God" that concludes, "That God, Jesus and the Bible always appeared as P.Ya is an unique intrinsic characteristic of the pattern."
If there are sections of your paper that people should be ignoring then you should be very clear about that.

--Percy
EvC Forum Director

This message is a reply to:
 Message 177 by Dubreuil, posted 04-18-2015 12:20 PM Dubreuil has not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 187 of 393 (756381)
04-19-2015 10:49 AM
Reply to: Message 184 by Dubreuil
04-18-2015 6:54 PM


simplifying the elements
RAZD writes:
Because of the way you define appearance\observation as including being named, it seems that it would not be possible to affect a person without identifying the person being affected and counting that as an appearance\observation, yes?
Yes, a person must be identified first. But there can also be references with "he" or "she" that can create +/- without an appearance, if it is clear who "he" or "she" is.
If it is clear who they are, then are they not identified\observed\appeared? I suppose they could be in the immediate previous event for the reference to work.
The only way I can see having a P.(A)+ or P.(A)- is for them to be preceded by the appearance of P.(A) ... and I think this is what NoNukes was driving at:
NoNukes writes:
The individual items of the pattern are not independent. Some items cannot happen unless other items have happened.
For example? There is no appearance that can't happen until an other appearance has happened, for example.
P.(A)+ or P.(A)- are not appearances and they have to be preceded by the appearance of P.(A): *P.(A) ... either in the same event (n) or the immediately previous event (n-1) (so "it is clear who "he" or "she" is")
RAZD writes:
Thus we could define elements as
  1. (*P.(A), not affected) = element 1
  2. *P.(A) becomes positively affected: P.(A)+ = element 2
  3. *P.(A) becomes negatively affected: P.(A)- = element 3
  4. (*P.(B), not affected) = element 4
  5. *P.(B) becomes positively affected: P.(A)+ = element 5
  6. *P.(B) becomes negatively affected: P.(A)- = element 6
    etc
It would be:
  1. P.(A) appears: *P.(A) = element 1
  2. P.(A) becomes positively affected: P.(A)+ = element 2
  3. P.(A) becomes negatively affected: P.(A)- = element 3
  4. P.(B) appears: *P.(B) = element 4
  5. P.(B) becomes positively affected: P.(A)+ = element 5
  6. P.(B) becomes negatively affected: P.(A)- = element 6
    etc
Curiously, I was trying to isolate the elements so that they are not dependent on another element preceding it, and that was why I was grouping the dependent elements with their precedents and redefining that as a distinct elements within the events to avoid that problem:
  1. P.(A) appears and is not affected: *P.(A)&‘ = element 1
  2. P.(A) appears and becomes positively affected: *P.P(A)&P.(A)+ = element 2
  3. P.(A) appears and becomes negatively affected: *P.P(A)&P.(A)- = element 3
  4. P.(B) appears and is not affected: *P.(B)&‘ = element 4
  5. P.(B) appears and becomes positively affected: *P.P(B)&P.(B)+ = element 5
  6. P.(B) appears and becomes negatively affected: *P.P(B)&P.(B)- = element 6
    etc
In this system any reference to "he" or "she" is counted as an appearance (which is valid because we know who the "he" or "she" is in order to assign the +/- effect).
An alternative would be to change the +/- aspects to generic marks M+ and M- that don't depend on specific previous appearances of specific people ... and with this system your previous example would be rendered as
P.Da: "I feel so sick"
P.Ri: "Does this medicine help?"
P.Da: "Yes, it helps a lot. I feel better now"
P.Ri: "I will go to P.WeC. This medicine will also heal him."
becomes: *P.Da, M-, *P.Ri, *P.Da, M+, *P.Ri, *P.WeC, M+
My reason for this is that it is necessary to properly calculate independent possibilities and probabilities. And I think this approach would be more consistent with your other "mark" elements.
It is also nearly impossible that there is no element within a episode that repeats itself. In this example:
*P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+
appeared *P.Da and *P.Ri already repeatedly. *P.Da and *P.Ri would triggered new events repeatedly.
Again if we counted *P.(A) as "one or more appearances of P.(A)" within an event then it simplifies things and the example would become:
*P.Da, M-, *P.Ri, M+, *P.WeC, M+
Or verbally: Data appears, bad happens Riker appears, good happens, Wesley appears, good happens.
And this would mean
  1. each "element" is independent of any other element
  2. the appearance of an element within an event is independent of the number of appearances
  3. there would be 13 people elements and 14 mark elements for a total of 27 elements
  4. and adding one for "something else" (MSE) would cover any ignored parts and bring the total elements to 28.
With this system all possibilities are described and we can then calculate probabilities.
Enjoy

we are limited in our ability to understand
by our ability to understand
RebelAmerican☆Zen☯Deist
... to learn ... to think ... to live ... to laugh ...
to share.


