Actually, experimentation will confirm the classic solution to the problem. If you have any programming skill, try writing a simple program to randomly choose a "car" door, randomly choose an "opened" door, and randomly choose a "selected" door. Have the program check to make sure it doesn't open the "car" door, then tally the results from 100 guesses always switching, and 100 always not switching.
If not, just play a mock-up game with a friend and keep score. Do it twenty times or so each way to get a decent number of results.
You will find that approximately 1/3 of the time without switching, and 2/3 of the time with switching you will get a car, which is exactly what is predicted through probability theory.
There are a number of ways of figuring the problem. Probably (haw haw) the simplest way is to draw out a diagram of possible results. As I said, experimentation will confirm the theory.
If Monty is unaware of the location of the car and chooses a door totally at random, then you will indeed have a 50% chance either way. This is because 1/3 of the time, the game will end early when Monty reveals the car door by mistake.
It has nothing to do with misdirection, and everything to do with probability. Remember, it only works out this way if Monty is aware of the contents of the three doors, and offers the switch every time (there are variations where he offers the switch only when you've selected a goat or a car, in which case the probability of winning through switching would be 1 and 0 respectively).