No, Rags is right. Take a look at this statistics wheel that I found. http://math.ucsd.edu/~crypto/Monty/images/wheel.jpg

In the wheel, the very middle circle represents the door in which the car is behind. It can either be door 1, door 2, or door 3, right? Now, take a look at the second circle. It represents the door that the player can choose. Once again, the player can either choose door 1, door 2, or door 3. In this case, the probability that the player will immediately choose the correct one is 1 out of 3, or 1/3. Finally, take a look at the last circle. This circle represents the door the can be opened. As you can see, the blue spaces represents the door that has the car behind it. So if the player was right from the very beginning, then the person will open either of the other two doors in which there is a goat behind it. If this was the case, then switching would be a bad option, and if you did do this, you would lose. Now, let's assume that from the start, the player chose an incorrect door with a goat behind it. As you can see, if that happens, there would only be one incorrect door left, and that is the door in which the person will open. If this is the case, switching would be the best option, and you would win if you did. Take a look at the amount of red spaces, which signify an incorrect door. There are four of them. Therefore, from the start, you have a 2 out of 3, or 2/3 chance of choosing a wrong door. And if you do choose a wrong door, then switching would mean that you would win the car, right? So, you have a 2/3 chance of winning if you switch. Also take notice as how there are twice as many red spaces as there are blue spaces. This makes sense because the odds of winning if you switch is double the odds of winning if you don't switch: they are 2/3 and 1/3 respectively.