I understand the common Monty Hall Problem and why switching provides a 2/3 chance of winning, but I'm having trouble wrapping my head around how the probabilities work when multiple players are involved, as their probabilities seem to be contradictory.
Let's say we have two players and four doors. Each door has a 1/4 probability of hiding a prize. Let's say contestant 1 chooses door A and contestant 2 chooses door B. Door D is then revealed to be empty, leaving A, B, and C. From contestant 1's perspective the odds are 1/4 for door A, 3/8 for door B, and 3/8 for door C. But from contestant 2's perspective the odds are 3/8 for door A, 1/4 for door B, and 3/8 for door C.
What are the actual odds for each door hiding the prize, does it change depending on if they know each other's choices, and does it differ from the normal example of the problem? If so why, and if not, why not?