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 It is all just terminology 

Perhaps all this argument is about nothing but terminology.  Does anyone disagree with any of the following statements?  It is assumed throughout that the car is placed randomly, the players initial pick is random, and that the host always opens an unchosen door to reveal a goat.

1) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 3, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

2) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 2, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

3) The probability of winning by switching, given that the player has picked door 1 and the host has opened one of the unchosen doors, is 2/3 for all host preferences. (Morgan call this the unconditional case, maybe this is where the confusion arises).

4) The probability of winning by switching given that the player has picked a door and the host has opened one of the unchosen doors is 2/3 for all host preferences.

5) The probability of winning by switching on average for player who switch is 2/3 for all host preferences.

  
You're right. And notice that in case 3, opening one of the unchosen doors is not a condition, because it is part of the rules of the game. The only condition, if you would consider the full model, there is the initial picking of door 1. In case 4 there is no condition at all in the full model and hence it is the same as case 5. But ... notice that in the MHP a specific door has been chosen and a door opened.  

I continue here. The difficulty is what we do mean by probability. If you mean an average over all cases in which any door has been opened, there is, as I mentioned, no condition. If you mean: a door has been opened, but I do not know which one, and I want to average over all cases concerning the same door, the answer depends on the door and is a number between 1/2 and 1.   

I do not quite agree with you.  If we average over all cases where a door has been opened, but you do not know which one, the answer is 2/3 if we include all possible host strategies in our average.  The only way to get an answer other than 2/3 is to average over cases where the host has a fixed strategy and the same door is opened every time.   

Nijdam, do you see the connection with my simple problem in which a prize is placed behind one of two doors and the player then picks a door?  If the player picks randomly the probability of winning is 1/2.  If the player has picked just one specific door then the probability of winning depends on the initial prize placement.   
  
Who else agrees?   

I agree with #1 and #2.  I find #3, #4 and #5 somewhat ambiguous without further clarification, for example I don't know precisely what you mean by ""for all host preferences"".  I would agree with the following revised versions:

By 'for all host preferences' I mean regardless of whether the host policiy is to: always open door 2 where possible, always open door 3 where possible, choose randomly where there is a choice, or employ any other method of deciding which door to open within the stated rules.   
  
3) The probability of winning by switching, given that the player has picked door 1, is 2/3.

4) The probability of winning by switching, given that the player has picked a door, is 2/3.

5) The probability of winning by switching on average for players who switch is 2/3.

These are clearly ""unconditional"" statements that say nothing about the player deciding after the