FrediFizzx wrote:Yeah, Yeah, Yeah! Another freakin' update. Now, with theoretical support from the Product Calculation for -a.b. Pretty freakin' awesome!!! 1jillion gazillion trials!
Cloud File.
https://www.wolframcloud.com/obj/fredif ... rum-pc2.nb
Direct Files.
download/newCS-33-S3quat-3D-forum-pc2.pdf
download/newCS-33-S3quat-3D-forum-pc2.nb
Enjoy the simulation that is a Bell and Gill theory killer!!!! It just keeps getting better and better!!
.
Gordon Watson wrote:Fred, given your simulation of N trials over N random particle-pairs, how many data-points appear in your plot? Is it always N? Thanks; Gordon
FrediFizzx wrote:Austin Fearnley wrote: ... [incomprehensible nonsense snipped] ... Also, are your graphs showing that Bell's Inequalities are broken? Every once in a while you state that nothing can break the Bell Inequalities. So what is the conclusion?
The graphs DO NOT show that the inequalities are broken. The current graphs show that both Bell's junk physics theory and Gill's junk theory are broken. Shot dead to a million pieces that never can be put back together.
FrediFizzx wrote:This expression seems a bit odd to me.
In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.
Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786
That was for 10,000 trials. For 5 million trials,
Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929
Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.
FrediFizzx wrote:FrediFizzx wrote:This expression seems a bit odd to me.
In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.
Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786
That was for 10,000 trials. For 5 million trials,
Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929
Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.
Ok, now for the next part of this.
QM assigns for those 4 outcome probabilities,
Again, in a simulation with many trials, we have to average and over all the (a-b) angles. Lo and behold, when we do that we obtain,
,
,
Because .
So, it seems to me that all of the parts of the original E(a, b) expression are all equal to 1/4. Analytically-wise.
FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation?
Joy Christian wrote:FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation?
No, it doesn't. I can think of dozens of functions for which would be true.
Consider, for example, .
This function gives .
Therefore, following your logic, .
.
FrediFizzx wrote:Joy Christian wrote:FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation?
No, it doesn't. I can think of dozens of functions for which would be true.
Consider, for example, .
This function gives .
Therefore, following your logic, .
.
Well, that is only true if . So, we still have . Try again.
Joy Christian wrote:Your logic does not work. I gave an example of why your logic does not work.
Joy Christian wrote:FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation?
No, it doesn't. I can think of dozens of functions for which would be true.
Consider, for example, .
This function gives .
Therefore, following your logic, .
Joy Christian wrote:Here is another function which does the same thing:
,
implying that
.
In fact, there are many functions that would do the same thing.
FrediFizzx wrote:Joy Christian wrote:Here is another function which does the same thing:
,
implying that
.
In fact, there are many functions that would do the same thing.
Hmm... that doesn't look right substitution-wise.
,
So,
,
,
.
Your example is still not working for me. Try again.
Joy Christian wrote:I give up, because I don't understand what setting means within your logic. To me, it means , always, throughout all experimental trials.
Keeping mind that P(++) and , etc. are actually averages.
FrediFizzx wrote:FrediFizzx wrote:FrediFizzx wrote:This expression seems a bit odd to me.
In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.
Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786
That was for 10,000 trials. For 5 million trials,
Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929
Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.
Ok, now for the next part of this.
QM assigns for those 4 outcome probabilities,
Again, in a simulation with many trials, we have to average and over all the (a-b) angles. Lo and behold, when we do that we obtain,
,
,
Because .
So, it seems to me that all of the parts of the original E(a, b) expression are all equal to 1/4. Analytically-wise.
Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation? Keeping mind that P(++) and , etc. are actually averages.
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