Can someone finally explain the monty hall problem in a way that makes intuitive sense

The 100-door version is the key to understanding. **100 doors version:** 1. You pick 1 door (1% chance of car) 2. Monty opens 98 doors that all have goats 3. One door remains unopened (besides yours) Your initial pick has 1% chance. The remaining door has 99% chance. Would YOU switch? OF COURSE! **Now reduce to 3 doors:** 1. You pick 1 door (33.3% chance) 2. Monty opens 1 door with a goat 3. One door remains Your initial pick has 33.3% chance. The remaining door has 66.7% chance. **Why its NOT 50/50:** The probability doesnt redistribute equally between the two unopened doors because Montys action gives you information. He KNOWS where the car is and deliberately avoids opening it. His knowledge affects the conditional probability. $$\Pr(\text{car behind door 1} \mid \text{Monty opens door 2}) = \frac{\Pr(\text{Monty opens door 2} \mid \text{car behind door 1}) \cdot \Pr(\text{car behind door 1})}{\Pr(\text{Monty opens door 2})}$$ If the car is behind your door (door 1), Monty can open either goat door with 50% probability each. If the car is behind door 3, Monty MUST open door 2 (cant open the car). This asymmetry creates the 2/3 probability. Whats stopping you from accepting it? Is it the feeling that "two doors left = 50/50"? Because thats the gamblers fallacy in disguise.