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This message is a reply to:
 Message 184 by Dubreuil, posted 04-18-2015 6:54 PM Dubreuil has not replied

  
New Cat's Eye
Inactive Member


Message 188 of 393 (756386)
04-19-2015 12:02 PM
Reply to: Message 181 by Dubreuil
04-18-2015 3:42 PM


Cat Sci writes:
It is not the odds of getting the pattern, it is the odds of its existence.
Yeah, that's what I thought.
Your whole argument is flawed: You don't figure out why and how a pattern exists by calculating the odds of it existing.
You are right about this.
Thanks for admitting it. That's a huge chunk of your argument.
Cat Sci writes:
Your pattern could just be a natural result of the TV series making process, you have not eliminated that possibility and that the chance of it happening is very low doesn't either.
For the origin I will point back to something I already said. From [Msg=162]:
quote:
1. Do you agree there is an coincidental contribution?
2. Do you agree that a coincidental contribution will change the row of appearances?
3. Do you agree that a change in the row of appearances will cause the pattern to not fit sometimes?
4. Do you agree that if the pattern doesn't fit that often, then the pattern will have only a low residual uncertainty like 1:10^2?
If all this questions are answered with Yes, then the involvement of chance precludes a pattern with a residual uncertainty of 1:10^7 because: 1.->2.->3.->4.
But I've already accepted that the pattern didn't come about by chance. I've also pointed out that this doesn't mean that it didn't occur naturally.
Your only argument against that has been incredulity.
Cat Sci writes:
You also haven't refuted this counter argument:
I did. From [Msg=120]:
That does not refute the couter argument.
Try again?
quote:
The complexity of the pattern could just be an artifact of the way in which you are notating the conditions compounded with the way that you are looking for the patterns.
There can be actually patterns like this with a residual uncertainty of 1:10 or 1:10^2. For example there are at least three persons discussing with each other in the first 30 seconds. This could probably happen in 9 out of 10 times. The found pattern has a residual uncertainty of 1:10^7, not 1:10. And the found pattern is more complex and includes complex pattern on its own.
Okay, but at this point I still think the complexity is a result of your method.
And the odds of it occurring don't really tell us anything about it.
So you haven't really said much yet.
Edited by Cat Sci, : No reason given.

This message is a reply to:
 Message 181 by Dubreuil, posted 04-18-2015 3:42 PM Dubreuil has not replied

  
Dubreuil
Member (Idle past 3041 days)
Posts: 84
Joined: 04-02-2015


Message 189 of 393 (756389)
04-19-2015 12:40 PM


NoNukes writes:
So the fact of an appearance increases the probability that some other things will occur like "becomes negatively affected" prior to the occurrence of the corresponding disappearance
Yes, if someone appears, then there can be higher possibility that he becomes negatively affected then. This pattern could have a certainty of maybe 30%. This is a residual uncertainty of 7:10. It is not a question that patterns are existing. There can be a lot patterns with a certainty of 90% or even 99%. But this pattern has a certainty of 5.3 sigma or 99.99999%. Your arguments should be more mathematical to refute a pattern with a high certainty like the found pattern. Only the appearances are not noted.
The next post could be helpful for the other answers you wrote.
RAZD writes:
If it is clear who they are, then are they not identified\observed\appeared?
Only if they are named, start to speak or appear. "he" or "she" is not a name.
RAZD writes:
I suppose they could be in the immediate previous event for the reference to work.
Yes. For example:
P.Da: "I feel so sick"
P.Ri: "Does this medicine help?"
P.Da: "Yes, it helps a lot. I feel better now"
P.Ri: "I will go to P.WeC. This medicine will also heal him."
P.Da: "Go to P.Tr instead. It will help her, but it doesn't help humans. He will certainly remain sick."
is: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+, *P.Da, *P.Tr, P.Tr+, P.WeC-
There can be a few events between * and +/-.
RAZD writes:
Or verbally: Data appears, bad happens Riker appears, good happens, Wesley appears, good happens.
This creates problems. For example:
P.Da: "P.Wo, you just won all my chips in this poker game"
It is good for P.Wo, but bad for P.Da. It would be difficult to objectively evaluate what is "good" and what is "bad". It can be evaluated more easily how it affects every person. An other example:
P.anevilperson: "I feel so sick"
This is P.anevilperson-, but is it M+ or M-? To feel sick could be a bad thing to happen, but it could be good that an evil person is affected by this.
RAZD writes:
And this would mean
  1. each "element" is independent of any other element
  2. the appearance of an element within an event is independent of the number of appearances
  3. there would be 13 people elements and 14 mark elements for a total of 27 elements
  4. and adding one for "something else" (MSE) would cover any ignored parts and bring the total elements to 28.
Only 27 elements is not much. The pattern would probably not be distinct and you would need M+ and M- at every event from E1-E15.
RAZD writes:
With this system all possibilities are described and we can then calculate probabilities.
I have calculated probabilities in the next post.
Cat Ski writes:
Thanks for admitting it. That's a huge chunk of your argument.
I only referred to the 1:10^7 probability. Not to the 1:10^3 probability or the arguments below.
Cat Ski writes:
But I've already accepted that the pattern didn't come about by chance. I've also pointed out that this doesn't mean that it didn't occur naturally.
Your only argument against that has been incredulity.
The residual uncertainty of 1:10^7 showed that it didn't come about by chance. The four questions you are referring to, show that the involvement of chance precludes any pattern with a residual uncertainty of 1:10^7. Any naturally imprinted pattern, for example imprinted by writers, would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance.
Your only argument against the four questions was to ignore them and to not answer them.
Cat Ski writes:
That does not refute the couter argument.
You refer to a question asked by RAZD 100 messages ago. If it would have not refuted his counter argument, then he would have mentioned it. Ask and formulate your own questions.

