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November 25th, 2016, 03:52 AM   #11
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Originally Posted by Scott Mayers View Post
I'd prefer a link first to Bell's actual theorem then, ...
The paper has a full specification of the theory.

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... not the literal experiments. What I DO understand is the LOGIC, which my proof above satisfies. You start simple BEFORE you introduce the other issues. And if it can't pass the simple logic, then the rest is just a smokescreen to make it seem more legitimate than it actually is.
Fair enough, but I actually find your reasoning just as difficult to follow as the paper.

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I don't get why some think it satisfactory to demand one go more complex before starting simple. If you can't pass the first stage in normal logic, the rest is just 'art' .....or politics, something I'm guessing likely has more to do with this.
I'm not demanding it, it is just a suggestion because I generally prefer people to start with work on the theory as specified by others and then move onto analogies/other situations if there is confusion in order to facilitate communication, rather than start with the analogies and then apply them to the genuine case later. This is because it isn't clear from the outset whether those analogies are relevant to the problem being assessed, especially to someone who isn't an expert (like myself).

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I mean no offense, but if you want to help, begin from where I'm seeing it to show WHAT you see is in error in the logic. Math IS logic too but you don't begin to punch numbers in to a calculator unless the actual logical motivation suffices.
I don't think I can help much to be honest because the violations of Bell's inequality and the EPR paradox are not my area of expertise, but I'll have a go if I have some free time.

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Thank you for the links though. I'll keep them on hand for a time when and if I can see there is something more to it than I already understand.
No worries
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November 25th, 2016, 07:33 AM   #12
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Okay... I have spent a little bit of time reading through your posts. The first post seems fine, but your second one is really confusing:

Quote:
Originally Posted by Scott Mayers View Post
First, imagine have a pair of gloves in which you wrap in two separate boxes, mix them up randomly, and then send one a great distance away. If you open one box to discover you have a left glove, you know the other has a right one and vice verse. This is basically what might happen if have some "entangled" particle that originates in some specific location and gets sent away in two different directions. What quantum mechanics is claiming is that the boxes themselves contain BOTH a left and right version 'entangled' UNTIL you open one of the boxes. Before opening any box, they each are both 1/2 a left and right glove. But as soon as you open it, the probability "collapses" and becomes only one of them. Then the other one, no matter how far away in space it is, "knows" which one is being observed and 'becomes' the opposite.
This is a really good analogy for explaining what entangled photons are for the lay-person

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Brian opted to treat each box's content as just having either both a spherical light that flashes the SAME color of two kinds, red and blue. He had Mulder, from the "X-files" send a whole bunch of these boxes to his most skeptic partner, Scully, with a note to have her call him when she receives them.

Scully gets these boxes of which each are numbered with matching versions that Mulder keeps. So one pair of boxes might be labeled #1, with one that Scully has and one that Mulder keeps. He tells her that he wants to prove that entangles boxes with these spheres break the speed limit of light and informs whatever one discovers in one to be equal in kind to the other. But he tells her that the spherical light inside randomly flashes red or blue. Scully objects that this could prove anything because each pair can be originally 'programmed' to flash in sync from the time when Mulder sent the boxes. As such, each pair would have a 'HIDDEN FACTOR' that pre-assigns what the other box will be by default and not because they actually communicate instantaneously to the other.
This part seems fair. It's precisely the kind of question that is being addressed by Bell's work... are the entangled photons exchanging some information between themselves when the state of one of them is observed? Either the information was set to begin with ('glove' analogy) or some sort of spooky action at a distance occurred. It it either one or the other.

This is where I lose you...

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So Mulder offers an altered experiment. He sends the set of boxes with each box having 3 doors on it that both can open to reveal the flashed sphere. (note that the sphere only flashes once per random selected opening not to be repeated again. So imagine that each 'top' has a door, one on a 'side', and another on the 'front' [don't worry about how this is arranged]. We label one door with "T", and the others, "S" and "F" respectively.

The given colors flashed are Red and Blue as possibilities (and probabilities that each have of 1/2 for any one of them).

For each box, the experiment is to make a list of the colors one observes flashed upon RANDOMLY selecting ANY door EACH of Mulder or Scully selects 'silently' (without each other knowing which door they choose).

Beginning with the first box, say, Mulder might randomly decide to open "T", sees a blue flash, then records 'blue' at the beginning of the list; Scully opens her first box by randomly picking "F" and sees a red flash and does the same.

As a DEFAULT, if they choose the same identical door, they already know that the colors flashed would be identical. This is no surprise because Scully already suspects this is the WAY these flashes were programmed.
You have P(X=R) = 1/2, P(X+B) = 1/2 and now introduce a new probability P(Y=T) = 1/3, P(Y=S) = 1/3 and P(Y=F) = 1/3.

