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You are wrong here, the problem statement states "Monty Hall opened a door and it had a Goat", so the situation you describe where he selected a car didn't happ
by username90 7y ago
You are wrong here, the problem statement states "Monty Hall opened a door and it had a Goat", so the situation you describe where he selected a car didn't happen. Since he didn't select a car you now have 50% chance behind both doors. But you are right that you have 1/3 chance from the beginning, but the situation described in the problem is not at the beginning.
- henrikschroder 7y agoNo, that is not how the original Monty Hall problem was formulated. The original said: "the host, who knows what's behind the doors, opens another door, which has a goat." Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities. And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car. It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.