Preview
Some of the confusion with the Monty Hall problem comes because it is not always stated that the host must always reveal a goat from the two doors that the player did not pick and offer the switch. Generally, it is only said that he knows the locations of the contents, and did the revelation this time. This leaves the possible interpretation that he only did it because the contestant had chosen the winner door, and the host is trying to make him change so he loses.
Now, there are those who argue that since it is established that he knows the locations of the contents, then it is not necessary to say that he always executes that action as a protocol, but it is enough to know that he did it this time, because for every possible extra behavior, there is always a counterpart that cancels it out and makes the probabilities $2/3$ again.
It's a reasoning of symmetry similar to the one used to argue that the chances that the host opens any of the other two doors are $1/2$ when the player has picked the winner option, despite nothing in the rules state that he must choose one of them at random. Someone could argue that he could have the preference of always opening the lowest numbered one of those that are available for him, but others could also argue that he could have the opposite preference of always opening the highest numbered one. That is, any argument that could be said about one door could also be said about the other, so the average of all possible preferences is that he is $1/2$ likely to reveal each door.
This symmetry seems to be accepted despite the set of all the possible preferences are not formally described.
Question
In the same way, despite for some of the host's behaviors it could be better to switch and for others not, can we apply a reasoning of symmetry similar to the described above to say that, when we don't know the current behavior, the average of all possible ones is $2/3$ chance of success by switching?
Can we be sure that for any possible behavior we can find an oppposite one such that, when combining the two, the probability to win by switching inside the group is always $2/3$? (And also that the relation will always be One-to-One, because if it isn't, we can't make the cancellation).
Example
Suppose that the only three possible behaviors were:
$A)$ He always opens a goat door regardless of what the player picked.
$B)$ He only opens a goat door when the player has picked a goat door.
$C)$ He only opens a goat door when the player has picked the car door.
There may be reasons to argue that cases $B)$ and $C)$ don't need to be equally likely, but if we bypass that part for now and assume for simplicity that they are in fact equally likely, then the overall probability to win by switching is $2/3$, because inside the group of $B)$ and $C)$, the opportunity would only be offered $2/3$ of the times that we are in case $B)$, and $1/3$ of the times that we are in case $C)$, meaning that switching ends up winning $2/3$ of the combined opportunities. When we join it with the proportion of case $A)$, that is also $2/3$ as has been discussed several times, the overall average remains $2/3$.
This does not need that the three possible behaviors are equally likely, as long as the pair of opposite behaviors $B)$ and $C)$ are.
Possible conterarguments
My concern is if for each possible behavior it will always be possible to find an opposite one that is equally likely to it. I was thinking about these two, that may seem opposite to each other because the door to reveal is never the same, but the probability inside the combined group is not $2/3$ but $1/2$:
$D)$ The host always tries to reveal the lowest numbered option of those that the player did not pick, but if it has not a goat, he does not do anything and the game ends.
$E)$ The host always tries to reveal the highest numbered option of those that the player did not pick, but if it has not a goat, he does not do anything and the game ends.
The probability to win by switching inside that group is $1/2$ because the opportunity would be offered everytime that the player has picked the car $(1/3)$, but half of the times that the player has picked a goat $(1/2 * 2/3 = 1/3)$, so each strategy would be equally likely.
If the probability of being in that group is greater than zero, then it would lower the average $2/3$ that the other possibilities combined have.
But maybe I'm doing it wrong when grouping those two behaviors. Maybe the opposite of $D)$ shouldn't be $E)$ but any other one.
And as someone pointed out in the comments, another possible protocol that seems to not have a counterpart is:
$F)$ He opens a door the player didn’t choose at random and if it has a goat, he offers the chance to switch to the other unopened door.
Although this protocol would not make use of the host's knowledge of the locations, so it could be said that it shouldn't be taken into consideration. But however unreasonable it may be, it has a chance to exist.