I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking wrong.
So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.
That's not correct, because if you agree that one kind of result tends to happen more than the other, then it is easier that the single game that you are currently playing belongs to the larger group. When you play once, it is like if you were randomly choosing a game of all the possible ones that you could play, in which case you are more likely to choose one of those that are more numerous.
It's the same reasoning that you would apply in other aspects of life. Imagine you have to travel to some town and there are only two possible roads: A and B. The stats tell that on road A there tend to occur 100 fatal accidents per year, while on road B only one accident has been registered in the last 10 years.
The causes of that disparity may be multiple, like, for example, that road A has many curves with precipices while B does not. But you don't even need to know those causes; the starts themselves tell you that as result it is easier to get a falal accident on road A than on road B. But, of course, it always exists the possibility that you survive in any of them, and also that you have an accident in any of them.
So, if you were to travel through one of them just once and had to choose, I don't think you would say that those stats don't matter only because you will travel once. You would prefer to take the safer road B.
In Monty Hall game, deciding to stay would be like taking road A, and deciding to switch would be like taking road B.
Moreover, if the game consisted in that instead of revealing a goat, the host gave you the opportunity to reject your first choice and instead check inside the other two doors and take which you prefer from them, it is obvious that it would be better to switch to the other two (unless you think 2 is not greater than 1), and that's true regardless of if you are playing just once or multiple times. Anyway, by checking those two doors you would necessarily find at least an incorrect one, as there is only one prize in total.
If you notice, as the host knows the locations of the contents and always removes a losing door from those that you did not pick, then it is like if he was doing that work for you of checking inside those other two doors, eliminating from them the incorrect one that you would have found anyway, and leaving closed exactly which you would have picked if you were who had checked inside those two doors.
So, if you agree that it would be better to switch to the other two doors, you must also agree that it is better to switch to the other single one that the host is offering, as both ways win in exactly the same cases.
The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, not all. Once a goat is revealed, you could only be in the 1/3 case of when the player's choice is correct, or in the 1/3 case of when the other door randomly left closed is correct, so each represents 1/2 of this subset, not one 1/3 and the other 2/3.
Moreover, to understand why with a random revelation of the goat the chances of each option are 50%, you first need to understand the real reason why they are not 50% in standard Monty Hall problem.
It is because when the player has picked a goat door, the host is restricted to reveal specifically which has the only other goat, but when the player has picked the car door, the host is free to reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case.
For example, if you select door 1 and he reveals door 2, it was 100% sure that he would have taken #2 if the correct were #3, as he wouldn't have had another option. Instead, we couldn't ensure that he would have opened door 2 in case the correct were yours, as he could have preferred to reveal door 3 in that case (each of them would have had 50% chance of being removed). So, from the times that you start selecting door 1, it tends to happen with twice the frequency that he opens door 2 once the correct is #3 than once the correct is #1.
Instead, if the host does not know the locations and his revelation is random, he cannot make that distinction of revealing one door more than the other depending on the prize location, precisely because he does not know where it is. If you pick door 1, you cannot say that he will open specifically door 2 with more frequency when door 3 is the correct than when door 1 is the correct. If he decides to open door 2, that choice is independent of where the car is.
You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation instead of the host. By the end both are doing it without knowledge so the results shouldn't tend to be different. For example, you could pick door 1 and then decide to reveal door 2. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, selecting door 1 and then revealing door 2 is the same as picking both door 1 and door 3 at once, and then discarding the other. So, if door 2 results to have a goat, which one do you think is which should have 2/3 probabilities of having the car, door 1 or door 3? Consider that the doors don't "know" if you picked them at once or one after the other. The result would have been the same if you had first declared door 1 as your staying option and number 3 as your switching one, or viceversa. So, neither of them can be more likely than the other.
Yes, you could still win 2/3 of the time by switching in all games, only that with those rules the probability for each remaining door once a goat is revealed would be 1/2, as you stated. So, you could also win 2/3 of the time by staying with your selection everytime a goat is shown and only switching to the car when it is shown.
But what I think that confuses most people is why one option has more probabilities than the other when there are two remaining doors and the car must be behind one of them, which does not occur with those conditions. It is needed that the host did it deliberately avoiding to reveal the car.
This only occurs because the host knows the locations of the contents and cannot reveal the player's door and neither which has the car. So the other that he leaves closed is always the best possible one that could be found among the rest, because if any of them has the prize, that is precisely which he will avoid to discard.
If you notice, the switching door is equivalent to the selection that a second player that was cheating would make, a cheater that checked inside all the doors that the first did not pick, and took which preferred from them. In this way, it is obvious that the second player will win as long as the first starts failing, which occurs 99 out of 100 times in the 100-doors version. And the revealed doors are like which neither of them picked.
So, when there are two doors remaining, you know that one is the first player's selection and the other is the cheater's selection, where the cheater was more likely to get the correct.
In Monty Hall Game you improve your odds to be 2/3, not 1/2. If you think the staying door wins 1/3 of the time and the switching door 2/3, where is the car located in the 1/6 missing?
So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.