The Monty Hall Problem (win a goat or a car)

There is a puzzle used in game shows known as The Monty Hall Problem. It's been around for a while but over lunch yesterday someone explained it to me for the first time and 3 out of 4 of us argued convincingly the same answer. And it was wrong. Here's the problem:

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?

The most logical answer to me was no it doesn't matter if you switch or not because you never knew what was behind the first door you chose anyway. It seemed to me that the problem was no different than having two choices, a goat or a car and randomly choosing one. But this is all wrong!

Since I'm not good at math I could only loosely follow the explanations for why I was wrong. So, naturally, I wrote some code to see it in action (bear with me, I spent all of 5 minutes on it):

import random
from decimal import Decimal
choices = ['goat','goat','car']
tries = 99000
switch_correct, stay_correct = 0, 0
for num in range(tries):
    doors = [c for c in choices]
    first = doors.pop(random.randint(0,len(doors)-1))
    for i,val in enumerate(doors):
        if val == 'goat':
    switched = doors[0]
    if switched == 'car':
        switch_correct += 1
    elif first == 'car':
        stay_correct += 1
print "stay: you win %s%% of the time" % (Decimal(stay_correct) / Decimal(tries) * 100)
print "switch: you win %s%% of the time" % (Decimal(switch_correct) / Decimal(tries) * 100)

I found the result astonishing:

stay: you win 33.5% of the time
switch: you win 66.5% of the time

The wikipedia link above explains why this is but it is still incredible to me, like a magic trick.