Why isn't the best case classical solution to the CHSH game 100%?

Question

[Edit 2] I would prefer to just forget that I had ever asked this question (because I was so wrong it's embarrassing), but for the sake of people who possibly make the same mistake I did, I'll try to fix the question so it accurately represents what I was trying to ask. In the end, it wasn't a constraint in the original theory that prevented the use of my strategy, it was that my "100% chance to win" was just a mathematical goof on my part, and the actual win chance for my strategy was 75%, just like it was supposed to be.

First off, please forgive my ignorance when it comes to physics and Quantum Mechanics, I'm just a programmer that likes to pretend he understands science.

There is an experiment that is used to prove that there are no "hidden variables" at work in quantum weirdness, but that particles in a superposition of states really do collapse only at the moment of measurement.

The experiment uses a game known as the "CHSH game" (summarized at http://www.americanscientist.org/issues/num2/2014/4/quantum-randomness/1) which is basically

With cooperating players "Alice" and "Bob":

1. After agreeing on a strategy they are separated and cannot communicate with each other in any way
2. They are each given a sealed envelope that randomly contains either a red card or a blue card
3. After being separated, they each open the envelope and must then raise either one or two fingers
4. They win if both cards are red and they raise a different number of fingers, or either card is blue and they raise the same number of fingers.

In the game, the best classical strategy for winning is for the players to ignore their cards and both raise one finger, which gives a 75% chance of winning. However, if they each have one of a pair of entangled particles and measure them slightly differently depending on which card they chose, they can increase their chance of winning to ~85.4%, which is only possible because the method the first person used to measure their particle changed the results that the other person would get when they measured their particle. The difference in the two chances of winning is known as Bell's Inequality, and is used as proof that there's actual quantum weirdness going on.

But it seems to me that the best classical strategy for the game (as summarized above) isn't to always raise one finger, but for Bob to always raise 2 fingers, and for Alice to raise 1 finger when her card is red, and 2 fingers when her card is blue. This results in a 100% chance to win.

This is such a simple strategy that there must be some condition in Bell's theory that prevents it from being used. Or perhaps the summary of the CHSH game in the article linked above skipped over some well-known context to the game that prevented it's use.

So what are the conditions in Bell's theory that prevent this 100% win strategy from being used in the CHSH game?

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Freudian Slip

Comments

Good question, but it doesn't actually have anything to do with physics since you're just asking about the classical game. We can migrate this question to another site where it would be more appropriate (so don't repost it elsewhere).

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David Z

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Nope, it still belongs in Physics - the assumption is that the original theory (I think it's "Bell's Inequality") contained some qualifying assumptions and/or conditions to the game that were missing in the summary I had read to exclude the possibility that I had raised.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Freudian Slip

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Bell's inequality has to do with physics, yes, but the question you're asking has nothing to do with Bell's inequality or quantum mechanics at all. It's purely a logic puzzle. (You could remove all references to QM from this question without changing what it's asking.)

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David Z

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Good point. I was unclear in what I was expecting as an answer. Ultimately, I was looking for the part of Bell's inequality or the CHSH game that was was missing from my "executive summary" of the CHSH game that prevented my "obvious" (and completely flawed) counter-example from being considered as the best-case classical solution. The fact that the answer turned out to have nothing to do with physics and was just "there actually is a lose condition and you just didn't see it" is purely accidental.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Freudian Slip

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This question appears to be off-topic because it is about game strategy.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Brandon Enright

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This is getting silly. How can a question about the CHSH game be off topic on a physics site? The idea that questions have to be "about physics" has always been a bit flaky (i.e. blatantly subjective), but if you're going to draw the line here it's just ridiculous.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Nathaniel

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The CHSH game has been invented by physicists to study a physics question (nonlocality). If you don't like to speak about games, call it the CHSH thought experiment!

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Frédéric Grosshans

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@Nathaniel I would respond that the idea that the guiding principle for on-topicness here should be anything other than "about physics" is ridiculous. Though it's counterproductive to use words like "ridiculous" because clearly what is ridiculous to one person is not to another. Anyway, let's not get into this further here. Maybe on Physics Meta, if you feel strongly enough about it.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David Z

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@DavidZ I feel strongly about it, so I posted a question on meta.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Frédéric Grosshans

Answer 1

If Bob's card is blue and Alice's is red, then your strategy leads to a loss. (One card is blue, but they raise different numbers of fingers.) So your strategy matches the best possible 75% win rate but doesn't beat it.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David Z

Comments

Yeah, chalk that one up to Excel being the wrong tool to check assumptions with.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Freudian Slip

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How is/was Excel the problem?

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David James

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It was a total bone-headed goof on my part. I used relative cell references instead of absolute references in my result cells when referring to a set of strategy cells. So when I copied-pasted the result cells to represent different conditions they started referring to the previous results instead of the strategies. I realized it about 0.5 sec after posting. Worst is I knew this already because of this blog post.

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user Freudian Slip

Answer 2

Here are all possibilities:

| C1 | C2 | F1 | F2 | O |
| -- | -- | -- | -- | - |
|  R |  R |  1 |  1 | L |
|  R |  R |  1 |  2 | W |
|  R |  R |  2 |  1 | W |
|  R |  R |  2 |  2 | L |
|  R |  B |  1 |  1 | W |
|  R |  B |  1 |  2 | L |
|  R |  B |  2 |  1 | L |
|  R |  B |  2 |  2 | W |
|  B |  R |  1 |  1 | W |
|  B |  R |  1 |  2 | L |
|  B |  R |  2 |  1 | L |
|  B |  R |  2 |  2 | W |
|  B |  B |  1 |  1 | W |
|  B |  B |  1 |  2 | L |
|  B |  B |  2 |  1 | L |
|  B |  B |  2 |  2 | W |

Legend:

* C1 = card 1 (Red / Blue)
* C2 = card 2 (Red / Blue)
* F1 = fingers 1 (1 or 2 fingers)
* F2 = fingers 2 (1 or 2 fingers)
* O = outcome (Win / Loss)

The strategy of "always raise 1 finger" (independent of the color of your card) has a 75% probability of winning:

| C1 | C2 | F1 | F2 | O |
| -- | -- | -- | -- | - |
|  R |  R |  1 |  1 | L |
|  R |  B |  1 |  1 | W |
|  B |  R |  1 |  1 | W |
|  B |  B |  1 |  1 | W |

This post imported from StackExchange Physics at 2014-06-28 21:05 (UCT), posted by SE-user David James