However, demonstrating that P(X) is dependent on P(Y) (i.e. the probability of seeing red or blue depends on which door was chosen) has nothing to do with the EPR paradox, Bell's theorem or non-locality. That is, even if they were dependent ('programmed'), you still can't tell whether the information related to the entangled states was placed at the beginning when the states were defined or whether there is spooky action at a distance.

This is mentioned in the review paper linked from the previous thread (Brunner et. al 2014). See equations (1) to (5) in that paper.
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November 26th, 2016, 11:02 PM   #13
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I started to respond but need to respond later. Thank you so far.
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November 27th, 2016, 04:44 PM   #14
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Since I'm having a simultaneous discussion on this, the following is also a response here:

Okay, I know PRECISELY where the error is but I will reduce this to a simple problem first that may seem unrelated:

Let's say we have a box to which a coin is placed inside.

You shake the box and we 'agree' to try to just determine whether when we open it if it will be heads.

(1) If we just do this ONCE, what are the odds prior to opening the box?
(2) And what are the odds AFTER we open the box?

In (1), it seems obvious to assert we have 1/2 a chance for it to be heads. But we don't assume that it is both a head and a tail simultaneously. We could continue shaking it but still require placing it down prior to opening it when we decide to. In entanglement, there are two boxes being shook but their entanglement means they are both being shook exactly the same way and so still holds to it having a connection to the time we began shaking them (a hidden factor). That is, if we could 'freeze frame' two boxes being shook in exactly the same way from a common initial state AND we could 'see' how the coins are oriented in space at any point in time, the coins would still be in an exact corresponding relationship, not conflicting ones. So the 1/2 probability does not require being a function of the coin itself but to our lack of knowledge of what the result will be of the two possibilities.

In (2), we should have either a 1 or 0 chance by the time we place it down just prior to opening it.

Using these facts, we know that we also could have opted to pick tails instead. As such, the same probabilities apply as above.

But then the new question is,

(3) Do we treat the collective probabilities to mean we have a 1 or 1/2 probability?

That is, we could ADD each by treating the collective odds for EACH side of the coin. But we could also multiply the odds of 1 x 1/2 to assert that we have 1/2 probability for getting either.

If we treat them collectively by adding, this treats these as two distinct events because we don't assert (heads OR tails) as a result, but (heads) XOR (tails) to which both have identical possible odds. You are not allowed to repeat this and so if you WERE to even ask what the probability of both are, you only get heads (1/2) or tails (1/2), correct?

---
I'm anticipating you agree unless you don't understand what I'm saying so far.

As such, now let me demand that we always MUST win. If we guess heads, the box MUST have heads. This is to equate the condition of the entangled condition of measuring the identical properties simultaneously. We know by default that if we learn of the velocity at one point, this assures we know the velocity of the other point entangled with it. It is only if we try to determine two distinct qualities, like velocity AND position.

Now the question is

(5) If we pick heads, then when we open the box, if it is heads, we 'stay' because this is what we expected. But if it is tails, because we demanded that it be heads, do we just pretend that it is heads?

We might think of this as having a coin with a head on each side (no tails). Then the question is do we treat the odds as reduced to 1/2 or 1?

Because the coin is 'fixed' at heads, this seems like it should reasonably be 100% to what we expected. But we could also use a regular coin with its heads and tails but just treat them both as 1.

This is the problem of the probabilities involved.

When we have Mulder initially select Top with a Blue flash about to be revealed, AND Scully picks the Top as well, the probabilities are either 100% or 50% or 50% independently. Even though we might think these are the same by adding the other two, they are not. When we use the other odds for other doors, we treat these as 50%(Blue) or 50%(Red). So I'm saying that we must consider whether

50%(Blue)
50%(Blue)

should be treated twice to represent each possibility independently? OR....

100%(Blue)

collectively? These would also equate to the same with Red. And so these could also be replaced with meaning also SAME:

50%(Same)
50%(Same)

or

100%(Same)


The distinction will make the outcome of "sames" in the actual experiment either FIT identically with the experiments or different. [(50% experiment = 50% probability) versus (50% experiment != 67% probability)]

To me, this HAS to be what maps to experiment because, while the 'value' of the times when each person picks the same door is identical, if the odds doesn't treat the events as distinct, this would be like cheating where we either allow a two-headed coin OR to opting to 'ignore' the value in times we pick the same door simply because we already KNOW the value is the same.

