Let’s play a game. We form two teams of three, the “players” (Andy, Bob, and Charles) and the “interrogators”.^{[1]} The rules of this game are as follows:

- Either all players are asked for the value of X, or one player is asked for the value of X while the two other players are asked for the value of Y.
- The possible values of both X and Y are +1 and −1.
- If all players are asked for the value of X, they win if (and only if) the product of their answers equals −1. Otherwise they win if (and only if) the product of their answers equals +1.

Once the questions are asked, the players are no longer allowed to communicate with each other. Prior to that, they may work out a strategy. Is there a fail-safe strategy? Can they make sure that they will win? Ponder this before you proceed.

The obvious strategy is to use pre-agreed answers.

Let’s call them X_{A}, X_{B}, X_{C}, and Y_{A}, Y_{B}, Y_{C}.

Now try this: Assign values (+1 or −1) to the following variables in such a way that the product of the three X values equals −1 while the product of the Y values in two of the three columns equals the X value in the remaining column — or else explain why this can’t be done.

X_{A} X_{B} X_{C}

Y_{A} Y_{B} Y_{C}

Here is why it can’t be done. The winning combinations satisfy the following equations:

X_{A} X_{B} X_{C} = −1

X_{A} Y_{B} Y_{C} = +1

Y_{A} X_{B} Y_{C} = +1

Y_{A} Y_{B} X_{C} = +1

Because the squares of the Y’s are equal to 1, the product of the left-hand sides of the last three equations is

X_{A} X_{B} X_{C} (Y_{A})^{2} (Y_{B})^{2} (Y_{C})^{2} = X_{A} X_{B} X_{C},

while the product of their right-hand sides is +1. Obviously these three equations cannot be satisfied as long as the first equation holds. Upshot: pre-agreed answers offer no fail-safe strategy.

And yet there is such a strategy.

1. [↑] Vaidman, L. (1999). Variations on the theme of the Greenberger-Horne-Zeilinger proof, *Foundations of Physics* 29, 615–630.