It's a 1/2 chance at the point where you're considering the switch.
If you start the game with the intention of switching, you have a 2/3 chance of being successful because the winning strategy is to miss the car on your first pick.
If you start the game with the intention of switching, you have a 2/3 chance of being successful because the winning strategy is to miss the car on your first pick.
Regardless of your intended plan, you only had a 1/3 chance of picking correctly the first time, so switching gets you a 2/3 chance.
When you choose initially, you have a 1/3 probability of getting the right one, leaving a 2/3 probability that the car is on one of the other two.
The host reveals one of the other two. So that 2/3 probability applies to the remaining door. Here is a short C implementation that made it very clear to me...
#include <stdio.h>
#include <stdlib.h>
int doround () {
int car = rand() % 3;
int firstchoice = rand() % 3;
// host reveals one of the goat doors
if (car == firstchoice) {
// you changing to the other door after the reveal is a loss
return 0;
}
if(car != firstchoice) {
// you changing to the other door after the reveal is a win
return 1;
}
}
int main(int argc, char** argv) {
int wins=0;
for (int round=0; round < 1000; round++){
wins+=doround();
}
printf("Worked in %d of %d rounds\n", wins, rounds);
}