Does this logic sink in better?
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November 27th, 2016, 05:08 PM   #15
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By the way, this is EXACTLY the problem of the Monty Hall problem. If you are aware, the intuitive guess of 1/2 is correct IF YOU TREAT THE GAME UNIQUE. Only where you expect to play more games does the 2/3 probability apply and 'fairness' (non-randomness) is defaulted to.

In this, the only difference is that EACH event IS independent and must act as the 1/2 probability for each try. This is because you cannot reopen the same box twice.

There are other problems dealing with randomness that is also not understood. If nature is literally random, you cannot actually 'determine' the outcome as 'fair' (balanced). If nature is fair, then you cannot determine the outcome as random.

Ironically, this is EXACTLY the Heisenberg principle applied to the logic of interpreting whether something is random or fair. And by this alone, it is sufficient to then assert that not even an experiment that tries to use randomness can be determined valid!
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November 28th, 2016, 01:31 AM   #16
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Quote:
Originally Posted by Scott Mayers View Post
By the way, this is EXACTLY the problem of the Monty Hall problem.
It doesn't seem to be. Monty provides extra information by deliberately opening an empty box. If Monty doesn't know what each box contains, he can bluff and might open an empty box by luck, but that wouldn't change the odds that the contestant uses. Whether more games will be played is irrelevant.
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November 29th, 2016, 02:56 AM   #17
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Here first is the layout of the problem diagrammatically. On the right

In the attachment you should see that on the right side, where Scully's options are, the problem lies in whether one interprets the door one opens that are the same as two distinct probabilities (1/2 each) or one whole one (1). I apologize if you can't see the image well on this site. But I see it as two distinct probabilities which then give the logic using Bell's Theorem here require expecting 1/2, NOT 2/3.

skipjack (and others wanting to investigate this),

I'll give you a challenge of which I just posted on another site:

Let's say that I invite you over for coffee with myself and a girlfriend of mine. By the time you get here, my girlfriend had already been here and we were already in the midst of some discussion.

At the kitchen table you see two boxes sitting there and wonder what we've been discussing. I decide to give you a quick update on what I was doing:

I tell you that in one box I have placed a toy Matchbox car in one of the boxes and my girlfriend, Shannon, has just picked a box prior to you walking in and I have just asked her if she'd like to switch to the other box. Without telling you what else we've been doing, I assure you ONLY that in one of those boxes is that car and the other is empty. She has randomly picked a box but when I asked her if she'd like to switch, she asked me why that should matter? I tell you that I told her that she would actually improve her chances to do so.

Given ONLY what I've told you, should you logically take her side or mine? [You cannot ask for more information either.]
Attached Images
File Type: jpg QM Bell'sTh illust.jpg (83.6 KB, 7 views)

Last edited by Scott Mayers; November 29th, 2016 at 03:06 AM.
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November 29th, 2016, 04:38 AM   #18
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This is like the 5th or 6th analogy now, so it just keeps getting more and more confusing!

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Originally Posted by Scott Mayers View Post
Here first is the layout of the problem diagrammatically. On the right

In the attachment you should see that on the right side, where Scully's options are, the problem lies in whether one interprets the door one opens that are the same as two distinct probabilities (1/2 each) or one whole one (1). I apologize if you can't see the image well on this site. But I see it as two distinct probabilities which then give the logic using Bell's Theorem here require expecting 1/2, NOT 2/3.
Yeah.... I don't understand that at all. If you open a door on a box but the flash (red or blue) is not dependent on the door at all, you'll find that the probability of getting blue or red is 50:50, which is more or less independent of the probability of opening a door. That is

$\displaystyle P(A \cap B) = P(A) P(B)$

where P(A) is the probability of getting a flash of blue or red and P(B) is the probability of getting a particular door after picking one at random.

However, even if you performed the experiment and found that the choice of door did matter, that has nothing to do with with entanglement or Bell's theorem because you could just have two probability distributions that are dependent on each other in a 'classical' manner, that is:

$\displaystyle P(A \cap B) \neq P(A) P(B)$

but

$\displaystyle P(\lambda A \cap \lambda B) = P(\lambda A) P(\lambda B)$

I think something was specified to that effect in the review article. Maybe I just don't know enough about entanglement, but it isn't clear to me why the Monty Hall problem is even relevant to Bell's theorem because a satisfactory solution to the Monty Hall problem is derived using 'classical' statistics.

From what I understand of Bell's theorem, an inequality can be created based on the probability of certain states of individual particles relative to the state of the two-particle system as a whole and that inequality is broken, showing that it doesn't make sense to try and describe the two-particle system using some form of combination of individual states. Like I said earlier, that's probably where you should look if you want to find problems with Bell's theorem.

