Monty Hall Problem simulation gives weird answers

Question

I tried to simulate the Monty Hall problem in Python, and ended up with 66% probability after a few attempts. However, the first incorrect attempts were more interesting, as they consistently gave a probability of 55% and 44% respectively.

Here's the code:

import random
from sys import argv

def hall_old():
    choices = [0]*3
    choices[random.randint(0,2)] = 1
    guess = random.randint(0,2)
    reveal = (random.randint(1,2)+guess)%3 # This gives ~55%
    reveal = 2 - guess # This gives ~44%
    if choices[reveal]==1:
        return hall()
    return choices[3-guess-reveal]==1

def hall(): # This gives 66%
    choices = [0]*3
    choices[random.randint(0,2)] = 1
    guess = 0
    for i in range(1,3):
        if choices[i]==0:
            reveal = i
            break
    return choices[3-guess-reveal]==1

def main():
    times = int(argv[1]) # times game is played
    correct = 0
    for i in range(0, times):
        if hall_old():
            correct += 1
    print str(correct * 100 / times) + '%'

main()

From what I understand, the function hall_old(), should still give 66% probability when run but for some reason, if you change the way you select the variable reveal then the probability changes. Can anyone explain why this is occurring?

Answer 1

Given a perfect PRNG, guess in hall_old should be a uniformly distributed random number $\in\{0,1,2\}$.

With reveal = (random.randint(1,2)+guess)%3, it should accordingly be that reveal is uniformly distributed $\in\{0,1,2\}\setminus\{\text{guess}\}$. Note that in this case 3-guess-reveal is simply the remaining third number (why non call it other?). In $\frac13$ of all cases, you call hall(), which itself seems to return true with a probability $\frac 23$. In the remaing $\frac23$ of all cases (namely, when reveal is not the win), you return if choice[other] is true, which shold be so with probability $\frac12$. Hence we expect something like $$\frac13\cdot \frac23 +\frac23\cdot \frac12=\frac 59. $$

If you use reveal = 2-guess instead, again in $\frac13$ of all cases you return hall() and in the other $\frac23$ of cases, you return if \choice[1]is true (as guess+reveal is definitely $2$). Note that this never the case if guess is $= 1$ (as it would already have been caught by the first if). Hence we expect something like $$ \frac 13\cdot \frac 23+\frac 23\cdot \left(\frac 13\cdot 0+\frac 23\cdot \frac12\right)=\frac 49.$$