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Example: Logic & Probability — The Monty Hall Deception

Domain: Mathematics · Probability Theory · Bayesian Reasoning
Difficulty: Moderate
Frameworks Used: First Principles Decomposition, Adversarial Verification

💬 User Prompt

"We have three doors. Behind one is a car, behind the others are goats. You pick Door 1. The host, who knows what's behind the doors, opens Door 3, which has a goat. Should you switch to Door 2?"

Phase 1: Problem Intelligence 🔍

1.1 Restatement

This is the classic Monty Hall Problem — a conditional probability puzzle where the host's informed action changes the sample space. The question is whether switching doors after a reveal increases the win probability.

1.2 Objective

Determine the mathematically optimal strategy (switch or stay) and prove it rigorously.

1.3 Constraint Space

Dimension Value
Hard Constraints Standard Monty Hall rules: host always opens a losing door, host never opens your door, host never opens the car door
Unknowns None — this is a fully specified problem

1.4 Hidden Assumptions

  • 🟢 The host must open a door (not optional)
  • 🟢 The host knows what's behind each door (informed action)
  • 🟢 The car placement is uniformly random
  • 🔴 Dangerous if violated: If the host opens a door at random (and it happens to be a goat), the analysis changes completely (it becomes 50/50). The word "who knows" in the problem statement confirms informed action.

Phase 2: Decomposition Engine 🔧

Sub-problems

  1. Prior Probabilities: What are the initial odds?
  2. Information Update: How does the host's action change the probability distribution?
  3. Bayesian Update: Apply Bayes' theorem to compute posterior probabilities.

Phase 3: Systematic Reasoning 🧮

Step 1: Enumerate All Scenarios

Car Location You Pick Host Opens Switch Result Stay Result
Door 1 Door 1 Door 2 or 3 🐐 LOSE 🚗 WIN
Door 2 Door 1 Door 3 (forced) 🚗 WIN 🐐 LOSE
Door 3 Door 1 Door 2 (forced) 🚗 WIN 🐐 LOSE

Step 2: Count the Outcomes

  • Switching wins: 2 out of 3 scenarios → P(win|switch) = 2/3
  • Staying wins: 1 out of 3 scenarios → P(win|stay) = 1/3

Step 3: Bayesian Proof

Using Bayes' theorem where H = host opens Door 3, C_i = car behind Door i:

P(C₂|H) = P(H|C₂) × P(C₂) / P(H)

P(H|C₁) = 1/2  (host can open Door 2 or 3)
P(H|C₂) = 1    (host MUST open Door 3, since you picked Door 1)
P(H|C₃) = 0    (host can't open Door 3 if car is there)

P(H) = P(H|C₁)P(C₁) + P(H|C₂)P(C₂) + P(H|C₃)P(C₃)
     = (1/2)(1/3)   + (1)(1/3)       + (0)(1/3)
     = 1/6 + 1/3 + 0 = 1/2

P(C₂|H) = (1 × 1/3) / (1/2) = 2/3 ✅
P(C₁|H) = (1/2 × 1/3) / (1/2) = 1/3 ✅

Verification via Monte Carlo Simulation

import random

def monty_hall(switch, trials=1_000_000):
    wins = 0
    for _ in range(trials):
        car = random.randint(0, 2)
        pick = 0  # always pick Door 0
        # Host opens a door that is not the car and not the pick
        doors = [d for d in range(3) if d != pick and d != car]
        host_opens = random.choice(doors)
        if switch:
            pick = [d for d in range(3) if d != pick and d != host_opens][0]
        if pick == car:
            wins += 1
    return wins / trials

print(f"Stay:   {monty_hall(switch=False):.4f}")  # ~0.3333
print(f"Switch: {monty_hall(switch=True):.4f}")   # ~0.6667

Simulation confirms: switching wins 66.7% of the time.

Phase 4: Multi-Path Analysis 🔀

Strategy Win Probability Proof Method
Always Switch 2/3 (66.7%) Enumeration, Bayes, simulation
Always Stay 1/3 (33.3%) Enumeration, Bayes, simulation
Random Choice 1/2 (50.0%) Expected value of 50/50 coin flip

Three independent proof methods (enumeration, Bayesian, empirical simulation) all converge on the same answer. This is a high-confidence result.

Phase 5: Failure Mode Detection 🛡️

When This Analysis Breaks

The 2/3 result depends on specific rules. It changes if:

Rule Violation New Probability
Host opens randomly (not informed) 50/50
Host can open the car door Problem changes entirely
Host doesn't always open a door Need game theory analysis
More than 3 doors P(switch) = (n-1)/n for n doors

The key insight is: the host's knowledge is what makes switching valuable. Without informed action, there is no information transfer.

Phase 6: Confidence Scoring 📊

Metric Score
Problem Understanding 100
Solution Correctness 100
Completeness 95
Evidence Quality 100
Overall Confidence 99

This is a mathematically proven result verified by three independent methods.

Phase 7: Final Output 📋

📌 Executive Summary

Always switch. Switching gives you a 2/3 (66.7%) chance of winning. Staying gives you 1/3 (33.3%). This is proven by enumeration, Bayesian inference, and Monte Carlo simulation.

🎯 The Key Insight

Your initial pick has a 1/3 chance of being right. The other two doors collectively hold 2/3 of the probability. When the host eliminates one of those doors (guaranteed to be a loser), the entire 2/3 collapses onto the remaining door. The host's informed action is what creates the asymmetry.

⚠️ Caveats

  • This only applies when the host always opens a losing door and knows where the car is
  • In the "ignorant host" variant (host opens randomly), the answer is 50/50

📊 Confidence: 99/100

This is a settled mathematical proof. The only source of error would be misunderstanding the problem rules.