If I'm thinking about this correctly, regardless of how the host chooses which door to open, there's a 1/3 chance the player initially picks the door with the car behind it, in which case they shouldn't switch doors and the door opened by the host will have been one of goat doors regardless of whether he was choosing at random or not (assuming he was choosing between the two doors not already chosen by the player). In the case that the player initially picks one of the goat doors, which has a 2/3 chance, if the host is picking a door randomly, there's a 50% he chooses a goat door, whereas there's a 100% chance of that of he knows where the car is and is purposely avoiding that door. But the way the problem is usually stated, we already know the door opened by the host was a goat door, so it seems that there should be 2/3 probability of the player choosing the car door if they switch to the unopened door, regardless of how the host decided which of the two unopened doors to open.
And yet, I've heard many people say that the key to understanding this problem is to realize that the host is intentionally opening a door with a goat and that, of he were instead choosing an unopened door at random, it really wouldn't matter if the player switches doors or not. Wikipedia seems to back this up:
For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation, the following two questions have different answers:
- What is the probability of winning the car by always switching?
- What is the probability of winning the car by switching given the player has picked door 1 and the host has opened door 3?
The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty's preference for rightmost doors means that he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3).
This isn't quite the same distinction as whether the host is choosing randomly or not, but the core thing I'm not understanding is why it matters at all how he chooses which unopened door to open, since we already know which one it is. I'm trying to figure it out by relating it to other conditional probability scenarios where the prior probability is clearly relevant, such as the textbook example of a medical test with a high accuracy rate where the probability that a positive result is correct depends on the prior probability the patient had the disease. I understand Bayes' theorem and medical test examples and the like fine. But I'm not seeing a clear way to make an analogy to variations of the Monty Hall problem where the host isn't purposely never choosing the door with the car.