4 ms·

> 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 correla

by 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.
- ynniv 3y ago
  > I'm pointing out that classical (non-quantum) scratch-off lottery tickets can have the property you described. No, if A always scratches the left, and B alternates between left and right, your classical trick of having some tickets with two winners and some with two losers won’t produce the correct win distributions for B’s choices. The CHSH game is an attempt at explaining the experiment. The core of CHSH is a formula describing an expected distribution: `S = E(a, b) - E(a, b') + E(a', b) + E (a', b')`, where classical explanations can achieve an S with magnitude approaching 2. Experiments arriving at an S > 2 are said to support the quantum theory.
  - amluto 3y ago
    > No, if A always scratches the left, and B alternates between left and right, your classical trick of having some tickets with two winners and some with two losers won’t produce the correct win distributions for B’s choices. My proposal was intended to be independent of A and B’s strategy. As a fully worked-out example, suppose the tickets are printed like this: With probability 42.5%: all positions win With probability 42.5%: all positions lose With probability 7.5%: Both of A’s options win and both of B’s options lose With probability 7.5%: Both of A’s options lose and both of B’s options win. This follows the rules of the game, and it has the following properties. A wins with probability 50% regardless of their strategy, as does B. And A and B get the same outcome 85% of the time. For the CHSH inequality (following the pretty-bad Wikipedia article), each E term is the probability that A and B both win when they use the indicated strategies, which is 0.425. So S=0.85, which is well within the limit, as it should be since these are extremely boring classical tickets.
    - ynniv 3y ago
      My point wasn't to derive the best classical strategy, but to describe the problem in simple terms that people who aren't familiar with quantum mechanics could understand. QM is a very academic field, verging on intentionally obtuse, and that keeps non-experts from participating. Cozy if you're entrenched, but not the best for fresh approaches.

by ynniv 3y ago

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.

ynniv 3y ago
> I'm pointing out that classical (non-quantum) scratch-off lottery tickets can have the property you described. No, if A always scratches the left, and B alternates between left and right, your classical trick of having some tickets with two winners and some with two losers won’t produce the correct win distributions for B’s choices. The CHSH game is an attempt at explaining the experiment. The core of CHSH is a formula describing an expected distribution: `S = E(a, b) - E(a, b') + E(a', b) + E (a', b')`, where classical explanations can achieve an S with magnitude approaching 2. Experiments arriving at an S > 2 are said to support the quantum theory.
- amluto 3y ago
  > No, if A always scratches the left, and B alternates between left and right, your classical trick of having some tickets with two winners and some with two losers won’t produce the correct win distributions for B’s choices. My proposal was intended to be independent of A and B’s strategy. As a fully worked-out example, suppose the tickets are printed like this: With probability 42.5%: all positions win With probability 42.5%: all positions lose With probability 7.5%: Both of A’s options win and both of B’s options lose With probability 7.5%: Both of A’s options lose and both of B’s options win. This follows the rules of the game, and it has the following properties. A wins with probability 50% regardless of their strategy, as does B. And A and B get the same outcome 85% of the time. For the CHSH inequality (following the pretty-bad Wikipedia article), each E term is the probability that A and B both win when they use the indicated strategies, which is 0.425. So S=0.85, which is well within the limit, as it should be since these are extremely boring classical tickets.
  - ynniv 3y ago
    My point wasn't to derive the best classical strategy, but to describe the problem in simple terms that people who aren't familiar with quantum mechanics could understand. QM is a very academic field, verging on intentionally obtuse, and that keeps non-experts from participating. Cozy if you're entrenched, but not the best for fresh approaches.