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Say I have three objects A,B and C, each behind a door(door 1, 2, and 3). I don't know which object is behind which door. All permutations are equally likely so for every door, it has a 1/3 chance of holding each object. After certain events I can conclude that door 3 cannot hold object C. To get the new probabilities, I listed all permutations and eliminated those which door 3 holds object C, counted, and got the following:

Door 1: P(A) = 1/4, P(B) = 1/4, P(C) = 1/2

Door 2: P(A) = 1/4, P(B) = 1/4, P(C) = 1/2

Door 3: P(A) = 1/2, P(B) = 1/2, P(C) = 0

Is there a way to calculate the new probabilities by just looking at the current probabilities?

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    This is the Monty Hall problem. It has engendered huge amounts of debate because the phrase "after certain events I can conclude that door 3 cannot hold object C" has many distinct interpretations, leading to different solutions. If you wish to distinguish your question from the MH problem, then please state specifically what those "certain events" might be. – whuber Jan 06 '22 at 22:53
  • This is the monty hall problem, which I have written about succinctly on my blog. – Demetri Pananos Jan 06 '22 at 22:54

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