I know how to prove this statement by Bayes theorem when we have 3 doors with one car. But is it possible to solve this problem using Bayes theorem for 1000 doors?
In case with 3 doors I considered 4 events: $A, B, C$ are such events: car is in the 1st, the 2nd and the 3rd doors respectively. $OB$ is event that door number 2 will be revealed. So, applying Bayes formula we get
$$P(C\mid OB) = \frac{P(OB\mid C) \cdot P(C)}{P(OB)}$$
where $P(OB\mid C)$ is probability that we are looking for (do we have to change dor that we chose at first ($A$) to another ($C$), to make it clear let us imagine that if it is possible B will always be revealed (if there is no prize). $P(OB\mid C)$ is probability that door 2 will be revealed considering the fact that the actual prize is in third door. $P(C)$ is probability that prize is in the 3rd door. And P(OB) is probability that door number 2 will be revealed.
*If you are able to prove that we should change door only without using Bayes theorem, that would be helpful too