The Monty Hall Problem

You are on a TV show. You are given a choice of 3 doors. Behind one door is a car; behind the others, goats. You are to pick a door, winning whatever is behind it. So you pick one, say door 1. Since there are 2 goats behind 3 doors, whatever you choose, there is at least one goat left behind the 2 other doors. The host, Monty Hall, who knows what's behind them, opens one of these 2 other doors, say door 3, to reveal a goat. He then asks you, "Do you want to stick to your choice, or do you want to switch to door 2?" The question really is to ask, "Does switching increase your chance of winning the car?"
When you pick door 1, the chance of winning the car is 1/3 (i.e. the chance of having the car behind door 1 is 1/3 since there is 1 car in 3 possible doors). And then Monty eliminates door 3 for you by revealing a goat behind door 3. Intuitively, you think you then get a better chance of winning the car as the chance of having the car in door 1 is now 1/2 (since there is 1 car in only 2 possible doors) and likewise in door 2 is also 1/2. So you think it makes no difference whether to switch or not. So you choose to stick with door 1. Right? WRONG!

Let's go back and examine the probabilities in more details.

When you pick door 1, your chance of having the car behind door 1 is 1/3. There's no question about it. Let's now group doors 2 and 3 together. Your chance of having the car NOT behind door 1 (i.e. behind either door 2 or 3) is 2/3 since the total probability must equal to 1. By eliminating door 3 from the scene (by Monty), your chance of having the car behind door 1 is NOT increased from 1/3 to 1/2 as intuitively thought. Your chance of having the car in door 1 is still 1/3 as when you make your choice, you choose 1 out of 3 possibilities (doors 1, 2 and 3). As the saying goes, the dice has been cast. The probability will not, and cannot be altered. Only IF you are free to choose at this point, you are then to choose 1 out of 2 possibilities (doors 1 and 2), and the chance of having the car behind door 1 is indeed 1/2. So by eliminating door 3 from the scene, the effect is ONLY simply changing our group consisting of doors 2 and 3 to a group consisting of door 2 only. Hence your chance of having the car NOT behind door 1 (now behind door 2 only) is still 2/3. In other words, AT THIS POINT, the chance of having the car behind door 2 is 2/3, not 1/2 as intuitively thought. So after Monty eliminates door 3 for you by revealing a goat behind door 3, the chance of having a car in door 1 is 1/3, and in door 2 is 2/3.
Hence, YES, you WILL increase your chance of winning by switching. IN THEORY, YOU SHOULD ALWAYS SWITCH.