# This Quantum World/Game

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## A quantum game

Here are the rules:[1]

• Two teams play against each other: Andy, Bob, and Charles (the "players") versus the "interrogators".
• Each player is asked either "What is the value of X?" or "What is the value of Y?"
• Only two answers are allowed: +1 or −1.
• Either each player is asked the X question, or one player is asked the X question and the two other players are asked the Y question.
• The players win if the product of their answers is −1 in case only X questions are asked, and if the product of their answers is +1 in case Y questions are asked. Otherwise they lose.
• The players are not allowed to communicate with each other once the questions are asked. Before that, they are permitted to work out a strategy.

Is there a failsafe strategy? Can they make sure that they will win? Stop to ponder the question.

Let us try pre-agreed answers, which we will call XA, XB, XC and YA, YB, YC. The winning combinations satisfy the following equations:

${\displaystyle X_{A}Y_{B}Y_{C}=1,\quad Y_{A}X_{B}Y_{C}=1,\quad Y_{A}Y_{B}X_{C}=1,\quad X_{A}X_{B}X_{C}=-1.}$

Consider the first three equations. The product of their right-hand sides equals +1. The product of their left-hand sides equals XAXBXC, implying that XAXBXC = 1. (Remember that the possible values are ±1.) But if XAXBXC = 1, then the fourth equation XAXBXC = −1 obviously cannot be satisfied.

The bottom line: There is no failsafe strategy with pre-agreed answers.

1. Lev Vaidman, "Variations on the theme of the Greenberger-Horne-Zeilinger proof," Foundations of Physics 29, pp. 615-30, 1999.