1) Sorry, but ignore Chris/Alice which are at distant locations from each other and replace with Chris/Bob which are at the same location. This is a serious misunderstanding of the experimental setup! In an actual experiment there are just Bob and Alice. Bob has two choices that differs from...
Hello DrChinese
Why would you include Chris/Bob? It is irrelevant to the experiment! If Alice chooses 0, then there are 3 choices 0,120, and 240 for the distant location. Alice is at a distant location from Bob and Chris. Agree? When the angles are different we are only interested in comparing...
Rolling dice will work. The probability you select same angle is 1/3 and selecting a different angle is 2/3. The probability of getting (+, +) or (-, -) for all runs is P(S) = (1/3)(1) + (2/3)(1/4). I have demonstrated the unit circle can reproduce the quantum theory prediction of 1/4 from the...
Strilanc
My post is clear. I will respond to why only two sequences repeat and not the others. Take the triplet (+, - ,-), now switch the first two to get (-, +, -).
You get a different sequence. Same with (-, +, +) you get (+, -, +). Changing Bob and Chris choices gives different sequences...
I disagree. It will give the same result for A, B, and C independently choosing an angle. I stated this in the post. No what order A, B, and C choose an angle only 6 sequences with two that repeat are possible. Rotation was mention only to demonstrate keep the points 120 apart and not a...
I would like to reopen a discussion on assertions made by David Mermin such as: “There is no conceivable way to assign such instruction sets to the particles from one run to the next that can account for the fact that in all runs taken together, without regard to how the switches are set, the...
Actually what you have shown (in a concise and eloquent manner using probabilities) is that the last two conditions are impossible. At least one of the three must be + or -, but all three simultaneously cannot have the same sign at the specified angles.
Bill, your #346 and #347 posts are excellent summaries of your argument. They tie in nicely with your post #270 which reveal that Bell-Type inequalities cannot apply to both scenarios.
Sure that is possible. He indeed emphasized the same-switch/same-color occurrence, the randomness of the data, and the other merely went unnoticed. Testing Bell’s inequality has taken top billing, but these simple data averages are equally important. However, I haven’t found any references to...
I understand your point. However, toss a coin a 1000 times and suppose you get 480 heads and 520 tails which gives the 0.48 probability. For practical reasons or convenience (limited space) you decide to use 50 tosses to show that coin tossing is random or 0.50. Now a data set typical of the...
The instructions are an inherent property of the trig functions and the unit circle, the principles of trigonometry dictate the possible outcomes and there are only 6 not 8 permutations. This was one key point that I was trying to make. Nearly all previous discussions including Mermin insisted...
Thanks for the comments. I certainly agree with you about n=45 trials. Number of trials was not the issue. This was fully understood before deciding to post. However, 100s of n=45 trials and Mermin selects a set that he clearly states as typical (his words not mine), but contradicts the main...
Dr. Chinese’s Challenge: 0,120,240 Data Set
Data Features (Quantum Theory):
P(B|A) = 1; P(B|Aʹ) = .25; P(B) = .50 Eq. (1)
“A” means “same setting”, “Aʹ” means “different setting”, and “B” means “different outcome”, “Bʹ” means “same outcome”. Here is a quote from David Mermin’s paper (Is the...
I may be wrong, but this does not seem correct. There are only two outcomes for this case:
1) Alice-green, Bob-red
2) Alice-red, Bob-green.
So let A=Alice gets green and B=Bob gets red. From the generalized conditional probability rule:
P(A|B) = P(A)*P(B|A) where P(A) = 0.5 and P(B|A) =...
Let's use Alice and Bob as the two different people and calculate the probability using the conditional probability formula.
P(Alice-green,Bob-red)=(.05)(1)=0.5 and P(Alice-red,Bob-green)=(0.5)(1)=0.5 and the probability that Alice and Bob get different colors is 0.5+0.5=1. Standard...
Doesn’t Bell do the exactly the same. That is, Bell presented his own proof and then he used quantum predictions as a counterexample to disprove it. Then concluded the locality assumption was impossible or false.
Could you convincingly explain what assumption Bill made that is impossible or...
Bill, you asked when comparing the theoretical and the experimental,
Why the violation?
1) The machine selects two of the coins at anyone time instead of three. This has been discussed before.
2) The fact that the results for <ab>,<bc>, and <ac> = -1. Not sure of the deeper meaning of this...
You basically have taken up David Mermin’s challenge (Is the moon there when nobody looks? Reality and the quantum theory) to find an instruction set to account for the experimental results. Everyone recognizes the obvious instruction set, that is, the source guarantees the correlation (or...
Hi Gordon,
Re: Post # 287
I am glad to see that someone has incorporated Malus in a classical way and compared to the qm predictions. If possible, could you summarize and/or elaborate in more detail?
I assume then you hand-picked your data set. And you may be correct about the inequality I chose to violate not being applicable in one sense. However, when all three data pieces (a,b,c) are used then I believe no matter how you pick the data pairs or write the inequality there can be no...
This data set could have been generated by tossing two coins at a time. Right!
The number of mismatches are:
nab=2
nbc=0
nac=1
nbc+nac≥nab is violated. How do you explain the violation. Is coin tossing non-local?
