A person has two choices half way through the game:

1) change
2) do not change

let us analyse the game for each of two players.

Firstly, the easy one. Somebody who does not change. He has already decided not to change and can therefore 'skip' any door being revealed. Therefore the decision is random selection out of three.

P(choosing the car with initial door selection)= 1/3
therefore
p(win car)=1/3

I am about to demonstrate that the alternative strategy, changing, gives a higher probability of winning. 'Not changing' is now NOT an option.

As before, there is a 1/3 probability of choosing the car first. However, for this individual, choosing the car would lead to finally picking a goat after swapping. EVERY OTHER initial choice of door leads to winning as Monty will reveal the other goat, leading the player straight to the car. The only way of not being led to the car is to choose the car the first time!!

P(selecting car first time)=1/3
and
P(being shown one of the two incorrect doors)=1

thus
P(swapping from incorrect door to correct door)=2/3

Hence this is the better strategy as it doubles the expectation of winning.

This result becomes clear as soon as you consider new rules as if swapping was required, or the converse.