Monty's knowledge has no effect on the problem or the player's choices. What has an effect is the information the player learns from seeing the revealed door. If the player sees a goat behind the revealed door - that is the relevant information. Whether Monty was surprised or not by the revealed goat is irrelevant.
And yes, your simulation should include that Monty always shows a goat. That is given in the problem statement.
I feel like I've said this thirty times now, I'm specifically discussing a misinterpretation of the problem where Monty does not know what he's opening, so he sometimes opens a prize door. A game where Monty always opens a goat door is the original, correct Monty Hall Problem, and switching is the correct strategy.
If you are talking about some other scenario - then why? I've just explained what this thread is about. Why would you invent an alternate scenario, not tell anyone about your alternate scenario, and argue that correct interpretations of the scenario I laid out in [2] are wrong because they don't fit with your interpretation of your private variation?
Bayes’ theorem is the relevant piece of information here. It tells us the relationship between a conditional probability and the information you have. Specifically, the probability of P(door 3 has the prize | host opened door 2 and found a goat) depends in a very important way on P(host opened door 2 and found a goat). You are acting as though it does not, that only the specific current state matters. This is the cause of your confusion: everyone else is working out the conditional probabilities and you are declaring them irrelevant.
The key is that if the host doesn’t know where the prize is, he can’t avoid picking the door its behind. This changes the probability that he would open that door and reveal a goat in the first place, and how that relates to the various places the prize could be. It would be really strange if the host’s behavior were in some way controlled by the location of the prize and this didn’t give you additional information on that location.
To spell it out:
You choose door 1, the prize is behind door 3, host opens door 2
Prop A: the prize is behind door 3
Prop B: the host opens door 2 and found a goat
P(A|B) = P(B|A)P(A) / P(B)
P(A) = 1/3 in either scenario
In regular Monty Hall, the host will open door 2 always if the prize is behind 3, and half the time it is behind 1. So:
P(B) = (1/2)(1/3) + 1/3 = 1/2
P(B|A) = 1
Thus P(A|B) = (1/3)/(1/2) = 2/3. You’ll recognize this as the answer to Monty Hall. Note that conditional probability is foundational to how we got that answer.
Let’s try with the new rules. The host can pick either door, whether it has a goat or not. He can do this because he doesn’t even know which door is which. The host opens 2 half the time. 2/3 of the time he picks that, the prize is not behind it, so he finds a goat.
P(B) = (1/2)(2/3) = 1/3
P(B|A) = 1/2
P(A|B) = (1/3)(1/2)/(1/2) = 1/2
If you scroll down to "Monty Fall," the description of the problem is "The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car" and the solution is "Switching wins the car half of the time." (This is the result demonstrated in my simulation)