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> 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 alte

by 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

> 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.