this post was submitted on 26 Mar 2024
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[โ€“] [email protected] 1 points 7 months ago (1 children)

Or there are 2 possibilities and then you introduce a 3rd door that is never correct?

Yes that one. Similar to the one you did with 100 doors, just in opposite direction.

[โ€“] [email protected] 1 points 7 months ago

Do you know the third door is never correct? Because then the probability doesn't change.

Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn't a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)

Scenario 2: If you think it's possible that 3 could be correct (but it actually never is) then, no, you wouldn't want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there's no way to know this ahead of time (because as soon as you know you shouldn't switch bc 3 is the wrong door, then you're back in scenario 1)

Scenario 3: For completeness, let's say the 3rd door can be correct sometimes. Then it doesn't matter if you switch or not. It's a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn't matter at all and the second choice is the 'real' choice bc that's the only time you're able to choose from all real possibilities.

The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)