Last Updated: January 09, 2017

The Monty Hall Problem in Python

Again, this is another coding exercise in Dream.In.Code. Well, it is actually a challenge but since my submission pales compared to the others, I don't prefer to call it as one.

What is the Monty Hall 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 [but the door is not opened], 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?

Basically, you need to prove that switching doors when asked to switch will improve your chances by 2/3 as opposed to the 1/3 chance if you don't switch doors.

Here's what I cooked up:

Door Creation

def addDoors():
picker =  randint(0,1)
if picker == 1:
        door1,door2,door3 = "dragon","goat","goat"
else:
        door1 = "goat"
        picker = randint(0,1)
        if picker == 1:
                door2,door3 = "dragon","goat"
        else:
                door2,door3 = "goat","dragon"
return [door1,door2,door3]

Goat Door Opener

def openGoatDoor(doors):
    door = ""
        while door != "goat":
        doorNumber = randint(0,2)
        door = doors[doorNumber]
    return doorNumber

The Monty Hall Simulation

def montyhall(doorNumber,switchOk):
    doors = addDoors()
    goatDoor = addDoors()
    while goatDoor == doorNumber:
        goatDoor = openGoatDoor(doors)
    if switchOk == "yes":
        doorNumber = [y for y in range(2) if y not in [doorNumber,goatDoor]]
        return doors[doorNumber[0]]
    else:
        return doors[doorNumber]

The Tester

def testMontyHall(playTimes,switch):
    wondragon,wonGoat = 0,0
    for i in range(playTimes):
        prize = montyhall(randint(0,2),switch)
        if prize == "dragon":
                wondragon += 1
        else:
                wonGoat += 1
    print "Dragons:",wondragon
    print "Goat:",wonGoat

The Results
>>> testMontyHall(1000000,"no")
Dragons: 332961
Goat: 667039
>>> testMontyHall(1000000,"yes")
Dragons: 666757
Goat: 333243

And this concludes The Monty Hall show. Hope you enjoyed it.

#python

#the monty hall problem

#exercise

#challenge

Written by Julius Santos

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1 Response

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randrewnichol

Here's mine, yours is a bit fancier!
Functions:

#monty_hall.py
def gen_doors():
    doors = [0,1,2]
    car = random.randrange(3)
    doors[car] = "Car"
    goats = list(range(3))
    del goats[car]
    for i in goats:
        doors[i] = "Goat"
    return doors

def random_play(doors = ["Goat","Car","Goat"], switch = True):
    goats = []
    [goats.append(i) for i in range(3) if doors[i] == "Goat"]
    pick = random.randrange(3)
    if switch == True:
        if pick in goats:
            return "Car"
        else: return "Goat"
    else:
        if pick in goats:
            return "Goat"
        else: return "Car"

#sim.py
import monty_hall
plays = 100000

# With Switch:
n_goats = 0
n_cars = 0
for i in range(plays):
    if monty_hall.random_play(monty_hall.gen_doors(), True) == "Goat": n_goats += 1
    else: n_cars += 1
print("With Switching:\nGoats: ", n_goats, "\n", "Cars: ", n_cars, "\n")

n_goats = 0
n_cars = 0
# Without Switch:
for i in range(plays):
    if monty_hall.random_play(monty_hall.gen_doors(), False) == "Goat": n_goats +=1
    else: n_cars +=1
print("Without Switching:\nGoats: ", n_goats, "\n", "Cars: ", n_cars)

[Out]
With Switching:
Goats: 33210
Cars: 66790

Without Switching:
Goats: 66812
Cars: 33188

over 1 year ago ·