But I'm still having difficulty comprehending the idea that there is a 66% chance of getting it right if you switch. This problem is essentially a choice between two doors, and each has an equal chance of having the item you want (each has 33% initially).
Given the fact that one door is shown to be empty, why would switching alter the fact that there are only two possible choices, and each had an equal chance of holding the prize?
Look at it this way: you are given three doors, door a, door b, and door c. You KNOW that door c will always be opened, no matter what you do, and it will be shown to be empty by the host. The chance of holding a prize, therefore, is equal between door a and door b, right?
The real choice behind this problem is not the initial choice between three doors, but actually the final choice between door a and door b, each with an equal chance of holding the prize. Regardless of what happens to door c, it does not change the probability of the prize distribution in door a and door b. I don't understand how showing door c to be empty skews the chance that one of the other doors will hold the prize.