# "quantum" card game

1. Dec 15, 2015

### jerromyjon

Hi all, I've been pondering this for some time and have decided to ask for help. My goal is to create a card game with "simulated" quantum entangled card deals, in which there must be 3 individual properties? For example red/black face, red/blue deck, high/low value. The goal would be to use the enhanced quantum probability between you and your partners chance of declared choice and perhaps choosing the angle of the next card deal you bid on? Obviously I can't get real cards to present quantum probability but I suppose worst case scenario I can have a chart of calculated odds for the minimal degrees of freedom necessary to gain an advantage against a computer team seated 90 degrees offset (east/west vs north/south)?

My goal is to get a better understanding of the inequality so any simulation of this nature would certainly save me the trouble! I am sure once I have a complete understanding I could make the card game as well for those who enjoy cards and quantum physics as much as I do!

2. Dec 15, 2015

### Khashishi

Sounds cool! One way of interpreting quantum mechanics is with negative probabilities.

3. Dec 15, 2015

### jerromyjon

That's where it gets fuzzy. I know 180 degree rotation would induce perfect correlation, while 90 degree rotation induces 50/50 odds? And various angles would induce positive and negative influences?

The overall realistic expectation I have is that you could a) play poorly and lose. b) use classical probability rules like any other card game and win 50/50 against the same c) use quantum probability to your advantage and always win more than a classic team d) win 50/50 with quantum advantage. Does that sound realistic?

4. Dec 15, 2015

### stevendaryl

Staff Emeritus
Something that might seem a little artificial in a card game is the measurement choice aspect. To make it similar to the EPR experiment, you would have two players, Alice and Bob. Each would be given a choice of one out of three "measurements" to perform, and each measurement would have two possible results. So there are 6 (not just three) possible outcomes. Neither player can perform more than one measurement.

So for example, maybe Alice and Bob are each sent 3 cards. One with a red back, one with a blue back, and one with a green back. The player picks a color, and then flips that card over. Then he sees either a red suit (heart, diamond) or a black suit (club, spade). So there are 6 possible values: (red-back/red-suit), (red-back/black-suit), (blue-back/red-suit), (blue-back/black-suit), (green-back/red-suit), (green-back/black-suit).

To make it like EPR, you would say something like: (this is taken from spin-1/2 EPR, with the three back colors corresponding to three measurement axes: 0o, 120o, 240o, measured clockwise in the x-y plane starting with the y-axis)
1. No matter which back is chosen, each player has a 50/50 chance of getting a red suit or a black suit
2. If both players choose the same back color, they always get opposite results.
3. If players choose different colors, they get the same result 75% of the time, and opposite results 25% of the time.
There is no way for a dealer to achieve these statistics without cheating.

5. Dec 15, 2015

### jerromyjon

That's where I'd have to simulate sets of possible deals until all possible deals have been exhausted listing if any fit the range of predicted and verified outcomes?

6. Dec 15, 2015

### Staff: Mentor

You don't have to, John Bell did that for you decades ago. With classical cards you cannot get these probabilities. You need "quantum cards".
Not sure how to make a nontrivial game out of it.

You might be interested in quantum werewolf (that is an actual game).

7. Dec 15, 2015

### jerromyjon

That certainly looks interesting, but seems a little less revealing than what I'm looking for, that is something more for when I would be "fluent" in quantum probability and could utilize it to my advantage, perhaps? I think the game I need must display the results openly to provide the level of intimacy I desire, to add another minute to learn, lifetime to master game with a unique and powerful aspect.

I think that 2 decks in my mind would have to be enough simply because you never get more than 2 out of 3 observations, in order to realize actual quantum statistical effects as opposed to simply classic outcomes? I mean its certainly eluding me for starters how many cards would need to be dealt to have the effects consistently and provably advantageous. I mean if I started with double deck pinochle deck and "stacked the deck" is there enough degrees of freedom to express the odds any quantum probability could exhibit or is that what Bell excludes the possibility of?

Last edited: Dec 15, 2015
8. Dec 15, 2015

### Mentz114

The only way to get the quantum statistics from a card game is if the cards can change their face value after being dealt. Cheating, in fact.

9. Dec 15, 2015

### Staff: Mentor

Make a computer game out of it. Classical cards just don't work, they are classical.

I never played quantum werewolf, but as far as I heard it is quite chaotic. You can be something like 30% dead, 20% wolf and 10% seer.

10. Dec 15, 2015

### jerromyjon

Yes, of course. I'm trying to figure out how to go about it, I mean achieving "outcomes" (computer generated) that realistically portray true quantum nature.

11. Dec 15, 2015

### jerromyjon

That's analogous to saying measurement results change after they are recorded? We only see the correlation after the result which have been recorded are compared? I'm confused.

12. Dec 15, 2015

### Staff: Mentor

Measurement results don't change.
But you cannot fix the measurement results before you decide what will be measured.

13. Dec 15, 2015

### PeroK

Cards have the problem that all observables are compatible. You really need cards where if you know it's a spade, then the denomination is probabilistic, and if you then determine the denomination the suit becomes probabilistic. In other words, there is always uncertainty about every card in your hand.

14. Dec 15, 2015

### Mentz114

You can't get the quantum limits with classical objects. 'Quantum cards' might not have a face value until they interact.

15. Dec 15, 2015

### jerromyjon

I let out that possibility in the wrong context earlier, I was thinking about that in the beginning with the decks having a hidden "non-local" value (red/blue deck) I also thought initially I would need to use entanglement as in an orientation hidden from the other team. I obviously don't have the whole setup down yet.