Replies to this message:
 Message 191 by New Cat's Eye, posted 04-19-2015 12:56 PM Dubreuil has not replied
 Message 194 by RAZD, posted 04-20-2015 9:23 AM Dubreuil has not replied
 Message 195 by RAZD, posted 04-20-2015 9:48 AM Dubreuil has not replied

  
Dubreuil
Member (Idle past 3041 days)
Posts: 84
Joined: 04-02-2015


Message 190 of 393 (756390)
04-19-2015 12:44 PM


Approximating the probability
RAZD mentioned the idea to approximate the probability about the pattern by accounting possibilities of single appearances. I have done this now. The calculations below are not part of the paper, but they support the results.
For E1 there are 25 occurrences that fit with the pattern (P.Al, P.BW, P.Da, P.LF, P.Pi, P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, P.Al-, P.BW+, P.Tr+, P.WeC-, P.BeC, P.Ri, P.Ya, M4, P.BW-, P.Da-) and 26 occurrences that break the pattern (*P.En, M3, M10, M11, M12, M14, P.Al+, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Pi-, P.Ri+, P.Ri-, P.Tr-, P.WeC+, P.Wo+, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya-). Appearances occur about 10 times more often than affected persons or M's. Therefore the 25 occurrences that fit with the pattern will happen more often than the 26 occurrences that break the pattern. There are (*=appearances) *: 12; +/-: 6; M's: 7 that fit with the pattern and *: 1; +/-: 20; M's: 5 that break the pattern. * will have a value of 10 to occur and +/- and M's a value of 1. Because appearances occur about 10 times more often than affected persons or M's. This is summarised to the values 133 (12*10 + 6*1 + 7*1) and 35 (1*10 + 20*1 + 5*1). Therefore an occurrence at E1 has a probability of p=133/(133+35)=133/168=0.792 to fit with the pattern. The probability for all other events for single occurrences are:
E1:
fit with the pattern: *:12; +/-:6; M's:7
doesn't fit with the pattern: *:1; +/-:20: M's:5
probability for the next single occurrence to fit with the pattern: p=133/168=0.79
E2:
fit with the pattern: *:13; +/-:5; M's:5
doesn't fit with the pattern: *:0; +/-:21: M's:7
probability for the next single occurrence to fit with the pattern: p=140/168=0.83
E3:
fit with the pattern: *:13; +/-:11; M's:11
doesn't fit with the pattern: *:0; +/-:15: M's:1
probability for the next single occurrence to fit with the pattern: p=152/168=0.90
E4:
fit with the pattern: *:10; +/-:4; M's:6
doesn't fit with the pattern: *:3; +/-:22: M's:6
probability for the next single occurrence to fit with the pattern: p=110/168=0.65
E5:
fit with the pattern: *:11; +/-:5; M's:7
doesn't fit with the pattern: *:2; +/-:21: M's:5
probability for the next single occurrence to fit with the pattern: p=122/168=0.73
E6:
fit with the pattern: *:12; +/-:5; M's:7
doesn't fit with the pattern: *:1; +/-:21: M's:5
probability for the next single occurrence to fit with the pattern: p=132/168=0.79
E7:
fit with the pattern: *:12; +/-:12; M's:9
doesn't fit with the pattern: *:1; +/-:14: M's:3
probability for the next single occurrence to fit with the pattern: p=141/168=0.84
E8:
fit with the pattern: *:12; +/-:16; M's:10
doesn't fit with the pattern: *:1; +/-:10: M's:2
probability for the next single occurrence to fit with the pattern: p=146/168=0.87
E9:
fit with the pattern: *:13; +/-:18; M's:9
doesn't fit with the pattern: *:0; +/-:8: M's:3
probability for the next single occurrence to fit with the pattern: p=157/168=0.93
E10:
fit with the pattern: *:13; +/-:14; M's:6
doesn't fit with the pattern: *:0; +/-:12: M's:6
probability for the next single occurrence to fit with the pattern: p=150/168=0.89
E11:
fit with the pattern: *:13; +/-:12; M's:6
doesn't fit with the pattern: *:0; +/-:14: M's:6
probability for the next single occurrence to fit with the pattern: p=148/168=0.88
E12:
fit with the pattern: *:13; +/-:14; M's:7
doesn't fit with the pattern: *:0; +/-:12: M's:5
probability for the next single occurrence to fit with the pattern: p=151/168=0.90
E13:
fit with the pattern: *:12; +/-:17; M's:7
doesn't fit with the pattern: *:1; +/-:9: M's:5
probability for the next single occurrence to fit with the pattern: p=144/168=0.86
E14:
fit with the pattern: *:13; +/-:21; M's:9
doesn't fit with the pattern: *:0; +/-:5: M's:3
probability for the next single occurrence to fit with the pattern: p=160/168=0.95
The shortest possible pattern is: E3->E9->E12->E13->E14->E15
The probability for it to fit would be: 0.90*0.93*0.90*0.86*0.95=0.615
But there is almost never a row of appearances that includes only one occurrence in every event. For 1x01-1x10 (Appendix A) there were 218 occurrences at 66 events, E15 not included. This are averagely 3.3 occurrences per event.
The probability to fit with two occurrences at every event would be: 0.90^2*0.93^2*0.90^2*0.86^2*0.95^2=0.378
The probability to fit with averagely 3.3 occurrences per event is: 0.90^3.3*0.93^3.3*0.90^3.3*0.86^3.3*0.95^3.3=0.201
The pattern can start at E1, E3, E4 and E5. The probabilities for the likeliest fits with the pattern are:
E3->E9->E12->E13->E14->E15: p=0.201
E3->E9->E10->E12->E13->E14->E15: p=0.139
E3->E9->E11->E12->E13->E14->E15: p=0.132
E3->E9->E10->E11->E12->E13->E14->E15: p=0.090
E1->E2->E3->E9->E12->E13->E14->E15: p=0.050
E1->E2->E3->E9->E10->E12->E13->E14->E15: p=0.034
E1->E2->E3->E9->E11->E12->E13->E14->E15: p=0.033
E1->E2->E3->E9->E10->E11->E12->E13->E14->E15: p=0.022
E5->E6->E7->E8->E9->E12->E13->E14->E15: p=0.003
E5->E6->E7->E8->E9->E10->E12->E13->E14->E15: p=0.002
E5->E6->E7->E8->E9->E11->E12->E13->E14->E15: p=0.002
E5->E6->E7->E8->E9->E10->E11->E12->E13->E14->E15: p=0.001
E4->E5->E6->E7->E8->E9->E12->E13->E14->E15: p=0.001
E4->E5->E6->E7->E8->E9->E10->E12->E13->E14->E15: p=0.0005
E4->E5->E6->E7->E8->E9->E11->E12->E13->E14->E15: p=0.0005
E4->E5->E6->E7->E8->E9->E10->E11->E12->E13->E14->E15: p=0.0002
The other fits are negligible. The overall probability that the pattern fits is:
p=0.201+0.139+0.132+0.090+0.050+...=0.711
This is the probability for the pattern to fit with single occurrences only. There can be simultaneous occurrences as well. For simultaneous appearances there are 12+11+10...=78 possibilities for two persons to appear simultaneous. If one of this two persons is P.Ya, then there are at E1 5 simultaneous appearances that fit: P.Ya with P.Al, P.BeC, P.LF, P.Ri, P.WeC and 7 simultaneous appearances that doesn't fit: P.Ya with P.BW, P.Da, P.En, P.Pi, P.Tr, P.Wo, P.WSA. For a single appearance the ratio of appearances was 12/1. For two simultaneous appearances the ratio is 5/7. The ratio to fit with the pattern will further decrease, if three or four simultaneous appearances are examined.
Therefore the probability for single occurrences and simultaneous occurrences is: p<0.711
Approximately every 20th occurence is a simultaneous occurrences, therefore its only a small effect. The theoretical result fits well with the experimental result. The pattern was tested on a random data source and the probability that the pattern fits was 0.625. Only for the actual beginning the rules of chance are suspended. The result is a stable statistical anomaly. The involvement of chance would normally prevent a pattern with a residual uncertainty higher than 1:10^2. The pattern even includes a reference about a triune God with a residual uncertainty of 1:10^3. This results are really intruiging, if you only wanted to write a paper about how to quantise nontrivial patterns first.
Edited by Dubreuil, : No reason given.

