Is this a good or bad 'simulation' for Monty Hall? How come?

Question

Through trying to explain the Monty Hall problem to a friend during class yesterday, we ended up coding it in Python to prove that if you always swap, you will win 2/3 times. We came up with this:

import random as r

#iterations = int(raw_input("How many iterations? >> "))
iterations = 100000

doors = ["goat", "goat", "car"]
wins = 0.0
losses = 0.0

for i in range(iterations):
    n = r.randrange(0,3)

    choice = doors[n]
    if n == 0:
        #print "You chose door 1."
        #print "Monty opens door 2. There is a goat behind this door."
        #print "You swapped to door 3."
        wins += 1
        #print "You won a " + doors[2] + "\n"
    elif n == 1:
        #print "You chose door 2."
        #print "Monty opens door 1. There is a goat behind this door."
        #print "You swapped to door 3."
        wins += 1
        #print "You won a " + doors[2] + "\n"
    elif n == 2:
        #print "You chose door 3."
        #print "Monty opens door 2. There is a goat behind this door."
        #print "You swapped to door 1."
        losses += 1
        #print "You won a " + doors[0] + "\n"
    else:
        print "You screwed up"

percentage = (wins/iterations) * 100
print "Wins: " + str(wins)
print "Losses: " + str(losses)
print "You won " + str(percentage) + "% of the time"

My friend thought this was a good way of going about it (and is a good simulation for it), but I have my doubts and concerns. Is it actually random enough?

The problem I have with it is that the all choices are kind of hard coded in.

Is this a good or bad 'simulation' for the Monty Hall problem? How come?

Can you come up with a better version?

Answer 1

Your solution is fine, but if you want a stricter simulation of the problem as posed (and somewhat higher-quality Python;-), try:

import random

iterations = 100000

doors = ["goat"] * 2 + ["car"]
change_wins = 0
change_loses = 0

for i in xrange(iterations):
    random.shuffle(doors)
    # you pick door n:
    n = random.randrange(3)
    # monty picks door k, k!=n and doors[k]!="car"
    sequence = range(3)
    random.shuffle(sequence)
    for k in sequence:
        if k == n or doors[k] == "car":
            continue
    # now if you change, you lose iff doors[n]=="car"
    if doors[n] == "car":
        change_loses += 1
    else:
        change_wins += 1

print "Changing has %s wins and %s losses" % (change_wins, change_loses)
perc = (100.0 * change_wins) / (change_wins + change_loses)
print "IOW, by changing you win %.1f%% of the time" % perc

a typical output is:

Changing has 66721 wins and 33279 losses
IOW, by changing you win 66.7% of the time

Answer 2

You mentioned that all the choices are hardcoded in. But if you look closer, you'll notice that what you think are 'choices' are actually not choices at all. Monty's decision is without loss of generality since he always chooses the door with the goat behind it. Your swapping is always determined by what Monty chooses, and since Monty's "choice" was actually not a choice, neither is yours. Your simulation gives the correct results..