I understand your concern and I have tried not to incorporate my own ideas or theories in the postings.
The examples of experiments and ideas presented in this thread are not original to me. I must give credit to others, particularly the inventor of Boolean logic, George Boole. And of course...
First, someone just makes up the data! You have got to be kidding. Per the first coin toss example flip three coins: a,b,c. You record the sequence of heads and tails. You will never violate the inequality: nab(HT) + nbc(HT) ≥ nac(HT). Do you know this?
Repeat the second example several times...
I understand what you are saying. However, in the context of our discussions with the available knowledge about the origin of these types of inequalities is not very convincing. I am here to learn. Give me a reference where these violations have been considered and ruled out. Show how randomly...
Thanks Bill. The error in the inequality was mine and apologize for it. I should have written it as: nab(HT) + nbc(HT) ≥ nac(HT). This inequality is derivable and impossible to violate using three lists.
However, if you use 6 lists as in the EPR experiments (because only one angle can be...
ttn,
billschnieder has provided three experiments relating to this thread for you to comment on.
1. Post # 102 - coin toss experiment
2. Post # 107 - Lyon, Paris, Lille study
3. Post # 125 – cyclic dependency
Your failure to make a comment is very telling. Why not start with the coin toss...
I agree. Except that tnn stated that "Anyway, hopefully people will at some point get around to actually reading the thing and then raising questions about the proofs, arguments, definitions, etc." The factoring step for the derivation of the CHSH inequality has been questioned as to the...
May be the best way to resolve this debate is for someone to analyze the actual experimental data and report their findings. Or, if possible, post the data in a convenient form on this forum (or link) for others to examine. Once sorted, Bell's inequality may be tested using the 3 or 6 data...
Thanks for the comments.
The data shows that the events A and B are dependent not independent, an assumption made by Bell. The P(A)*P(B/A) ≠ P(A)*P(B). Can you exlain how Bell got it right using an invalid assumption?
In regards to Jaynes’ view: Bell incorrectly factored a joint probability; it may be informative to analyze the data set presented by N. David Mermin in his article: “Is the moon there when nobody looks? Reality and the quantum theory.” The following represents the summary of the data.
A =...
When all the relevant facts cannot be know, wouldn’t you agree that the main point is that in order to prove Bell’s inequality is applicable to the EPRB experiments, that it should be consistent with the known facts in a large number of random trials, without exception?
For someone trying to understand Bell’s theorem, I find it essential to read papers on both sides of the argument. Of course, the problem here is that probability theory is not easily understood by the average reader and the experts seem to be able to rationalize away the other person’s...
This is not true as clearly demonstrated in the spread sheet. Look at Trial #4, there is a count for cb(HH), but none for abc(HHT/TTH) or abc(TTH/HHT). There is simply no one-to-one mapping of the counts and the equation will be false because of Trials such as #4. Refer to the Simultaneous...
The reason the equation does not sum correctly is because of choosing two coins out of the three coins when tabulating the data. The equation is always valid when analyzing all three coins simultaneously.
For comparison, here is a second spread sheet for n=25 trials for a simultaneous 3-coin...
Coin Toss Simualtion
It is my intention to provide a simple coin toss simulation to clearly demonstrate that the derivation by Sakurai (http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality) is invalid and that Bell’s inequality is pointless. The coin toss experiment consists of n=100...
Apparently you have not read any of the papers listed by Bill Schneider. These simulations have already been published and shown to violate Bell's inequalities. You simply refuse to acknowledge their relevance to EPRB experiments.
Yes great and sorry for any confusion.
No you missed the meaning of a1 and a2. a1 is the sequence of values when Alice randomly chooses coin "a" and Bob randomly chooses coin "b". a2 is the sequence of values when Alice randomly chooses coin "a" and Bob randomly chooses coin "c". See Post #51...
There is not any communication between Alice and Bob in the Example 2 activity in OP. Alice randomly generates her own coin selections and measured outcomes without the knowledge of Bob’s coin selections or outcomes. Your suggestion that there is communication comes from not actually performing...
Not the case, however, by your own admission you apparently don't understand the OP or Post #27 which addresses the EPRB connection both which are self explanatory.
Sorry, (considering all the posts in this thread) the above paragraph is not worth responding to. It is clear to me you don't...
Yes I disagree, as stated in earlier post. Bell's theorem like any theorem can never be violated under the conditions of the theorem. It only takes one counter example to disprove a theorem. If any of the sample statistics violate the theorem then the theorem is disproved or the conditions of...
See post #27
If interested in the same answer when the settings are the same, just have Bob who is viewing from the bottom of the glass table record the opposite of the three coins that he observes. This will certainly guarantee the same outcome for each coin for Alice and Bob. In Example 1...
The law of large numbers will not change the fact that the sequences a1 and a2 are different, and a violation will still occur. (IMHO)The only way a violation would not occur is if a1 equals a2 and are identical.
Bell's theorem requires three sequences. More than three may result in a violation of the theorem that should be obvious given the two examples.
Coin tossing certainly relates to Bell's inequality. Read the post again. Bell's theorem pertains to any two-valued variables. Do you disagree...