This is not correct. In case Monty opens a random door which may or may not have the prize behind it, there is no reason to switch. Probability table is: 1/3 c

This is not correct. In case Monty opens a random door which may or may not have the prize behind it, there is no reason to switch. Probability table is: 1/3 chance you pick the prize on first choice. (And Monty opens a random goat-door) 2/3 chance you don't, in which case 1/2 of the time, Monty opens the prize door and you lose. The other 1/2 of the time Monty opens a random goat) So after Monty has opened a door there is now 3 possible scenarios: 1. 1/3 chance you picked the prize. Don't switch 2. 1/3 = 2/3 * 1/2 chance you didn't pick the prize and Monty opened the prize door. You lost. 3. 1/3 = 2/3 * 1/2 chance you didn't pick the prize and Monty opened a goat door. In this case, switch. The probability of 1 and 3 are equal and switching or not switching doesn't matter. Now if Monty knows where the prize is and always opens a goat door, things are very different. 1/3 chance you picked the prize. Monty opens a goat. You shouldn't switch. 2/3 change you picked a goat. Monty opens a goat. You should switch. Now 2/3 of the time you can switch and get the prize. So switching is correct. It all depends on the unstated assumption that Monty may actually have told you something, rather than taken a risk. And that's why this problem is mostly a good lesson in what can happen if you don't state your problem well.