Quote:
skipjack (and others wanting to investigate this),

I'll give you a challenge of which I just posted on another site:

Let's say that I invite you over for coffee with myself and a girlfriend of mine. By the time you get here, my girlfriend had already been here and we were already in the midst of some discussion.

At the kitchen table you see two boxes sitting there and wonder what we've been discussing. I decide to give you a quick update on what I was doing:

I tell you that in one box I have placed a toy Matchbox car in one of the boxes and my girlfriend, Shannon, has just picked a box prior to you walking in and I have just asked her if she'd like to switch to the other box. Without telling you what else we've been doing, I assure you ONLY that in one of those boxes is that car and the other is empty. She has randomly picked a box but when I asked her if she'd like to switch, she asked me why that should matter? I tell you that I told her that she would actually improve her chances to do so.

Given ONLY what I've told you, should you logically take her side or mine? [You cannot ask for more information either.]
It makes no difference because you have two boxes and no one knows the contents of the boxes, so it's 50:50 either way regardless of whether you've asked someone to switch or not.

Maybe you meant to say "One person is undergoing the classic Monty Hall problem with three boxes and is just about to make a decision of whether to switch to another box or not after being shown that one of the boxes is an empty box. You, unaware of this sequence of events, can choose whether to be updated by the current scenario and follow the opinion of the person doing the Monty Hall problem or just pick a box at random ignorant of the situation". In that particular instance, I would side with the person undergoing the Monty Hall problem every time because

$\displaystyle P(A=T|B=F) = 2/3$

whereas

$\displaystyle P(A=T) = 1/3$

so I'd take 2/3 success over 1/3 any day. This is basically the same as if I went and did the Monty Hall problem myself

[Three boxes are A, B and C. Contents are T ('True' or 'contains car') or F ('False' or 'empty')]

Last edited by Benit13; November 29th, 2016 at 04:51 AM.
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November 29th, 2016, 03:37 PM   #19
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Benit, you definitely missed my explanation. Click and enlarge the image I inserted to actually see how I laid it out in depth. I covered all odds literally. You have to look at the options Scully picks and you'll see that for each of Mulder's 6 options (1/3 door x 1/2 color) she has one possibility for each where she'd pick 'THE SAME' Door. In that case, it is either with a weight of 1 as a whole of 1/2 + 1/2 independently, even though the color is the same regardless. This is exactly the kind of error made in the Monty Hall problem.

As such I asked to step back and asked the smaller challenge to make a point.

You asserted that you would interpret my girlfriend correct to state that she doesn't need to switch because the odds are 1/2. You made my case for one independent case of the Monty Hall game. What you MISS, is that before you came in to my place, I had an original three boxes to which I asked Shannon to pick and then removed one just as you walked in the door. Without knowing what occurred by your perspective, you are CORRECT to interpret that Shannon doesn't need to switch.

But then if I told you about our prior part of the game, it is the Monty hall's first round with three boxes and my removal. Thus, by your faith in the Monty Hall's solution, this would be 2/3 for Shannon to switch. How can she both have an advantage to switch a 2/3 AND no difference at 1/2 for the same identical conditions that only differ on whether you OBSERVED the first part of the game. This is like a hidden factor in the Bell test.

Ignore other factors and just deal with this reduced example. Do you understand how it can be both 1/2 and 2/3 depending on nothing but when you observed the game? And you'd have to also guess as Einstein had, that there is a "hidden" factor, even if you don't know what, should the actual result turn out to be 2/3!!
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November 30th, 2016, 02:26 AM   #20
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But then if I told you about our prior part of the game, it is the Monty hall's first round with three boxes and my removal. Thus, by your faith in the Monty Hall's solution, this would be 2/3 for Shannon to switch. How can she both have an advantage to switch a 2/3 AND no difference at 1/2 for the same identical conditions that only differ on whether you OBSERVED the first part of the game. This is like a hidden factor in the Bell test.
No... Shannon has more knowledge than me, so Shannon's probability is P(A|B) = 2/3, whereas mine is just P(A)=1/2. The fact that $\displaystyle (PA|B) \neq P(A)$ allows Shannon's choice of box to be more informed than mine.

Quote:
Ignore other factors and just deal with this reduced example. Do you understand how it can be both 1/2 and 2/3 depending on nothing but when you observed the game? And you'd have to also guess as Einstein had, that there is a "hidden" factor, even if you don't know what, should the actual result turn out to be 2/3!!
You mean either 1/2 or 2/3 depending on the situation? Well, of course, but that's because of $\displaystyle P(A|B) \neq P(A)$, not because of anything from quantum mechanics; that happens in classical statistics too.

Last edited by Benit13; November 30th, 2016 at 02:29 AM.
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