# 6 A quantum game

Let’s play a game. We form two teams of three, the “players” (Andy, Bob, and Charles) and the “interrogators”. 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 XA, XB, XC, and YA, YB, YC.

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.
XA   XB   XC
YA   YB   YC

Here is why it can’t be done. The winning combinations satisfy the following equations:
XA   XB   XC = −1
XA   YB   YC = +1
YA   XB   YC = +1
YA   YB   XC = +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

XA XB XC (YA)2 (YB)2 (YC)2 = XA XB XC,

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.

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