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The Monty Hall Problem: Demystified

Reading time: 5 minutes
The Monty Hall Problem Demystified

By Michael Laracy, Founder

When I first came across the Monty Hall problem, I found it fascinating, but it also hurt my brain. I lost half a night of sleep thinking about it, yet it still didn’t make sense.  The next morning, I had to investigate the answer personally so that I could understand the assumptions of the problem, and either prove that the answer was correct (or that it wasn’t).

The Problem

If you’re not familiar with the seemingly simple Monty Hall problem, here it is: You are on a game show, and the host (Monty Hall) asks you to choose one of three doors. He tells you that behind two of the doors is a goat, and behind one of the doors is a car. You get to keep whatever is behind the door that you choose (a goat or the car; most people want the car). You choose one of the doors. Monty then opens up one of the two remaining doors, revealing a goat.  Monty asks you if you want to stay with your original choice, or switch your choice to the other unopened door.  What do you do?

The Monty Hall Problem

Simple answer or brain twister?

Should you stick with your original door choice, or should you switch to the other unopened door? You originally had a one out of three chance to have chosen the door with the car, and a two out of three chance to have chosen a door with a goat.  But Monty opened one of the other doors, revealing a goat behind it.  This is where most people get tripped up. It would appear that you have a 50% chance of winning the car, regardless of whether or not you choose to switch your choice of doors. It doesn’t matter if you switch doors or change; your chances of winning are the same, right?  Wrong. 

In reality, you increase your chances of winning if you choose to switch from your original door to the other unopened door.  By switching to the other door, you now have a 2/3 chance of winning the car (approximately a 66.67% chance).  If you stay with your original door choice, you have a 1/3 chance of winning (a 33.33% chance).  In other words, by switching doors, you double your chances of winning the car. 

WAIT.  WHAT?!  HOW?

It doesn’t seem right, does it?  I was initially convinced that switching to the other door would do nothing to increase the probability of winning the car (instead of a goat).  It bothered me so much that I decided to dig in and find an explanation.

Before reading on, think about it for a bit and see if you can wrap your mind around it. 

Standard Monty Hall Problem (Asymmetric Information)

A critical component to the standard version of the Monty Hall problem (and one which most people overlook) is that, as the host, Monty knows what’s behind each door.  That’s important.  You, as the contestant, have no information about what’s behind each door.  With a choice to switch doors, you essentially have a chance to choose two doors instead of just one door. Since Monty knows what’s behind each of the doors, he isn’t actually providing you with additional information by revealing one of the doors with a goat behind it. 

That’s because he will always have at least one goat door to show you, so he can always reveal a goat door. By offering you the chance to switch from your original choice to the other remaining door, Monty is basically offering you to go with your original chance of one out of three, or to change your door choice and get a two out of three chance of winning the car. Again, it doesn’t matter that he showed you a goat door. He’s always going to have a goat door to show you.

The Monty Hall Problem Probability of Changing Your Original Choice

A Modified version of the Monty Hall Problem (with Symmetric Information)

What happens if we change the rules/assumptions so that information is symmetric? In other words, what if Monty did NOT have more information than you?  If he didn’t know what was behind each door, and he randomly selected one of the two remaining doors, one-third of the time he is going to show you a goat door, and one-third of the time he’s going to reveal the door with the car behind it.  If he reveals the car door, the game ends, and you don’t get a chance to choose between your first pick and the other unopened door because Monty already revealed the winning door.  In those cases, you lose.  Both of the remaining doors have goats behind them. 

It’s interesting how this slight change to the rules impacts the probability behind your decision to switch doors. If you know that you are playing with this new set of rules, where Monty has no information about what’s behind each of the doors, when he does reveal a goat door, the decision to go with your first pick vs. switching to the other door would give you an equal probability of winning.  In other words, you now have a 50% chance of winning the car, regardless of whether you go with the original door that you picked or switch to the other door. 

The Monty Hall Problem with Symmetric Information

In the symmetric information version of the problem, you can see that when Monty reveals a door without a car behind it, you have a 50/50 chance of winning the car, regardless of whether you stick with your original door, or switch to the other door.  

Proving the Answer with Math

A written explanation didn’t quite satisfy me, though. I ran simulations in our analytics software, Construct, to prove to myself that switching doors would increase your chance of winning the car.

Testing the standard Monty Hall problem in Rapid Insight's data prep tool Construct

The top stream in the above workflow tested the standard Monty Hall problem (the asymmetric version). I ran 10,000 simulations. Lo and behold, the simulations showed that when contestants switched their door choice, they won approximately 67% of the time.  When they didn’t switch their door choice, they won only 33% of the time.  

The bottom stream tested the problem for when Monty has no information about what’s behind each door (the symmetric version of the problem). I ran another 10,000 simulations of the problem through this workflow. In these simulations, the game ended approximately 3,333 times because Monty revealed the door with the car behind it, resulting in an automatic contestant loss. In the 6,667 times that the game didn’t end automatically, the contestant won 50% of the time. 

Final Thoughts

If your brain still tells you that taking Monty up on his offer to let you switch your choice of doors DOES NOT increase your chances of winning, don’t feel bad. It’s a lot to wrap your head around. But if it’s any comfort, it’s not so much that your answer is wrong; rather, you have flawed assumptions. 

The key to understanding the problem is realizing that the information in the classic Monty Hall problem is not symmetric. Monty knows more than the contestants; he knows what’s behind each door. Most people forget that critical fact when they think about this problem. Statistically speaking, that gap between what you know and what Monty knows makes all the difference.

There are many other statistical problems like the Monty Hall problem.  I’ll plan to write about a couple of them in future posts.

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