Solving the Monty Hall Paradox
Apply George Pólya's four-stage mathematical methodology to dismantle cognitive illusions. From prior probability formulation and N-door generalization to Bayesian inference and empirical verification.
Establish initial distribution: P(car behind initial choice) = 1/3, while P(car behind other doors) = 2/3. The critical constraint is that the host has complete knowledge of prize locations and must always open a goat door.
P(Choice) = 1/3 | P(Remaining Doors) = 2/3 | P(Host reveals goat) = 1Establish Prior Probability Partition
Your chosen door holds P(Car) = 1/3. The remaining two doors collectively hold P(Car) = 2/3.
Identify Host Knowledge Constraint
The host is not choosing at random: he knows the prize location and is strictly bound to open a goat door.
Preserve Initial Choice Invariance
Because the host's door selection is constrained to the remaining doors, your initial 1/3 prior cannot change.
Simulate 3 doors vs 100 doors to verify the Bayes calculation live.
George Pólya's Four Stages of Problem Solving
Understand the Problem
Priors & The Host Knowledge Constraint
Devise a Plan
Generalization to N Doors (N = 100)
Execute the Plan
Bayes' Rule & Posterior Concentration
Look Back
Dismantling the 50/50 Illusion & Empirical Proof
Monty Hall Proofs & Logic
Progressive derivations from the classic 3-door case to 100 doors, generalized N-door systems, rigorous Bayes' theorem proofs, and peer-reviewed citations.
When a contestant selects a door, the sample space partition gives an initial probability of 1/3 for the chosen door and an aggregate 2/3 probability for the two unchosen doors. Because the host must reveal a goat behind an unchosen door, switching shifts that entire 2/3 probability mass onto the single unopened candidate.
// 1. Initial State (N = 3):
Let C be the door hiding the car, with C in {1, 2, 3}.
P(C = 1) = P(C = 2) = P(C = 3) = 1/3
// 2. Contestant Selection:
Contestant chooses Door 1.
P(win by staying) = P(C = 1) = 1/3
P(C in {2, 3}) = P(C = 2) + P(C = 3) = 2/3
// 3. Host Action & Conclusion:
Host opens an unchosen goat door (say Door 3).
The unchosen probability mass (2/3) collapses entirely onto Door 2:
P(win by staying) = 1/3 (approx. 33.33%)
P(win by switching) = 2/3 (approx. 66.67%)Verify With the Monte Carlo Engine
Run 1,000+ empirical simulations across 3 to 100 doors to observe convergence in real time.