Using Code to Intuit the Monty Hall Problem
June 23, 2020
The Monty Hall problem is a famous puzzle that shows how statistical facts can be unintuitive. It’s based on a game show in which the contestant selects one of three doors, receiving whatever prize is behind it. One door has a desirable prize behind it, which is usually a car, and the others have a gag prize, usually a goat. My father told me about this problem when I was probably under 10, and at the time I neither understood the math, nor why anyone would want a car when they could have a goat. So for the rest of this article, assume that the preferred prize is a goat, and the others are, say, bags of sand.
Anyway, after the contestant selects a door, the show host (originally named Monty Hall) opens another door, revealing a bag of sand. The contestant then has the option to switch to the remaining door. The question is: Is it better to switch or does it make no difference?
The answer, as you may have heard before, is that it’s better to switch, because the probability of getting the goat when you initially pick a door is ⅓, which doesn’t change until you switch. This is confusing, however, because it seems like after the host shows you the bag of sand, there’s one door with a goat behind it, and one door with sand behind, so the probability of winning should be ½, regardless of which door you pick. There are a zillion articles explaining why this isn’t the case, which comes down to the fact that this is a conditional probability, not just a choice between two random options. However, depending on your background, you may have a better intuition for code than for math, in which case the easiest way to understand the problem is by seeing it as a program.
Take the following Python script. If you’re new to Python, don’t be intimidated by the
comprehensions, which is where you have
brackets: they are not hard to understand and they will make your life much easier if you
learn to use them.
import random, statistics def simulate(switch): """Return True if the contestant wins the goat, False otherwise.""" # Shuffle the doors and pick one at random = ['goat', 'sand', 'sand'] doors random.shuffle(doors)= random.randint(0, 2) choice # Pick one of the doors with sand behind it to show = [i for i in range(3) if doors[i] == 'sand' and i != choice] remaining_sand = random.choice(remaining_sand) shown # If switch is selected, switch choice to the remaining door if switch: choice = [i for i in range(3) if not i in (choice, shown)] return doors[choice] == 'goat' = 100000 trials print("Switch: ", statistics.mean([simulate(True) for i in range(trials)])) print("Stay: ", statistics.mean([simulate(False) for i in range(trials)]))
Run the program, and we see that indeed, you win twice as often if you switch than if you stay:
Switch: 0.66462 Stay: 0.33362
But we already knew that. However, the key insight is on line 16:
if switch: choice = [i for i in range(3) if not i in (choice, shown)]
This is the line that changes the choice to the remaining door. If
False, it won’t run. Comment out line 16, and your IDE or your linter will tell
you something like this:
monty-hall.py:13:4: W0612: Unused variable 'shown' (unused-variable)
When the contestant doesn’t switch, the information provided by opening the door isn’t
used at all. In fact, the Python VM will probably skip over the lines computing
shown because it’s not used. In other words, choosing not to switch is
exactly the same as picking randomly from three options.
I decided to write this article because after writing the problem in code, the answer seemed completely obvious. Before, I was sort of able to understand the standard stats textbook explanation, but the answer still seemed sort of opaque. Hopefully, there will be a few other people out there to whom code is a more native language than math who will find this as clarifying as I did.