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 $\Pr\{H_3 \mid C_3, Y_1\}$?
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.
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.