4 ms·

My understanding is that it's like two of the same scratch-off lottery ticket. There are two spots you can scratch, only one has a prize, we don't know which on

by ynniv 3y ago

My understanding is that it's like two of the same scratch-off lottery ticket. There are two spots you can scratch, only one has a prize, we don't know which one that is but it's same for both tickets. The tickets are taken to separate rooms. The people in each room then pick a spot and scratch it. How likely is it that at least one of them will win? The odds of one scratch winning are of course 50%, and the odds of the other person winning are of course also 50%. The odds of both people winning are 25%, and both people losing also 25%. The CHSH experiment suggests otherwise: that when one person loses the other person is less likely to also lose, such that the likelihood of both losing is only 15%. How can this happen? I haven't a clue. I'm not a theoretical physicist, and I haven't personally conducted this (~ $5,000) experiment. But it's a result that deeply bothers me. If you build this system you can then scratch a whole lot of tickets. Each ticket has a 50% chance of a side having the prize, but as long as both parties are consistently scratching the same side of entangled pairs they will win more often than they lose, revealing whether or not they are in agreement. I have no idea how this could be true, but people keep running the experiment and keep getting similar results. And, it seems like this could result in faster than light communication, so there's probably a better ELI-5 out there.

amluto 3y ago
> The CHSH experiment suggests otherwise: that when one person loses the other person is less likely to also lose, such that the likelihood of both losing is only 15%. How can this happen? I haven't a clue. That’s easy, even without quantum mechanics. All you need is to have a different distribution of lottery tickets. Print tickets such that, if one wins, the other is more likely to win, which you can do by having the number of winning spots per ticket be random and correlated. For example, there could be a 50% chance that neither ticket has a winning spot and a 50% chance that each ticket has one winning spot. But you could imagine a pair of tickets with internal radios, such that, when you scratch one, it tells the other one, and together they simulate a general function where the joint probability of (win on ticket 1, win on ticket 2). If you set up the probabilities appropriately, then you can’t use the tickets to send a signal, and the result is referred to in the literature as “non-signaling boxes” (the magic tickets are the boxes). Quantum mechanical entanglement can do something like this, except that the probability distributions you can generate with entanglement are a subset of the more general non-signaling boxes.
- ynniv 3y ago
  > Print tickets such that, if one wins, the other is more likely to win, which you can do by having the number of winning spots per ticket be random and correlated. In order to refute this you need to get into the CHSH experiment's design, which doesn't use the raw detection rate but compares detections that should not have similar results. Imagine that both tickets have a prize on the right side, and both participants have agreed to always choose the left side. CHSH predicts that some of the time the prize will change sides: but, crucially, only from the losing side to the winning side, and only if the other ticket lost, and even if the tickets are revealed close enough in time that a signal could not propagate between them. The conclusion of CHSH is that detections are correlated to the angle between the sensors, and not the angle between the sensors and the source. This makes absolutely no sense. There is no such thing as a winning or losing side until it's been scratched, and even if the ticket were able to randomly switch from a loser to a winner, how would it ever know that the other ticket was also a loser? And if it's true that detection is correlated to the angle between detectors, it seems easy enough to build a statistical system around variable sensor angles that is able to communicate instantaneously. I might not have interpreted everything you said, either. Internet forums aren't the best way to discuss deep problems.
  - amluto 3y ago
    > In order to refute this Refute what? I'm pointing out that classical (non-quantum) scratch-off lottery tickets can have the property you described. This is only thematically related to the CHSH game, and I'm rather confused as to what you're trying to say. The Wikipedia articles about the CHSH inequality are IMO quite bad. The conclusion of CHSH has approximately nothing to do with measurement angles. Here's what CHSH is saying, based on one formulation, which is IMO far, far nicer than Wikipedia's and makes the same point. 1. There's a game that two players could play (with the help of a neutral third party). The neutral third party picks random bits x and y, tells player A the value of x and tells player b the value of y. Each player then chooses a play (A's play is a and B's is b) according to whatever strategy they like, except that they can't communicate. They are allowed to communicate with each other beforehand, though. You can look at sites like this for more details about the game works: https://circles.math.ucla.edu/circles/lib/data/Handout-2987-2567.pdf https://circles.math.ucla.edu/circles/lib/data/Handout-2987-... 2. If A and B are only allowed to choose random numbers (jointly before the game and separately once it starts) and do local calculation, then they cannot win with more than a certain probability. This is the CHSH inequality, and it's a statement about classical statistics. 3. If A and B are allowed to produce an entangled state before the game and each hold on to one part of the state, then they can make measurements on their parts once the game starts and use those measurements as part of a better strategy. This strategy wins with probability higher than the CHSH inequality allows.

