How to model and solve the variant of Monty Hall problem in which the host opens a door randomly? The Monty Hall problem (wiki) is described as follows:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

In the analysis, it reads,

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence doesn't hold.

I am interested in the variant in which the host opens a door randomly and try to formalize it. 
First, it is crucial to explicitly identify the assumptions for this variant:


*

*$A_1:$ The host must always open a door that was not picked by the contestant.

*$A_2:$ The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

*$A_3:$ The host does not know what lies behind the doors and opens one at random that happens not to reveal the car.


By Bayes' theorem, we can obtain that
\begin{align*}
    \Pr\{C_2 \mid H_3, Y_1\} = \frac{\Pr\{H_3 \mid C_2, Y_1\}}
      {\Pr\{H_3 \mid C_1, Y_1\} + \Pr\{H_3 \mid C_2, Y_1\} + \Pr\{H_3 \mid C_3, Y_1\}},
  \end{align*}
where, the event $C_i$ denotes that the car is behind the door $i$, $Y_1$ denotes that you pick door 1, and $H_3$ denotes that the host opens door 3 and happens to reveal a goat.
Therefore, the task is to assign probabilities to $\Pr\{H_3 \mid C_1, Y_1\}$, $\Pr\{H_3 \mid C_2, Y_1\}$, and $\Pr\{H_3 \mid C_3, Y_1\}$.
Questions:


*

*Are the three assumptions correct and sufficient to characterize the situation where the host opens a door randomly?

*Should the assumption $A_3$ enforce the requirement that "... happens not to reveal the car"? Similarly, should the event $H_3$ specify that "... happens to reveal a goat"?

*How to assign probabilities to $\Pr\{H_3 \mid C_1, Y_1\}$, $\Pr\{H_3 \mid C_2, Y_1\}$, and $\Pr\{H_3 \mid C_3, Y_1\}$?

 A: If the host opens a door without knowledge and behind that door a goat is found then the new situation is that you can choose from two doors that have equal probability (so both $0.5$) to hide a car.
To come to this conclusion plain thinking is enough and the rule of Bayes is not needed.
There is no profit in changing your original choice and no profit in keeping to it.
A: Your formulation is correct. 
Now $\Pr\{H_3\mid C_1,Y_1\}$ is simply the probability that the host opens door $3$ given $C_1$ and $Y_1$ (because you are given $C_1$, the host can't open door $3$ to reveal a car), and this is $1/2$ because he randomly chooses between the two doors you didn't open. Similarly $\Pr\{H_3\mid C_2,Y_1\}=1/2$.
However, $\Pr\{H_3\mid C_3,Y_1\}=0$ - if the car is behind door $3$, he can't open that door and reveal a goat. This gives $\Pr\{C_2\mid H_3,Y_1\}=1/2$ as expected.
A: Without loss of generality, you may simply label door 1 as whatever door you choose, door 2 as whatever door the host opens and door 3 as whatever door is left.  The only random variable is: behind which of these is the car?
So: Given that the host has not revealed the car (but choose without bias), what is the probability that the car is behind door 2?
Then by your formulation:$$\mathsf P(C_2\mid Y_1,H_3,(C_1\cup C_2))=\dfrac{\mathsf P(C_2\mid Y_1,H_3)}{\mathsf P(C_1\cup C_2\mid Y_1,H_3)}= \dfrac{1/3}{2/3}$$
A: What are problems like the Monty Hall Problem?
Here’s one: Originally urn 1 contains 1 black ball and 9 white balls. A ball is picked at random from urn 1 and placed in urn 2. Balls are then drawn from urn 1 (now having one less ball) without replacement: This illustrates/demonstrates the probabilities of Monty revealing the prize,and/or you getting it depending on your choice, as doors are shown not to have the prize behind them.
Probabilities of what color the ball is on each particular pick are shown below depending on what happened when the first ball was taken from urn 1 and on the previous pick(s). Note N is the number of the “pick” after you’ve chosen which ball to put in urn 2 and the probabilities in a particular row are the apriori probabilities of getting a particular color ball on the pick you are making. The last column represents(9/10 * the probabilities in the columns for when urn 2 contains a white ball) Note that as you go down the yellow shaded column those probabilities are essentially conditional probabilities, in that they reflect having the event(s) in the cells immediately above not occur on the previous picks,i.e. not getting a black ball=not revealing the prize and, correspondingly, the event(s) to the immediate left occur. urn 2 having the black ball corresponds to you having made the correct choice for the prize. we are interested in the probability of getting the prize(black ball) as Monty opens doors equivalent now to picking from urn1. Monty is not allowed to touch urn 2. Of course if he were, the probability he would reveal the prize would be 1/10.
This is exactly what happens in the Monty Hall problem. Say instead of 3 doors, there were 10 doors, with the prize behind only one. A door is first chosen, but then eliminated from further participation in the selection process. The last row in the table above, highlighted in green, represents the probabilities associated with you choosing the last ball from urn 1 or, equivalently, choosing the final remaining unopened door in the Monty Hall problem, instead of your original choice.
Just for fun let’s make the same table for urn 1 originally containing 3 balls, 2 White, 1 Black. We have:
So, the black ball (the prize) has a 2/3 probability of being behind the remaining unopened door, and if there were 10 doors to begin with, a 9/10 chance of the same. And not a word has been spoken about whether Monty knows where the prize is or not. Only tabulating the probabilities for getting the prize (or not) as doors besides the one you originally chose are opened…
A: It does not matter if Monty knows where the prize is or not.
Trying to work what Monty knows or doesn’t know into the problem is a tempting (apparently) but needless distraction.
This is all there is to the Monty Hall problem:
What’s the probability the prize is behind the doors you didn’t pick? Does this probability depend upon whether those doors are opened simultaneously or in sequence? I think you might agree that it doesn’t. So as doors (say there are N doors to start with, not necessarily limited to 3) are opened, the probability that the prize is in the group of doors you didn’t pick remains the same. Which means that the probability it’s behind the door you did pick also remains the same. That’s all you need to know.
