It's 50/50, either you pick the car or you dont

"Behind one is a car" well then obviously, I choose one

The key is that tripping is accidental. There was a 1/3 chance that when Monty trips, he reveals a car behind door 2 and the game is ruined.

It so happens there wasn’t a car in this instance, but because it wasn’t intentional, there is still a 50-50 split.

Suppose Monty Hall shows you three doors, and behind one is a car. You pick one to open, say door 1, and Monty trips and accidentally reveals that door 2 has no car behind it. What is the probability you get the car if you switch to door 3?

Just curious to see how intuitive this problem is.

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