The Monty Hall Problem: Proof Through Modeling

The Problem

A game show has three doors. One hides a car; the other two hide goats.
You pick a door. That is your original choice.
Monty, who knows where the car is, opens a different door to reveal a goat.
You can stay with your original door or switch to the remaining closed door.

1

Your pick

Stay

2

Switch

🐐

3

Monty opens

Your Choice

Which is the better strategy:

The Best Choice

Before telling you what the best solution is, I'll tell you that there are at least four ways of demonstrating the best solution. The first way is by showing you what happens when the game is run 1000 times with your choice (), then 1000 times with the other choice (), and see which has the better success rate. I'll tell you right now that neither choice will win all of the time, but one choice definitely improves your odds of winning.

Simulation

Press the button to play 1000 games using your chosen strategy. A graph of the cumulative win rate will appear.

0

Results

This simulation shows that it is better to switch doors after Monty reveals the goat. On average, this strategy will win 66.6% of the time.
In addition to a simulation, there are three other ways of understanding this preferred choice.

Alternative Explanations

Information Explanation:
- Monty knows where the car is, and he always opens a door that hides a goat.
- That makes his choice a signal rather than luck. If he opened doors at random, he would sometimes reveal the car, and a goat reveal would tell you nothing new.
- When your first pick is wrong (2 times in 3), Monty has no choice: he must open the only other goat door, so the car is behind the door he leaves closed.
- When your first pick is right (1 time in 3), he can open either goat door, and the door he leaves closed hides a goat.
- His reveal therefore tells you something about the remaining door, and only Strategy B (Switching) uses this information.
Probability Explanation:
- Your first pick has a 1 in 3 chance of hiding the car, so the other two doors together hold a 2 in 3 chance.
- Monty's reveal can't change the odds on your door, because he can always open a goat door no matter where the car is.
- What it does change is where the other 2 in 3 sits: it was split between two doors, and now that Monty has opened one of them and shown a goat, all of it rests on the single door left closed.
- Staying keeps the 1 in 3. Switching claims the 2 in 3.
- Put another way, switching wins whenever your first pick was wrong, and that happens 2 times in 3.
Edge Case Explanation:
- Imagine 100 doors. You pick one, and Monty opens 98 goat doors, leaving yours and one other.
- Your pick was a 1-in-100 shot. The other closed door carries the other 99 in 100.
- Monty never opens the car door, so if the car was behind any of the 99 doors you didn't pick, it is the door he deliberately leaves closed.
- Staying wins only if you were lucky enough to pick the car on your first try, a 1-in-100 chance.
- The three-door game works the same way on a smaller scale, so the 2-in-3 edge for switching is a gentler version of this effect.
- Press the button below to watch it play out.
Your pick: 1 in 100 The other 99 doors, together: 99 in 100

Thanks for participating in this exercise. The Monty Hall Problem is an exercise that illustrates how we use additional information (in this case, the revealing of the goat door) to modify our probability estimates and our decisions.

To learn more about the Monty Hall Problem, see the following resources:

There is a whole body of work that examines how we use information in an incremental fashion to revise our thinking. One method is called Bayesian analysis.

I'm developing a similar simulation to help you understand this, but until it's ready, here are some good resources:

I hope you enjoyed this. Please provide feedback to me at jamullen03+biases[at]gmail.com