  
New Cat's Eye
Inactive Member


Message 191 of 393 (756391)
04-19-2015 12:56 PM
Reply to: Message 189 by Dubreuil
04-19-2015 12:40 PM


The residual uncertainty of 1:10^7 showed that it didn't come about by chance.
No, it doesn't. As you've admitted, the odds of a pattern existing don't tell is how or why it occured.
The four questions you are referring to, show that the involvement of chance precludes any pattern with a residual uncertainty of 1:10^7.
Obviously, the pattern is not based on chance. But your 4 questions do not eliminate any and all involvement of chance.
Any naturally imprinted pattern, for example imprinted by writers, would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance.
No, that's not true. You've not proved this.
And again, the odds of the pattern existing tell us nothing about how or why it happened.
Your only argument against the four questions was to ignore them and to not answer them.
You said, along with the questions, that if the answers to all questions were 'yes' then the pattern could not have come about by chance. I told you that I already accepted that the pattern did not come about by chance so your questions are irrelevant.
Where it sits: The pattern did not come about by chance, but it still could have occured naturally.
You refer to a question asked by RAZD 100 messages ago.
No, it was a statement from my Message 65:
quote:
The complexity of the pattern could just be an artifact of the way in which you are notating the conditions compounded with the way that you are looking for the patterns.
Its left unrefuted.

