It always surprises me when I talk about the Monty Hall problem, how many people argue with me about what the “right” answer is. It goes against most everyone’s ideas of what is right. Here’s the problem:
The Three Doors
You are presented with three doors, and are told that behind ONE of them, there is a shiny, new car. Behind the other two doors, however, there are goats. The object of this game, obviously, is to choose the door that has the car.
So make your choice. Do you choose Door #1, Door #2, or Door #3?
Once you have chosen a door, your gracious host, Monty Hall, chooses to open one of the doors you did not select. He will ALWAYS open a door that contains a goat. For the purposes of illustration, let’s assume you chose door #2 initially. In this case, then, Monty knows that there is a goat behind Door #1. So he opens it.
This is when the Monty Hall problem rears its ugly head. Once that door is revealed, he offers you a choice. You can keep the door you initially selected, or you can switch to the other unopened door. It’s entirely your choice. What would YOU do? Would you stay with your original door? Would you choose the other one? (Leave a comment on this post with your choice right now…before you continue reading…) In our example, again, we are going to stay with our initial choice.
After you’ve made your choice, Monty will open the door you DIDN’T select. Here’s a picture of how our scenario played out:
So the point of this whole post comes down to probability:
Why You Should Always Switch Doors
This is where the controversy comes in. Many people (probably including you) disagree with this. The common thought is that no matter what you do, your initial choice of the three doors can’t actually be improved. I’m writing this to tell you that you’re wrong.
In fact, you can DOUBLE your odds of winning by choosing the other unopened door. Let me illustrate the flow of choices you get to make, and how they improve with a “switch.” In this example, we assume that the “contestant” always chooses Door #1.
Recall back to how the problem works: You pick one of three doors. So your odds are 33% (1/3) that you’re going to pick the car. Said another way, there’s a 66% (2/3) chance that you’re going to pick a goat.
Here’s what that means, as you look at the image above: 2 out of 3 choices result in you choosing a door with a goat. Switching, when you have selected a goat initially, results in getting the car 100% of the time. So if you ALWAYS switch, you’ve got a 66% chance of winning the car. If you NEVER switch, which means you stick with your initial choice, you are forcing yourself to take a 33% chance of the car being behind your door.
Yes, that does still mean that 1/3 of the time you will still lose by switching, but I’ll take that over losing 2/3 of the time by NOT switching.
In the end, you DOUBLE your odds of winning the car by switching every time.
Let’s Discuss This Further
I’d like to think that this topic will generate some conversation. You probably have stared at the diagram, and yet, inside, you still disagree. You might say that switching is a 50/50 shot. That’s where I disagree. If you’re simply playing the percentages, switching will always give you the best results.
Leave a comment. Leave several. Let’s have a lively discussion on this one.
UPDATED: Marvin S. Schwartz, PhD. has written a simple C# application that lets you play this game. Here’s a link to his Monty Hall source code.