Probability about switching choices

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: ...

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?