This message is a reply to:
 Message 189 by Dubreuil, posted 04-19-2015 12:40 PM Dubreuil has not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 192 of 393 (756396)
04-19-2015 3:02 PM
Reply to: Message 184 by Dubreuil
04-18-2015 6:54 PM


Transitions
RAZD writes:
Would you be able to change that so each hypothetical episode matches your "pattern" and is unique?
This is actually not possible. Episode A-D shall have a E3->E4 transition. That are 4 episodes, but there are only 3 possible transitions: P.Al-, M4, M10. *P.Wo and *P.Al doesn't trigger E4 at E3 in the episodes A and B. There are other transitions like this:
E1->E2: 6 possible transitions
E2->E3: 13 possible transitions
E3->E4: 3 possible transitions
E5->E6: 6 possible transitions
E10->E11: 11 possible transitions
What are they? Let's take E1->E2 first. From Message 166 we have:
E1: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.BW, P.Da, P.LF, P.Pi,P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, P.Al-, P.BW+, P.Tr+, P.WeC-. Caveat: +/- cannot be observed without * appearance of individual.
In revised system A (per Message 187) this would be written:
E1: Elements are observed, either singly or in combinations and all with possible repeated appearances -- {*P.Al&‘}, {*P.Al&P.Al-}, {*P.BW&‘}, {*P.BW&P.BW+}, {*P.Da&‘}, {*P.LF&‘}, {*P.Pi&‘},{*P.Tr&‘}, {*P.Tr&P.Tr+}, {*P.WeC&‘}, {*P.WeC&P.WeC-}, {*P.Wo&‘}, {*P.WSA&‘}, {M1}, {M2}, {M5}, {M6}, {M7}, {M13}.
(19 independent elements)
In revised system B (per Message 187) this would be written:
P.Al, P.BW, P.Da, P.LF, P.Pi,P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, M-, M+
(17 independent elements)
But you take issue with M+ and M- :
Message 189: This creates problems. For example:
P.Da: "P.Wo, you just won all my chips in this poker game"
It is good for P.Wo, but bad for P.Da. It would be difficult to objectively evaluate what is "good" and what is "bad". It can be evaluated more easily how it affects every person.
Your call, but now you are stuck with system A to make your elements independent. I'll simplify the notation a little in the following discussion:
P.(A) appears and is not affected: *P.(A)&‘ = element 1
P.(A) appears and becomes positively affected: *P.P(A)&+ = element 2
P.(A) appears and becomes negatively affected: *P.P(A)&- = element 3
and a total of 51 independent elements plus one for "something else" (MSE)
RAZD writes:
If it is clear who they are, then are they not identified\observed\appeared?
Only if they are named, start to speak or appear. "he" or "she" is not a name.
But you know who they are to assign +/- and therefore they are identified\observed.
Again, from Message 166 we have:
E2: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.BeC, P.LF, P.Ri, P.WeC, P.Ya, M4, M5, P.BW-, P.Da-.
In revised system A (per Message 187) this would be written:
E2: Elements are observed, either singly or in combinations and all with possible repeated appearances -- {*P.Al&‘}, {*P.BeC&‘}, {*P.BW&P.BW-}, {*P.Da&P.Da-}, {*P.LF&‘}, {*P.Ri&‘}, {*P.WeC&‘}, {*P.Ya&‘}, {M4}, {M5}.
Comparing E1 and E2 with list of all elements per modified system A:
E1transitionE2
*P.Al&‘yescan stayyes
*P.Al&+nonono
*P.Al&-yesmust endno
*P.BeC&‘nonoyes
*P.BeC&+nonono
*P.BeC&-nonono
*P.BW&‘yesmust endno
*P.BW&+yesmust endno
*P.BW&-nobecomesyes
*P.Da&‘yesmust endno
*P.Da&+nonono
*P.Da&-nobecomesyes
*P.En&‘nonono
*P.En&+nonono
*P.En&-nonono
*P.LF&‘yescan stayyes
*P.LF&+nonono
*P.LF&-nonono
*P.Pi&‘yesmust endno
*P.Pi&+nonono
*P.Pi&-nonono
*P.Ri&‘nonoyes
*P.Ri&+nonono
*P.Ri&-nonono
*P.Tr&‘yesmust endno
*P.Tr&+yesmust endno
*P.Tr&-nonono
*P.WeC&‘yescan stayyes
*P.WeC&+nonono
*P.WeC&-yesmust endno
*P.Wo&‘yesmust endno
*P.Wo&+nonono
*P.Wo&-nonono
*P.WSA&‘yesmust endno
*P.WSA&+nonono
*P.WSA&-nonono
*P.Ya&‘nonoyes
*P.Ya&+nonono
*P.Ya&-nonono
M1yesmust endno
M2yesmust endno
M3nonono
M4nonoyes
M5yescan stayyes
M6yescan stayno
M7yesmust endno
M10nonono
M11nonono
M12nonono
M13yescan stayno
M14nonono
MSEyescan stayyes
Notice that I have added the "something else" marker MSE to both events, as it could be present in any event and not have been recorded.
There are other transitions like this:
E1->E2: 6 possible transitions
What are those 6 possible transitions? I get 13 to 15
Enjoy

