I'm gonna be dead honest, as interesting and intriguing as all this is, we are in a time of turmoil, war, and dishonesty amongst thieves if you get what I mean, meaningless math like this has no place in this day and age, but I pick door number 2 if that helps any lol
The Monty Hall Problem (counter intuitive math teaser)

Unless of course he's being a good host and picking a random door; than it would be 1/3 chance of being correct.

Yes, that is right Shadow. It's strange how such a slight difference can cause such a different result.

(Just a clarifier: I'm not arguing that what is stated is wrong, I've seen it before and therefore assume it to be true. I'm just looking for a more detailed explaination)
I don't see how the host knowing or not changes the probability. Don't you have a 1/3 chance of picking the car on your first guess whether or not the host knows where it is? And if he then reveals a goat, doesn't the other door still hold a 2/3 chance of holding the car?

That's true Ryan, but if the host reveals the car (remember, he doesn't know where anything is in your variant), then there's a 0% chance for the car obviously.
I've heard that in the actual episodes of Let's Make a Deal with Monty Hall himself, Monty Hall didn't know behind what door was what. Therefore, he occasionally revealed (or didn't reveal, but just got rid of the door without showing) the only real prize.
Shadow, if I were to explain the solution to you, I'd pretty much just be telling you what wikipedia.org says about it. Just search the Monty Hall problem on Wikipedia, and it will explain it to you quite well I think. If confusion still remains, Youtube.com should clear it up. And lastly, if some shred of doubt still remains, familiarize yourself with Bayes' theorem and work out the solution yourself.
I recently received Jason Rosenhouse's book, The Monty Hall Problem. It's a well known problem in mathematics (mainly probablitiy) that is extremely counter intuitive. It has fooled even bright professors and philosophers.
From Wikipedia:
'Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?'
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Because there are 2 doors left (doors 1 and 2), most people think there is a 50% chance that the car is behind either one of them. This is incorrect.
The correct answer is door #2 has a 2/3 chance of being the correct door while your original choice of #1 has a 1/3 chance.
Rosenhouse's book goes into quite a lot of detail with Bayes' theorem and the quantum Monty Hall problem, but Wikipedia gives a good enough explanation for those less familiar with math.
An interesting variant is when the host does not know which door contains what, but invariabley reveals one with a goat. In this situation, switching doors wins 50% of the time, as expected.