CHSH

In the CHSH game [1], Alice and Bob receive some bits and try to output bits that should satisfy a condition depending on the received information. This game is very similar to the Magic Square game -- players know incomplete information about the game, both players have different information available, and they are not allowed to communicate with each other once the game is in progress.

The game goes as follows: Alice receives a bit $x$ and Bob receives a bit $y$. They both output another bit, $a$ and $b$ respectively. They win the game if $x * y = a + b \mod 2$.

Now, the best classical strategy is for Alice and Bob to always output 0. In that case, they have a winning probability of 0.75 -- there are four possible combinations of $x$ and $y$ they can get, and only if they get $x=y=1$ would they lose.

However, as in the Magic Square game, Alice and Bob can come up with a better strategy if they share an entagled pair of qubits. By using a strategy that depends on the kind of measurements Alice and Bob do on their qubits, they can achieve a winning probability of roughly 0.85 (see more on this strategy in Lecture Notes on Quantum Computing, chapter 16.2).

In this application, you can see for yourself what Alice and Bob output based on the $x$ and $y$ inputs you define.

[1] J. F. Clauser, M. A. Horne, A. Shimony, and R. A. Holt. Proposed experiment to test local hidden-variable theories. Physical Review Letters, 23(15):880–884, 1969. DOI: 10.1103/PhysRevLett.24.549

You can define which bits $x$ and $y$ Alice and Bob receive.

The results show which kind of measurements Alice and Bob do on their qubits from the entangled pair and which bit they output.