###### 10.02.2017difficulty level - Q

## Playing cards with quantum entanglement

**by Gláucia Murta**

You have probably heard that entanglement is a very strong correlation way beyond anything we can conceive classically. However, as we’ve seen from Jeremy’s post , these strong correlations by itself do not allow us to send any information to the other part. So what can we use entanglement for?… To play games!

Let us consider that Alice and Bob are in a bar with two other friends and their friends propose to play a drinking game: the friends are going to be the referees, one referee for Alice and one referee for Bob. Each referee is going to randomly choose one between two cards, red or black (the ‘question cards’), and give the card to Alice and Bob respectively. Alice and Bob also have two cards in their hands (one red and one black ) that are going to represent their answers. After receiving the question cards, they decide upon one answer by putting the card facing down on the table. After both players have chosen their answers, the cards are turned up and the answers are revealed. Their challenge is to fulfill the following condition: every time they receive cards of the same color they have to reply with equal cards (i.e. both red or both black) and every time they receive cards of different colors they have to reply with different cards, as described in the following table:

Alice’s question | Bob’s question | Winning reply |

or | ||

or | ||

or | ||

or |

To make the game more fun they add a payoff: whenever Alice and Bob win they receive a beer from their friends, and if they loose they have to pay a beer to their friends. Should Alice and Bob agree to play the game?

An important point is that they don’t know which card the other has received, and unless they cheat (by blinking or trying any kind of communication) they cannot know it. And even though there is no fun in winning a game by cheating (at least I find it way more rewarding when one can win playing by the rules), in order to guarantee our characters’ good-nature let us stress that their friends are watching them during the game, therefore, any cheating attempt would be easily caught!

Now let us analyse one possible fair strategy: upon receiving the question cards from their friends they randomly decide on their reply. Note that, for every pair of possible question cards, there is two possible correct answers out of four (check Table 1), therefore by playing with the random strategy the probability of winning the round is one half. Then if they decide to play using the random strategy, they are going to have a lot of fun and after many hours in the bar they would have basically received as many beers as they would have payed.

But what if their friends propose that whenever Alice and Bob loose they have to pay four beers instead of one? Should they still play? If they do, it is better that they think about a smarter strategy if they want to keep having fun.

Let us analyse the following strategy: Alice and Bob always reply with cards of the same color as their question cards (i.e. upon receiving , Alice replies and if she receives she replies , and Bob does the same). Now we see that with this strategy Alice and Bob can win all the rounds! They should just be smart enough to come up with the best strategy.

After some hours loosing a lot of beers, their friends come up with a more interesting game, whose winning condition is described by the following table:

Alice’s question | Bob’s question | Winning reply |

or | ||

or | ||

or | ||

or |

For the reader familiar with nonlocality, you have probably recognized the famous CHSH game!

Remember that if Alice and Bob lose they have to pay 4 beers to their friends whereas if they win they only get 1 beer to share. So they better think about a good strategy.

Should they accept to play the game? And which strategy could they use?

With a bit of thinking we can come up with the following strategy: no matter the question cards they receive, they always reply red . Note that unless they both receive question they will succeed. So by playing many rounds of the game they are going to win about ¾ of them.

Actually one can prove that, from all possible strategies Alice and Bob can think of, the maximum probability of success in the CHSH game will be ¾.

Now let us calculate what would be the net gain attained by Alice and Bob by playing the CHSH game with the optimal strategy:

$latex 3/4 times (1 text{ beer}) + 1/4 times (-4 text{ beers})=-1/4 $

This value indicates that by using the optimal strategy the net result will be to pay one beer for their friends in every 4 rounds.

Doesn’t sound like an interesting game to play.

But what if they can make use of quantum mechanics?

Then they can turn the tables!

So finally the super strong correlations that we get with entanglement can play their role!

But you may wonder how can that be used? The idea is the following: Alice and Bob are going to use these strong correlations in order to construct their strategy, i.e. they are going to use the strong correlations observed in the outcome of measurements in an entangled state in order to decide on their answer cards.

So to make it (a little bit) more concrete, let us assume that Alice and Bob have access to diamonds with entangled NV centers (see Norbert’s post ) and that they have equipment to measure the electron spin of these NV center’s in different directions. Their quantum strategy will then be to associate each question card to a particular direction of spin to be measured (e.g. if Alice receives she makes a measurement of spin the x-direction, if she receives she measures the spin in the z-direction) and each reply card to one of the possible outcomes of the measurement (e.g. spin down= , spin up= ). (The CHSH inequality was used- or rather, the CHSH game was played- in the groundbreaking experiment performed in QuTech in 2015)

The point here is: if Alice and Bob are smart and experimentally skillful enough, they can reach a success probability of ~0.85! Calculating the net gain

$latex 0.85 times (1 text{ beer}) + 0.15 times (-4 text{ beers})=+1/4$

we see that now they have a net advantage of 1 beer per every 4 rounds!

This simple fact that quantum systems can give you an advantage over classical strategies in playing such games has profound fundamental implications. But this is a subject for a future post.

For now I hope at least you are convinced that you can use entanglement to have fun!

Finally, if you are wondering whether this is an over simplified game that is far from the card games we play in real life, check this out! One can actually improve your success probability in Bridge by using quantum strategies! And I’m sure one can find CHSH instances in many other card games.

Gláucia Murta is a theorist at QuTech. She works with quantum cryptography. Besides trying to circumvent the action of malicious people with quantum mechanics, she enjoys eating, betting, and (whenever possible) practicing a variety of dances.