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Door Decisions: The Surprising Solution to the Monty Hall Problem
Should You Switch?
Stephen Hill
February 01, 2024
Welcome to the “Illumined Insights” newsletter! Thank you so much for subscribing. This weekly newsletter touches on all things analytics and data science with a focus on areas such as data visualization, AI, and sports analytics.
This week we take a look at one of the classic problems from the world of analytics: the Monty Hall problem. I’ll also take a moment to introduce my new course offering: DataVizBoost.
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DALL-E 3 Image: Flat art style image of Monty Hall problem.
Welcome to this edition of the Illumined Insights newsletter, where we delve into one of the most intriguing puzzles in probability and statistics - the Monty Hall problem. Named after the host of the American television game show "Let's Make a Deal," this problem has baffled and divided both amateurs and professionals alike with its counterintuitive solution.
“Let’s Make a Deal” with Monty Hall (Image source: IMDB)
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, based on the television game show "Let's Make a Deal" and named after its original host, Monty Hall. The scenario presents a contestant with three doors: behind one door is a car (a prize), and behind the other two doors are goats.
The Three Doors of the Monty Hall Problem
The contestant is first asked to select one door, say No. 1, and the host, who knows what is behind the doors, opens another door, say No. 3, which has a goat. The host then asks the contestant if they would like to switch their choice to the remaining unopened door (No. 2). The question is: Should the contestant switch their choice? Does it matter?
Should the contestant switch to Door 2 or stay with Door 1?
At first glance, the problem may seem straightforward, but its solution is anything but. The debate at the heart of the Monty Hall problem lies in whether the contestant should stick with their initial choice or switch to the other unopened door to increase their chances of winning the car. What do you think?
Understanding the Basics
Before we dive into a statistical explanation, let's understand the basic premise of the Monty Hall problem. The crucial aspect of the problem is the host's behavior, which is to always open a door that was not picked by the contestant and always reveals a goat. This action changes the probability distribution over the doors, a fact that is central to the puzzle.
Many people argue that, after the host opens a door to reveal a goat, the probability of winning by either sticking with the initial choice or switching is 50-50. This intuition, however, fails to take into account the host's behavior and how it affects the odds.
A Statistical Perspective
From a statistical standpoint, the initial choice of the door has a 1/3 chance of being the winning door, and this probability remains unchanged after the host opens a door to reveal a goat. However, the probability that the car is behind one of the two doors that were not initially chosen is 2/3, and the host's action of opening one of these doors (always revealing a goat) effectively transfers this 2/3 probability to the remaining unopened door.
The decision tree below shows the result if the contestant chooses to not switch their door choice. Let’s look at the possibilities. There is a 1/3 chance that the contestant initially selected the door with the car. If he selects this door, then he wins. Otherwise he loses. So the probability of winning the car with the “not switching” strategy is 1/3 and the probability of losing is then 2/3.
Probabilities if the player does not switch doors
How does this change if the player adopts a strategy to always switch his door. If he originally selected the door with the car, then switching will cause him to lose. However, there was only a 1 in 3 chance of correctly picking the car at the beginning. Switching improves the probability of winning to 2/3.
Probabilities if the player chooses to switch doors
Still not convinced? This explanation makes even more sense if we consider a situation with more than three doors. What if we had 100 doors? The probability of picking the door with the car from the beginning would be 1/100. Assuming that Monty Hall opens all of the other doors except one, revealing a goat behind them all. Should we switch? If we don’t switch, our probability to win is 1/100. However, if we do switch, our probability of winning is 99/100. The choice is obvious.
Simulation Model Approach
To complement our theoretical understanding, let’s look at a simulation model of the Monty Hall problem. This model randomly generates 10,000 game scenarios, allowing us to observe the outcome of both sticking with the initial choice and switching. Through this empirical approach, we can illustrate the statistical reasoning behind the problem and demonstrate why switching is the better strategy. To see Python code (hosted in a Jupyter Notebook on Google Colab) for the simulation model visit here: Link. A bar chart of results from a simulation run is shown below (your results may vary slightly due to randomness).
There is also an excellent R Shiny application that allows us to simulate the Monty Hall problem (linked below).
Illumined Insights Book Recommendations
This week I recommend one of my personal favorite business/analytics books.
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