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This message is a reply to:
 Message 184 by Dubreuil, posted 04-18-2015 6:54 PM Dubreuil has not replied

  
Dubreuil
Member (Idle past 3041 days)
Posts: 84
Joined: 04-02-2015


Message 193 of 393 (756398)
04-19-2015 4:02 PM


Cat Ski writes:
No, it doesn't. As you've admitted, the odds of a pattern existing don't tell is how or why it occured.
It doesn't tell where it come from, but it tells where it doesn't come from. The probability for the pattern to occur coincidental was 0.625 for the random data test and <0.711 for the calculation, although the pattern fit at (00:00). The probability that this is the result of chance was calculated to 1:10^7. Because of this it is almost impossible that the pattern occured because of coincidental effects.
Cat Ski writes:
You said, along with the questions, that if the answers to all questions were 'yes' then the pattern could not have come about by chance. I told you that I already accepted that the pattern did not come about by chance so your questions are irrelevant.
That it could not have come about by chance was shown with the 1:10^7 probability. The four questions show that any naturally imprinted pattern would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance.
Cat Ski writes:
Any naturally imprinted pattern, for example imprinted by writers, would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance.
No, that's not true. You've not proved this.
Let's take a look to the four questions again:
quote:
1. Do you agree there is an coincidental contribution?
2. Do you agree that a coincidental contribution will change the row of appearances?
3. Do you agree that a change in the row of appearances will cause the pattern to not fit sometimes?
4. Do you agree that if the pattern doesn't fit that often, then the pattern will have only a low residual uncertainty like 1:10^2?
If all this questions are answered with Yes, then the involvement of chance precludes a pattern with a residual uncertainty of 1:10^7 because: 1.->2.->3.->4.
The pattern is really sensitive to coincidental effects. An example from [Msg=58]:
A row of appearance that doesn't fit:
*P.Al, *P.Pi, M14, P.Al-, P.Al+
But two rows of appearances that would fit:
*P.Pi, *P.Al, M14, P.Al-, P.Al+
M14, *P.Al, *P.Pi, P.Al-, P.Al+
Only the first three appearances were mixed here and it has already changed whether the pattern fits or not. Every coincidental contribution can change the row of appearances or simultaneous appearances and affects whether the pattern fits or not. Although coincidental contributions should have highly influenced any naturally imprinted pattern, the found pattern has a high likelihood to appear and a high residual uncertainty, against the coincidental contributions. If you want to keep discussing about this, then I insist that you answer all four question. That will simplify the discussion.
Cat Ski writes:
And again, the odds of the pattern existing tell us nothing about how or why it happened.
Yes, it is only an indication for Intelligent Design. Intelligent Design claims there is a intelligent cause in evolution and this pattern shows a signal in evolution-like processes. The residual uncertainty of 1:10^3 is an other reference and the four questions above can preclude an other natural origin.
Cat Ski writes:
quote:
The complexity of the pattern could just be an artifact of the way in which you are notating the conditions compounded with the way that you are looking for the patterns.
Its left unrefuted.
The distinctness of the pattern was actually created this way. From [Msg=166]: "The elements therefore have to be grouped in a way to create a pattern of the most possible distinctness". The pattern was created to fit with the first three seasons of the data source and was then tested on a data source it was not created for. The argument about this is that there is such a complex pattern at all. From [Msg=166]: "If you are interested about this, then the pages 11 to 13 of the paper could be revealing to you. It was tried there to add actual random data from episodes to the pattern. To make the pattern fit with this random data, it became a random pattern itself. Large gaps were removed and the patterns within the pattern had to be removed too. The pattern also didn't fit anymore with previous episode, for example 4x08 as shown at the end. To make the pattern fit again with 4x08, M3 must be added to E12 and M14 to E13. This would again remove large gaps and would make the pattern even more random.". For the random data source it was not possible to create a distinct pattern that fit with the additional 9 episodes. For the first 76 episodes it was possible to create a distinct pattern, but not for the 9 episodes out of the random data source. This shows a difference between the actual beginning and random data. But for the residual uncertainties of 1:10^3 and 1:10^7 is the actual complexity of the pattern unimportant. This additional complexity (E11=E13,...) is mainly a side note, not part of any calculation.
RAZD writes:
There are other transitions like this:
E1->E2: 6 possible transitions
What are those 6 possible transitions? I get 13 to 15
*P.BeC, P.BW-, P.Da-, *P.Ri, *P.Ya, M4
Every element that is present in E2 but not in E1 will cause a transition. An element that is present in E1 and E2 will not cause a transition. An element that is not present in E1 and E2 will break the pattern at E1.
E3->E4: P.Al-, M4, M10
E5->E6: *P.Al, P.Pi-, P.Ri-, *P.Tr, M1, M6
RAZD writes:
Caveat: +/- cannot be observed without * appearance of individual.
It can. For Example:
E5: *P.Pi /E6: P.Pi- /E7: P.Pi+, *P.Pi /E8: P.Pi-
RAZD writes:
But you know who they are to assign +/- and therefore they are identified\observed.
Yes, they are identified but it is not an appearance. An appearance for this pattern is defined as someone being named, start to speak or appear. And "he" or "she" is not a name.
You try to rewrite the pattern rules. For example that every affected person includes an appearance of the person. If you look at table 4 on page 5, then you will see there a lot - or + without *. If they are all replaced from "+" and "-" to "*, +" and "*, -", then an reappearance of this person would not cause a transition.
a row of appearances:
*P.Da, P.Da-, *P.Da, *P.WeC, *P.Da, P.Da-, *P.Da
With the actual rules:
E1: *P.Da /E2: P.Da- /E3: *P.Da /E9: *P.WeC /E11: *P.Da /E12: P.Da- /E13: *P.Da
For your rules (probably):
E1: *P.Da /E2: P.Da-, *P.Da, *P.WeC, *P.Da, P.Da-, *P.Da
You can test your revised system A and B, but they are completely different to the pattern introduced in the paper. Didn't you said you wanted to reproduce the work and not to create your own?

