Probability about switching choices This question is similar to the Monty Hall problem, but this problem I don't understand:

There are $99$ doors where $33$ doors have cars and $66$ have goats, and you can only choose one door to win a car. After you make your choice, $33$ other doors are opened to reveal goats. Does your probability of winning a car increase if you decide to switch your door choice?

So obviously in the beginning your one door has a $\frac{33}{99}$ or $\frac{1}{3}$ chance of having a car. But I can't tell if switching doors in this case will be better or not, since we don't know what the groupings of the doors are (otherwise we would've won by knowing which group the $33$ cars are in).
 A: Let $C$ denote the event that behind the first chosen door there is a car. 
Let $G$ denote the event that behind the first chosen door there is a goat.
Note that events $C$ and $G$ are mutually exclusive and are exhaustive.
Let $W$ denote the event of winning a car.
The probability of winning a car by not switching is:$$P(W)=P(C)=\frac13$$
The probability of winning a car by switching is:$$P(W)=P(W|G)P(G)+P(W|C)P(C)=P(W|G)\times\frac23+P(W|C)\times\frac13$$
Can you find $P(W|G)$ and $P(W|C)$ yourself?
A: After you pick a door and $33$ goats are revealed their will be $65$ doors remaining.  How many goats and cars are behind these depends on what lies behind the door you picked.
Case 1: Your first choice hides a car (this happens with probability $1/3$).  


*

*There will be $32$ cars and $33$ goats behind the other doors.


Case 2: Your first choice hides a goat (this happens with probability $2/3$). 


*

*There will be $33$ cars and $32$ goats behind the other doors.


So, using the law of total probability, the probability for picking a car if you switch is: ...
A: Simple argument in favor of switching doors: when 33 doors with Goats are removed the probability of winning a car on switching doors, or choosing a door from the remaining 65 doors (excluding the one chosen originally), is the probability of selecting a door with a Car from either (a) 32 doors with Cars + 33 doors with Goats; or, (b) 33 doors with Cars + 32 doors with Goats. This probability is at least $32/65\approx 1/2$.
As the probability of winning a car with the original selected door is $1/3$ (not changed when the 33 doors with Goats are removed) and $1/3<32/65\approx 1/2$, then switching doors is a no brainer. It is easier than the original Monty Hall problem.
A: yes. the probability increases no matter what was behind the door you picked initially because the total number of choices is lower after the switch therefore increasing your chance to win.
