November 25th, 2016, 03:52 AM  #11  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,836 Thanks: 592 Math Focus: Physics, mathematical modelling, numerical and computational solutions  The paper has a full specification of the theory. Quote:
Quote:
Quote:
Quote:
 
November 25th, 2016, 07:33 AM  #12  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,836 Thanks: 592 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
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:
Quote:
This is where I lose you... Quote:
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 nonlocality. 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.  
November 26th, 2016, 11:02 PM  #13 
Newbie Joined: Mar 2016 From: Saskatoon, Saskatchewan, Canada Posts: 29 Thanks: 1 Math Focus: Logic 
I started to respond but need to respond later. Thank you so far.

November 27th, 2016, 04:44 PM  #14 
Newbie Joined: Mar 2016 From: Saskatoon, Saskatchewan, Canada Posts: 29 Thanks: 1 Math Focus: Logic 
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 twoheaded 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? 
November 27th, 2016, 05:08 PM  #15 
Newbie Joined: Mar 2016 From: Saskatoon, Saskatchewan, Canada Posts: 29 Thanks: 1 Math Focus: Logic 
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' (nonrandomness) 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! 
November 28th, 2016, 01:31 AM  #16 
Global Moderator Joined: Dec 2006 Posts: 16,369 Thanks: 1172  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.

November 29th, 2016, 02:56 AM  #17 
Newbie Joined: Mar 2016 From: Saskatoon, Saskatchewan, Canada Posts: 29 Thanks: 1 Math Focus: Logic 
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.] Last edited by Scott Mayers; November 29th, 2016 at 03:06 AM. 
November 29th, 2016, 04:38 AM  #18  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,836 Thanks: 592 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
This is like the 5th or 6th analogy now, so it just keeps getting more and more confusing! Quote:
$\displaystyle P(A \cap B) = P(A) P(B)$ 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 twoparticle system as a whole and that inequality is broken, showing that it doesn't make sense to try and describe the twoparticle 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:
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 $\displaystyle P(A=TB=F) = 2/3$ 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.  
November 29th, 2016, 03:37 PM  #19 
Newbie Joined: Mar 2016 From: Saskatoon, Saskatchewan, Canada Posts: 29 Thanks: 1 Math Focus: Logic 
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!! 
November 30th, 2016, 02:26 AM  #20  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,836 Thanks: 592 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Quote:
Last edited by Benit13; November 30th, 2016 at 02:29 AM.  

Tags 
bell, error, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Bell's theorem: simulating spooky action at distance of Quantum Mechanics  humbleteleskop  Physics  17  July 25th, 2015 03:38 AM 
Interpolating (I think?) from a bell curve.  marlan  Probability and Statistics  1  March 10th, 2015 08:20 PM 
Something about Bell's Theorem  J Thomas  Physics  4  January 13th, 2015 09:47 PM 
Percentage error and Binomial theorem  Sam1990  Applied Math  1  October 18th, 2014 01:53 PM 
Taylor's theorem. Usage + Quantify error?  bentley4  Calculus  1  May 19th, 2011 10:23 AM 