16. Dec 15, 2015

### jerromyjon

That is what I'm implying. It's like a game of war with complex variables.

17. Dec 15, 2015

### Strilanc

Relevant: quantum game theory.

My expectation is that it will be quite difficult to come up with a fun card game that uses entanglement in a crucial way. It's also worth noting that secure multi-party computation, or just a trusted referee, are strictly more powerful than quantum effects as far as a game goes. SMPC is general enough to simulate all the relevant quantum aspects (though, unlike a real quantum system, SMPC requires rounds of fancy encrypted back-and-forth communication to do so).

18. Dec 20, 2015

### jerromyjon

That's #1 so I imagine only usefulness could be expected of it's mastery. SMPC seemed like a requirement in the beginning then I thought a "bid" as a message in fashion which allows a random element "flipping" its meaning would suffice with a choice of rotation (clock position perhaps?) but now again I see the multiple layers of constructive and destructive interference that one should have to capitalize on to increase their odds while reducing their opponents.

I'm leaning strongly towards the concept of being dealt a quantum "object" in the form of 2 cards, in which the opposite entangled pair of inherently separate cards are dealt to your partner and yourself, and you have a 2 bit message to use in the form of bidding a suit similar to pinochle. Flexibility in the rank of face value and the number of cards per face (multiple identical cards and AQJ1098/765432 and a possible random deck color inversion 50/50 chance the bid gets flipped to opposite suit. Entanglement swapping between the cards, so many possibilities! But, obviously, I need to start small with a simple 1 card basic version and produce a quantum card dealing program that exhibits inequality simply through the ranking and clock position that each card is dealt at by some sensible system which I have a feeling will need some type of classical random element (the suits) as opposed to the quantum probability of rank. That's where I am at now. How can I "display" quantum probability. The only simple way that makes logical sense to me is to say that perhaps K is dealt most likely and A is dealt least likely and 7 would be the "classical" 50/50 equality so that would be 45 degrees making 4 choices -22.5,0,22.5,45. These are the angles I've seen numerous times but I still don't see anything specific as to how and at exactly what amount do the values change to know I have an applicable example of actual quantum probability? I guess If it exceeds 2 times square root of 2 it is too much... I'm going to try something.

19. Dec 21, 2015

### Ilja

20. Dec 21, 2015

### jerromyjon

That is definitely the type of game I am interested in, alas I could not find a link to actually "play" it. The main problem I am stuck on is finding a description of the inequality. The only substantial quantitative bits I have found are "3/4 chance as compared to 2/3 chance" and "a total 2 x sqr(2) = ~2.8 as compared to 2".
Nevermind I think I found the solution.

21. Dec 23, 2015

### jerromyjon

"So close but no lumps" never felt so truer, the art of gravy is much simpler than the art of quantum physics... I feel so close but not quite sure if there is a way to smooth out the probabilities I'm dealing with. I'm down to the element "rho" which had the featured thread about pi, and how it relates to my problem of "adapting" a flat 2d square with 4 , 90 degree corners to 1/8 of a sphere with 3 , 90 degree corners (from equator to north pole, right turn back to equator, right turn back to start) so my latest idea is that I want to "flatten" out this this triangle by looking through the center of the curved face into the center of the sphere and relate it to the square so that a random point around the circle correlates to the "curve" inside or outside the square which would be rho2? No that's wrong I need square root of rho" squared somehow. Rho, rho, rho your square, gently, take great care, because your paddle's a triangle curved to interfere with a squared.

22. Dec 26, 2015

### jerromyjon

Sorry for the lapse of sanity, this is still making me crazy! I think the problem I am having most is understanding is the actual constraints of quantum behavior, so if someone could verify that I have these basic facts correct and if my simple logic fits to describe what is known. I am using cards simply as a familiar media for understanding probability. The simplest form of a quantum object that I can imagine is 3 cards which are of value red or black. Each of these 3 cards represents a measurement axis or detector angle or whatever example of a quantum variable this best applies to. To "initialize" these variables in a classic nature, take 1 black card and 1 red card, equate that with 50/50 probability, (A gets 1 and B gets the opposite card which we'll dispose of for this basic example) so now you have 3 cards with 50/50 chance of red or black individually which yields 8 cases: [1] R R R [2] R R B [3] R B R [4] R B B [5] B R R [6] B R B [7] B B R [8] B B B You choose any 2 of these cards in any of 8 cases (if the cards had literal faces on them as dealt) you get a match 50% of the time, but quantum probability is 1/3? This is where I lose clarity, but I have a concept of each case having it's counter case as a hidden set which gets mixed (R leaves BRB and B leaves RBR) to adjust the odds against a match. Does that make sense for "getting the correct quantum probability"? At the simplest level all I need to do is say you get x/x chance of getting say 1 red + 3 black = 1/4 odds of red when you'd get at least 1/3 chance?

Last edited: Dec 26, 2015
23. Dec 26, 2015

### eufisica

In 2012, with other Physics teachers, we built a card game called BOSEMON.
You can find the instructions and the deck here: https://sites.google.com/site/bosemongame/

Each card has the characteristics of each particle (like mass, charge, spin,...) and antiparticles are included in the game. The aim objectives are build the proton, families, hydrogen atom,...

We translated this educational game in several languages and it was implemented already in the high school.

24. Dec 26, 2015

### jerromyjon

...but has no quantum behavior.
What I am trying to get at seems rather simply like saying where you'd classically have 3 cards you really have 6 cards with a disproportionate number of matches or would a larger pool of cards be needed or am I missing something that makes this more complicated?

Last edited: Dec 26, 2015