There are 48 ✓ and 24 ✗ above, total is 8*9 = 72 possibilities.
So the expected fraction of difference is:
$$
\frac{2}{3}
$$
Simulation
import random
# Define the eight deterministic plans as dicts mapping axis→outcomePLANS = [
{1:+1, 2:+1, 3:+1}, # Plan 1: (↑↑↑) {1:+1, 2:+1, 3:-1}, # Plan 2: (↑↑↓) {1:+1, 2:-1, 3:+1}, # Plan 3: (↑↓↑) {1:+1, 2:-1, 3:-1}, # Plan 4: (↑↓↓) {1:-1, 2:+1, 3:+1}, # Plan 5: (↓↑↑) {1:-1, 2:+1, 3:-1}, # Plan 6: (↓↑↓) {1:-1, 2:-1, 3:+1}, # Plan 7: (↓↓↑) {1:-1, 2:-1, 3:-1}, # Plan 8: (↓↓↓)]
defsimulate_hidden_variable(runs=10000):
mismatches =0for _ in range(runs):
plan = random.choice(PLANS)
# randomly pick measurement axes i, j ∈ {1,2,3} i = random.randint(1,3)
j = random.randint(1,3)
a = plan[i] # Alice’s outcome b =-plan[j] # Bob’s outcome is opposite on same-axisif a != b:
mismatches +=1return mismatches / runs
print(simulate_hidden_variable()) # ≳0.56 (never reaches 0.50)
Quantum
Give $\uparrow$,
$$
p(\downarrow) = \cos^2(\frac{\theta}{2})
$$
Give $\downarrow$,
$$
p(\uparrow) = \cos^2(\frac{\theta}{2})
$$
When $\theta=0$, $p=1$ and when $\theta=120^{\circ}$, $p=\dfrac{3}{4}$
Situation 1
Bob 1 ↓
Bob 2 ↓
Bob 3 ↓
Alice 1 ↑
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↑
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↑
75 % ✓
75 % ✓
100 % ✓
Situation 2
Bob 1 ↓
Bob 2 ↓
Bob 3 ↑
Alice 1 ↑
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↑
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↓
75 % ✓
75 % ✓
100 % ✓
Situation 3
Bob 1 ↓
Bob 2 ↑
Bob 3 ↓
Alice 1 ↑
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↓
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↑
75 % ✓
75 % ✓
100 % ✓
Situation 4
Bob 1 ↓
Bob 2 ↑
Bob 3 ↑
Alice 1 ↑
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↓
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↓
75 % ✓
75 % ✓
100 % ✓
Situation 5
Bob 1 ↑
Bob 2 ↓
Bob 3 ↓
Alice 1 ↓
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↑
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↑
75 % ✓
75 % ✓
100 % ✓
Situation 6
Bob 1 ↑
Bob 2 ↓
Bob 3 ↑
Alice 1 ↓
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↑
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↓
75 % ✓
75 % ✓
100 % ✓
Situation 7
Bob 1 ↑
Bob 2 ↑
Bob 3 ↓
Alice 1 ↓
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↓
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↑
75 % ✓
75 % ✓
100 % ✓
Situation 8
Bob 1 ↑
Bob 2 ↑
Bob 3 ↑
Alice 1 ↓
100 % ✓
75 % ✓
75 % ✓
Alice 2 ↓
75 % ✓
100 % ✓
75 % ✓
Alice 3 ↓
75 % ✓
75 % ✓
100 % ✓
So the expected fraction of difference is:
$$
\frac{1}{2}
$$
Simulation
import random
import math
defsimulate_singlet(runs=10000):
mismatches =0for _ in range(runs):
# pick axes i, j ∈ {1,2,3}, represented by angles θ_i=0°,120°,240° angles = {1:0, 2:120, 3:240}
i, j = random.randint(1,3), random.randint(1,3)
θ = abs(angles[i] - angles[j]) %360if θ >180:
θ =360- θ
# Probability of “different” for singlet: P = ½(1 + cos θ) p_diff =0.5* (1+ math.cos(math.radians(θ)))
if random.random() < p_diff:
mismatches +=1return mismatches / runs
print(simulate_singlet()) # ≃0.50
Why
No local hidden-variable theory can reproduce the quantum correlations.
All eight of those “quantum-tables” look identical because, unlike the eight hidden-variable plans, quantum mechanics doesn’t pick one deterministic instruction set at each run—instead it gives one singlet-state rule for all runs.
Mathematically, the Bell-inequality violation tells us that no joint probability model of the form