What's wrong with this equal probability solution for Monty Hall Problem?

I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective gives me equal probability.

Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.

Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.

Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.

What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.

Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50\%$, whereas there is definitely a goat behind door $C$.

In particular, let $C_{\text{goat}}$ be the event that there's a goat behind door $C$, and $C_{\text{revealed}}$ be the event that door $C$ is revealed, and $A_{\text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{\text{car}}$ and $C_{\text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50\%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50\%$. Thus, one concludes that $$P(A_{\text{car}}|C_{\text{revealed}})=P(A_{\text{car}})=\frac{1}3.$$ The calculation you've done (annotated with a $\neq$ sign where things go wrong) is: $$P(A_{\text{car}}|C_{\text{revealed}})\neq P(A_{\text{car}}|C_{\text{goat}})=\frac{1}2$$

Therefore, showing an open door with a goat reveals nothing new about which door has the car.

Two-thirds of the time, our initial pick will be wrong. In that case, Monty cannot randomly pick a door out of the remaining two to reveal. He must pick the door that has a goat. This gives the additional insight into which door has a car.

Conversely, one-third of the time, our initial pick will be correct so Monty can then open any of the two remaining two doors. Switching in that case will then lead us to the wrong door. The good news is, that only happens one-third of the time!

Let me try to guess your birthday. My guess is April $2$nd.

Ignoring leap years, remove $363$ dates that are not your birthday and that are not April $2$nd (because I chose that), so that we are now left with only two options. You offer me to switch to the date that I didn't pick, which I do.

That I was right in my initial guess is unaffected by the dates that you subsequently remove.