In the Monty Hall problem, does it matter that the host knows which door the car is behind? If so, why?

Question

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:

1. What is the probability of winning the car by always switching?
2. 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.

Answer 1

There are 3 golfers, named A, B, and C. One of them is a master, the others are amateurs. The master will always outgolf the amateurs. You want to guess which one is the master.

Scenario 1: you are told that all 3 of them played a round of golf, and B came last. A and C are clearly equally likely to be the master. This is equivalent to the modified Monty Hall problem where Monty opens one of the goat doors at random and happens to pick B.

Scenario 2: you are told that only B and C played a round of golf, and B came last. The probability that C is the master is 2/3. This is equivalent to the original Monty Hall problem where Monty is only allowed to pick a goat door from among B and C and picks B.

Answer 2

It is easier to picture with 100 doors.

Scenario 1: You pick 1 random door and Monty opens 98 doors that he knows have goats behind them.

Scenario 2: You pick 1 random door and Monty picks 1 random door. (Note that this is the same as him randomly picking 98 other doors to open first, since picking the first 98 doors or last 1 door amounts to the same thing.)

In Scenario 1 there is a 99% chance that the car is behind the doors you did not pick. By eliminating 98 goat doors non randomly that still leaves a 99% chance that the door he did not open has the car and a 1% chance your door has the car. There were always going to be at least 98 doors in the unchosen group that had goats. Opening those doors on purpose does not change the original 99% chance that a car was behind your unchosen 99 doors. You should always switch here since there is a 99% chance it is behind the door you did not choose.

In Scenario 2 there is a 2% chance that your door or Montys door have the car. If you open the other 98 doors and there randomly happened to be goats behind them then there is still an equal chance that your door or Montys door have the car. There is no benefit to switching in this scenario since each door now has a 50% chance of having the car.

One way to help further visualize this is to imagine that you always pick door 99, Monty always picks door 100, and the doors are always opened in order. After doors 1 through 98 are opened with goats you have the option to switch with Monty and take door 100. There was always an equal chance it would be behind 99 or 100, and there is now still an equal chance it will be behind 99 or 100.

Monty choosing randomly is absolutely different from Monty knowing and opening all the goat doors first on purpose.

Answer 3

The point about Monty's door choices is that it provides additional information, but also that the information interacts with the first choice by the contestant and the place of the true position of the price.

We can have alternative situations where one or both of the interactions are not present.

This question relates to the question: Monty Hall problem and causality where we saw the first diagram on the left below, and relating to this question we can add the adaptations that are made in the second and third diagram:

^{The images above are directed acyclic graphs that express causal relationships. At the start of the graph are three independent variables, u_door_with_the_prize, u_quizmaster and u_first_choice. The arrows indicate which variables depend on one and another. The first variation has the arrow between the 'u_door_with_the_prize' and the 'door opened by the quizmaster' eliminated. The second variation has an additional relationship between the 'second choice' and the 'u_door_with_the_prize'.}

In the situation that you mentioned, Monty opens the doors independently from the the true position of the prize. That would be like the middle graph. Then, the strategy, to switch or not switch can not have any effect, because the final decision, however it is made, is causally independent from the actual position of the prize and therefore the strategy can't have a positive effect.

An exception would be when we allow for the strategy, and the second choice, to be in some way dependent on the position of the prize, and in that case we get the third graph. This would be the case when the strategy is not just to switch to the door that was left unopened by Monty, but when the strategy is also allowed to switch to other doors and in particular a door opened by Monty that might have the prize behind it.

This type of conditional strategy is similar to what is happening in the Wikipedia example where based on the two different type of situations (a plain strategy decided before knowing what Monty opens, or a strategy that relates to what Monty actually opens).