by ynniv 3y ago

amluto 3y ago

> The CHSH experiment suggests otherwise: that when one person loses the other person is less likely to also lose, such that the likelihood of both losing is only 15%. How can this happen? I haven't a clue. That’s easy, even without quantum mechanics. All you need is to have a different distribution of lottery tickets. Print tickets such that, if one wins, the other is more likely to win, which you can do by having the number of winning spots per ticket be random and correlated. For example, there could be a 50% chance that neither ticket has a winning spot and a 50% chance that each ticket has one winning spot. But you could imagine a pair of tickets with internal radios, such that, when you scratch one, it tells the other one, and together they simulate a general function where the joint probability of (win on ticket 1, win on ticket 2). If you set up the probabilities appropriately, then you can’t use the tickets to send a signal, and the result is referred to in the literature as “non-signaling boxes” (the magic tickets are the boxes). Quantum mechanical entanglement can do something like this, except that the probability distributions you can generate with entanglement are a subset of the more general non-signaling boxes.

ynniv 3y ago
> Print tickets such that, if one wins, the other is more likely to win, which you can do by having the number of winning spots per ticket be random and correlated. In order to refute this you need to get into the CHSH experiment's design, which doesn't use the raw detection rate but compares detections that should not have similar results. Imagine that both tickets have a prize on the right side, and both participants have agreed to always choose the left side. CHSH predicts that some of the time the prize will change sides: but, crucially, only from the losing side to the winning side, and only if the other ticket lost, and even if the tickets are revealed close enough in time that a signal could not propagate between them. The conclusion of CHSH is that detections are correlated to the angle between the sensors, and not the angle between the sensors and the source. This makes absolutely no sense. There is no such thing as a winning or losing side until it's been scratched, and even if the ticket were able to randomly switch from a loser to a winner, how would it ever know that the other ticket was also a loser? And if it's true that detection is correlated to the angle between detectors, it seems easy enough to build a statistical system around variable sensor angles that is able to communicate instantaneously. I might not have interpreted everything you said, either. Internet forums aren't the best way to discuss deep problems.
- amluto 3y ago
  > In order to refute this Refute what? I'm pointing out that classical (non-quantum) scratch-off lottery tickets can have the property you described. This is only thematically related to the CHSH game, and I'm rather confused as to what you're trying to say. The Wikipedia articles about the CHSH inequality are IMO quite bad. The conclusion of CHSH has approximately nothing to do with measurement angles. Here's what CHSH is saying, based on one formulation, which is IMO far, far nicer than Wikipedia's and makes the same point. 1. There's a game that two players could play (with the help of a neutral third party). The neutral third party picks random bits x and y, tells player A the value of x and tells player b the value of y. Each player then chooses a play (A's play is a and B's is b) according to whatever strategy they like, except that they can't communicate. They are allowed to communicate with each other beforehand, though. You can look at sites like this for more details about the game works: https://circles.math.ucla.edu/circles/lib/data/Handout-2987-2567.pdf https://circles.math.ucla.edu/circles/lib/data/Handout-2987-... 2. If A and B are only allowed to choose random numbers (jointly before the game and separately once it starts) and do local calculation, then they cannot win with more than a certain probability. This is the CHSH inequality, and it's a statement about classical statistics. 3. If A and B are allowed to produce an entangled state before the game and each hold on to one part of the state, then they can make measurements on their parts once the game starts and use those measurements as part of a better strategy. This strategy wins with probability higher than the CHSH inequality allows.