Replies to this message:
 Message 196 by New Cat's Eye, posted 04-20-2015 11:32 AM Dubreuil has not replied
 Message 197 by RAZD, posted 04-20-2015 11:41 AM Dubreuil has not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 194 of 393 (756446)
04-20-2015 9:23 AM
Reply to: Message 189 by Dubreuil
04-19-2015 12:40 PM


Patterns and Kaleidoscopes
RAZD writes:
And this would mean
  1. each "element" is independent of any other element
  2. the appearance of an element within an event is independent of the number of appearances
  3. there would be 13 people elements and 14 mark elements for a total of 27 elements
  4. and adding one for "something else" (MSE) would cover any ignored parts and bring the total elements to 28.
Only 27 elements is not much. The pattern would probably not be distinct and you would need M+ and M- at every event from E1-E15.
Curiously I'm still trying to figure out how you develop a pattern, and I would think a simpler system would be the first approach, only making it more complicated when the simple pattern fails.
In this case with generic M+ and M- I would think you would get more matches rather than less.
Going back to my hypothetical episodes in Message 183:
Events Episode A Episode B Episode C Episode D Episode E
Event #1 P.Al P.BW P.Da P.LF P.Tr
Event #2 P.BeC M5 P.LF P.Ya P.Ya
Event #3 P.En P.Pi P.Ri P.Tr P.Wo
Event #4 P.Wo P.Al M4 M10
Event #5 P.Da P.En P.Wo P.Ya
Event #6 M1 P.Ri P.Al M6
Event #7 P.BW P.Tr P.WeC P.Da
Event #8 M4 P.Wo P.BW M6
Event #9 P.Ri P.BeC P.En P.Ya M2
Event #10 P.WSA P.Pi P.Da
Event #11 P.WeC P.LF M7
Event #12 P.Tr P.Ya P.BeC P.BW M10
Event #13 M5 M6 M1 M7 P.LF
Event #14 M2 M3 M13 P.BeC P.Pi
Event #15 M14 M4 M12 P.En M4
When I look at the "pattern" in Message 166 for event #15 I see
Event #15: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.WSA, P.Ya, M4, M12, M14, P.Al+, P.BW-, P.Da+, P.En+, P.LF+, P.Pi-, P.Ri+, P.Tr-, P.WeC-, P.Wo-, P.Ya+.
it seems to me that I can make B15 be P.Da and then E is unique from A-C:
Events Episode A Episode B Episode C Episode D Episode E
Event #1 P.Al P.BW P.Da P.LF P.Tr
Event #2 P.BeC M5 P.LF P.Ya P.Ya
Event #3 P.En P.Pi P.Ri P.Tr P.Wo
Event #4 P.Wo P.Al M4 M10
Event #5 P.Da P.En P.Wo P.Ya
Event #6 M1 P.Ri P.Al M6
Event #7 P.BW P.Tr P.WeC P.Da
Event #8 M4 P.Wo P.BW M6
Event #9 P.Ri P.BeC P.En P.Ya M2
Event #10 P.WSA P.Pi P.Da
Event #11 P.WeC P.LF M7
Event #12 P.Tr P.Ya P.BeC P.BW M10
Event #13 M5 M6 M1 M7 P.LF
Event #14 M2 M3 M13 P.BeC P.Pi
Event #15 M14 P.Da M12 P.En M4
So now if I change D2 and either D5 or D9 and then either D6 or D8...
When I look at the "pattern" in Message 166 for events #2, 5, 8 and 9 I see
Event #2: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.BeC, P.LF, P.Ri, P.WeC, P.Ya, M4, M5, P.BW-, P.Da-.
Event #5: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Wo, P.Ya, M2, M4, M5, M7, M14, P.BW-, P.Pi+, P.Wo-.
Event #6: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.Tr, P.Wo, M1, M6, P.Pi-, P.Ri-.
Event #8: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.BW, P.Da, P.LF, P.Pi, P.Ri, P.Tr, P.Wo, M1, M4, M5, M6, M10, M13, P.Al-, P.BW+, P.En-, P.LF-, P.Pi-, P.Ri-, P.Tr+, P.WeC-, P.Wo+.
Event #9: Elements are observed, either singly or in combinations and all with possible repeated appearances -- P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.Ya, M1, M2, M4, M7, M11, M14, P.Al+, P.BW-, P.Da-, P.Pi+, P.Tr-, P.Wo-, P.Ya+.
It would appear that I can use P.Al or P.Ri or P.WeC, or M4 in the D2 position ... : I'll use P.Al for D2
... that I can use P.Pi or P.Ri or M2 or M4 in the D5 position and leave D9 as P.Ya, or leave P.Ya in the D5 position and use P.Al or P.Pi or P.WeC or M1 or M2 or M4 or M11 or M14 in the D9 position ... : I'll use P.Pi for D5 and leave D9 as P.Ya
... and that I can use P.Wo in the D6 position and leave M6 in the D8 position or leave M6 in the D6 position and use P.Pi or P.Ri or M1 or M5 or M13 in the D8 position ... : I'll leave M6 in D6 and use P.Ri for D8
Events Episode A Episode B Episode C Episode D Episode E
Event #1 P.Al P.BW P.Da P.LF P.Tr
Event #2 P.BeC M5 P.LF P.Al P.Ya
Event #3 P.En P.Pi P.Ri P.Tr P.Wo
Event #4 P.Wo P.Al M4 M10
Event #5 P.Da P.En P.Wo P.Pi
Event #6 M1 P.Ri P.Al M6
Event #7 P.BW P.Tr P.WeC P.Da
Event #8 M4 P.Wo P.BW P.Ri
Event #9 P.Ri P.BeC P.En P.Ya M2
Event #10 P.WSA P.Pi P.Da
Event #11 P.WeC P.LF M7
Event #12 P.Tr P.Ya P.BeC P.BW M10
Event #13 M5 M6 M1 M7 P.LF
Event #14 M2 M3 M13 P.BeC P.Pi
Event #15 M14 P.Da M12 P.En M4
So now I have 5 hypothetical episodes that fit your pattern ...
... and the question is why should I think there is a pattern here --
I do not see any equivalence between any elements in any one event
I do not see any pattern in the sequence of events for any episode
I do not see any sequence >2 elements in one episode that is repeated in any other episode (for example P.Ri comes after P.Al in episode A but before P.Al in episode C).
There is no pattern here that I can see: what should I be seeing that I am missing?
When you assemble mirrors and bits of colored plastic in a kaleidoscope you can create the appearance of colored patterns, where any one would be highly unlikely to occur, but the pattern is an artifact caused by the mirrors, not the colored plastic bits.
Enjoy

