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?
In the case where both contestants can see everything, it's more or less the same as a single contestant getting to select two doors. The probabilities under both doors effectively freeze, and the the $1/4$ for door D flows to door C, giving C a $1/2$ chance of having the prize, leaving A and B at their original $1/4$.
In the case where contestants can't see what door the other contestant picks, then you have the situation where their door has $1/4$ chance, and the other two remaining doors both have $3/8$. It's fine that both contestants have different probabilities for the doors; it results from both the contestants having different information about the probability distribution, since only they know which door they selected.