The experimental situation is the following:

The state ψ is arbitrary. The measurement device consists of two parts M_{1}, M_{2}, with three possible input values (control parameters) for each of the two devices a_{1}, a_{2} ∈ {v_{1},v_{2},v_{3}}. We measure two results y_{1}, y_{2} ∈ {1,-1}, one for each device M_{i}.

We measure the expectation value of the product f(y) = y_{1} ⋅ y_{2}. Following the definition of realism, a realistic explanation for the measurement results has the following form:

E(f, a_{1}, a_{2}) = ∫ y_{1}(x, a_{1}, a_{2}) ⋅ y_{2}(x, a_{1}, a_{2}) ρ(x) dx

Assume the two measurement devices M_{i} are located far away from each other, and the measurements (that means, the input of the a_{1} and the fixation of the y_{1}) happens in short enough time (say, less than a minute) at approximately the same time, for example on Earth and Mars, so that light signal need several minutes to reach the other device. (The velocities of Mars and Earth are small enough, so that relativity of simultaneity does not matter, but, for definiteness, we can use the system of rest of the Sun to define "the same time".)

Thus, a signal from M_{1} after the choice of a_{1} by the first experimenter reaches M_{2} only after y_{2} has been fixed, and reverse. That means, that, in case of _{2}(x,a_{1}, a_{2}) cannot depend on a_{1}, and, as well, y_{1}(x,a_{1}, a_{2}) cannot depend on a_{2}. Therefore, the formula reduces to

E(f, a_{1}, a_{2}) = ∫ y_{1}(x, a_{1}) ⋅ y_{2}(x, a_{2}) ρ(x)dx

This is already the main formula (2) used by Bell to prove his inequality. Everything philosophically or methodologically important or possibly problematic has already happened. The problematic assumptions — realism and strong Einstein causality — are already used. What remains is quite elementary math.

It should be noted here, that there are many different variants of the Bell inequalities, with different advantages and disadvantages. For practical tests, one needs variants which allow for uncertainties of the measurements or preparations — in exchange for increasing complexity. We, therefore, consider only the variant we have used in our game.

If nobody cheats in the game, it follows that we have the following property: If a_{1}=a_{2}=a, then

E(f,a, a) = ∫ y_{1}(x,a) ⋅ y_{2}(x,a) ρ(x)dx = 1,

and this is possible only if always y_{1}=y_{2}. Therefore, the two functions
y_{1}(x,a) and y_{2}(x,a) should be equal. This defines a common function y(x,a) ∈ {-1,1} so that

E(f,a_{1}, a_{2}) = ∫ y(x,a_{1}) ⋅ y(x,a_{2}) ρ(x)dx.

For such functions with values in {-1,1} we have always

y(x, v_{1}) ⋅ y(x, v_{2}) + y(x, v_{2}) ⋅ y(x, v_{3}) - y(x, v_{3}) ⋅ y(x, v_{1}) ≤ 1,

as can be easily seen: Each of the three products can be only ±1. To violate the inequatlity, all three product terms would have to give the value +1. But this is impossible, because it would follow that y(x,v_{1})=y(x,v_{2}), y(x,v_{2})=y(x,v_{3}), but y(x,v_{3})=-y(x,v_{1}), which is impossible.

It remains to integrate the inequality to obtain

E(f, v_{1}, v_{2}) + E(f, v_{2}, v_{3}) - E(f, v_{3}, v_{1}) ≤ 1,

which is the variant of Bell's inequality we need in our game.