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by our ability to understand
RebelAmerican☆Zen☯Deist
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This message is a reply to:
 Message 189 by Dubreuil, posted 04-19-2015 12:40 PM Dubreuil has not replied

  
RAZD
Member (Idle past 1405 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 195 of 393 (756450)
04-20-2015 9:48 AM
Reply to: Message 189 by Dubreuil
04-19-2015 12:40 PM


RAZD writes:
Or verbally: Data appears, bad happens Riker appears, good happens, Wesley appears, good happens.
This creates problems. For example:
P.Da: "P.Wo, you just won all my chips in this poker game"
It is good for P.Wo, but bad for P.Da. ...
P.Da M- P.Wo M+ or P.Wo M+ P.Da M- ... I don't see this as a problem
... It would be difficult to objectively evaluate what is "good" and what is "bad". It can be evaluated more easily how it affects every person. An other example:
P.anevilperson: "I feel so sick"
This is P.anevilperson-, but is it M+ or M-? To feel sick could be a bad thing to happen, but it could be good that an evil person is affected by this.
Well isn't that a problem that applies to other instances -- your subjective evaluation of good/evil, based on your worldview?
P.AEP M- M+
RAZD writes:
And this would mean
  1. each "element" is independent of any other element
  2. the appearance of an element within an event is independent of the number of appearances
  3. there would be 13 people elements and 14 mark elements for a total of 27 elements
  4. and adding one for "something else" (MSE) would cover any ignored parts and bring the total elements to 28.
Only 27 elements is not much. The pattern would probably not be distinct and you would need M+ and M- at every event from E1-E15.
Curiously I think the pattern would be more universal: what I am trying to do is figure out is how you develop the pattern, so I figure a simpler system would be easier to start with.
Enjoy

we are limited in our ability to understand
by our ability to understand
RebelAmerican☆Zen☯Deist
... to learn ... to think ... to live ... to laugh ...
to share.


Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click)

This message is a reply to:
 Message 189 by Dubreuil, posted 04-19-2015 12:40 PM Dubreuil has not replied

  
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