You are misunderstanding conditional probability. You keep doubling down on the idea that the alternative scenarios don’t matter, but this is mistaken. They ver

You are misunderstanding conditional probability. You keep doubling down on the idea that the alternative scenarios don’t matter, but this is mistaken. They very much matter. They are the crux of how Monty Hall “works”. 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