Quantum pseudo-telepathy describes the use of quantum entanglement to eliminate the need for classical communications.[1][2] A nonlocal game is said to display quantum pseudo-telepathy if players who can use entanglement can win it with certainty while players without it can not. The prefix pseudo refers to the fact that quantum pseudo-telepathy does not involve the exchange of information between any parties. Instead, quantum pseudo-telepathy removes the need for parties to exchange information in some circumstances.

This is a cooperative game featuring two players, Alice and Bob, and a referee. The referee asks Alice to fill in one row, and Bob one column, of a 3×3 table with plus and minus signs. Their answers must respect the following constraints: Alice's row must contain an even number of minus signs, Bob's column must contain an odd number of minus signs, and they both must assign the same sign to the cell where the row and column intersects. If they manage they win, otherwise they lose.

Alice and Bob are allowed to elaborate a strategy together, but crucially are not allowed to communicate after they know which row and column they will need to fill in (as otherwise the game would be trivial).

Classical strategy[edit]
It is easy to see that if Alice and Bob can come up with a classical strategy where they always win, they can represent it as a 3×3 table encoding their answers. But this is not possible, as the number of minus signs in this hypothetical table would need to be even and odd at the same time: every row must contain an even number of minus signs, making the total number of minus signs even, and every column must contain an odd number of minus signs, making the total number of minus signs odd.

With a bit further analysis one can see that the best possible classical strategy can be represented by a table where each cell now contains both Alice and Bob's answers, that may differ. It is possible to make their answers equal in 8 out of 9 cells, while respecting the parity of Alice's rows and Bob's columns. This implies that if the referee asks for a row and column whose intersection is one of the cells where their answers match they win, and otherwise they lose. Under the usual assumption that the referee asks for them uniformly at random, the best classical winning probability is 8/9.

Currently I don't understand the point. If their task is to simply name three bits each, why can’t they, if the judge tells them the third row and the third column, name the option not based on this picture, but so that everything matches? For example -1,+1,-1 in the third row and +1, +1, -1 in the third column?