3 ms·

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 succe

• cjbest 5y ago
  No, I don’t think this is right. 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.
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• Nursie 5y ago
  No, it's a 2/3 chance at that point. 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); }