Q:

A confusion about the Monty Hall problem

After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses. 
Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;
1-> car
2-> goat
3-> goat
or
1-> goat
2-> car
3->goat
Since, we do know that the car is not behind the door 3.
So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.
So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?
Edit:
In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.

A:

You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases. 
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine. 
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $\{2,3,4,5,6,7,8,9,10,11,12\}$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!

A:

We have $$\begin{array}{c|c|c}\#1&\#2&\#3&\mbox{Result at }\#1&\mbox{result otherwise}\\\hline c&g&g&win&lose\\\hline g&c&g&lose&win\\\hline g&g&c&lose&win\end{array}$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $\#2$ or $\#3$ "away", but the results won't change.

Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50\%$ for each door. you have $\#1$ your choice and host open $\#2$ to be $1/3\cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $\#1$ your choice and host open $\#3$ to be $1/3\cdot1/2=1/6$, together you gave $1/3$. 
But if you chose incorrect there is $100\%$ that the host will open a particular door: $\#1$ your choice and host open $\#2$ to be $1/3\cdot1=1/3$ and and host open $\#3$ to be $1/3\cdot1=1/3$, together it is $=2/3$.

