Figure 5
P values for the Mermin’s inequality39 with perfect random number generators. Mermin’s inequality is a tripartite inequality in which each party has two inputs and two possible outputs. It is an example of a non-bipartite inequality that has already been violated in the laboratory.55–57 The three parties Alice, Bob and Charlie receive three random chosen inputs x, y and z with the promise that the parity of the inputs is even—that is, that the inputs are limited to (0, 1, 1), (1, 0, 1), (1, 1, 0), (0, 0, 0), and produce outputs a, b and c, which can also be taken to be bits – that is, x,y,z,a,b,c∈{0,1}. The winning condition for Mermin’s inequality is that a⊕b⊕c=x∨y∨z. That is the game is won if the xor of the outputs equals 0 when (x, y, z)=(0, 0, 0) and if the xor of the outputs equals 1 in the remaining cases. Hence, we get sabc|xyz=a⊕b⊕c⊕1 when (x, y, z)=(0, 0, 0) and sabc|xyz=a⊕b⊕c when (x, y, z)≠(0, 0, 0). The winning probability for this game is p(win)=3/4,58 but note that in contrast with CHSH if Alice, Bob and Charlie share entanglement they can win with probability one. The curves show the P value as a function of S=8(c/n−1/2) for a fixed number of trials n (c is the number of wins). The three curves show from top to bottom the P value for n=150, n=200 and n=250. The P values are computed with the binomial upper bound (18).
