Updated: Nov 25, 2019
What's your favorite game show? You may know Who Wants to Be a Millionaire, Wheel of Fortune or 101 Ways to Leave a Game Show (my personal favorite), but the one I want to talk about is Let's Make a Deal and its long-running host Monty Hall.
Throughout his career, Monty Hall received almost legendary status for the various radio and TV shows he hosted, and today, one of probability's favorite paradoxes bears his name.
Here's the setup.
Monty Hall shows you three doors. Two of the doors have a goat behind them, and one has a car. Your goal is to pick the door with the car.
Monty gives you a chance to pick a door. Say you pick Door 1.
He then opens one of the doors with a goat to show you one door that doesn't have a car. Say he opens Door 2.
He then gives you a choice. Do you want to keep Door 1 or do you want to switch to Door 3?
Whatever is behind the door of your final decision you get to keep.
The scenario always plays out this same way: The player picks a door, Monty opens one of the two remaining doors and the player is given a choice whether to keep their door or switch. The question is: What should you do? Is there a benefit to keeping your door? Is there a benefit to switching?
Let's try it. Visit https://www.mathwarehouse.com/monty-hall-simulation-online/ to get a feel for the problem. What kind of probabilities emerge?
Let's check our intuition. Do you think keeping the door you chose versus switching to the other has any impact on your probability of winning? Most people assume it's a 50-50 chance. There are after all two doors remaining and the car was placed completely randomly, so many contestants on the show decide to keep their door because they assume their chances are equal and want to stick with their initial gut choice.
But I'll cut to the chase. The probability of winning if you switch is not 1/2, but 2/3! That means the probability of winning if you keep your door is only 1/3. That's right: somehow, switching your door gives you twice the chance as keeping it!
To understand how this works, realize that you initially had a 1/3 chance of choosing the right door, and Monty Hall opening a door does not affect the chances of your initial choice being right. Let's stick with our example: You chose Door 1 and Monty opens Door 2 to show you a goat. Think about it this way. You locked down your choice before he showed you a goat, so since there's a 1/3 chance of you choosing Door 1 in the first place, there's a 1/3 chance Door 1 has the car behind it. However, the chance of Door 2 containing the car is 0 (We know it has a goat behind it!) and probabilities sum up to 1, so Door 3 has a 2/3 chance of being the correct choice.
That's the way I first heard it explained to me, and though it does discuss what's going on here, that explanation misses something. The key to understanding the Monty Hall problem relies on something different: understanding that Monty's moves actually depend on your moves. Sometimes, he's forced to reveal a certain goat, and sometimes he's able to choose which door to reveal.
Let me show you what I mean. Let's assume the car is behind Door 3:
If you choose Door 1, Monty is forced to open Door 2, the only goat remaining.
If you choose Door 2, Monty is forced to open Door 1, the only goat remaining.
If you choose Door 3, you've chosen a car so Monty randomly opens another door.
In the first two scenarios, Monty's move is forced, and that means if you want the car, you're forced to switch over. 2/3 of the time, switching guarantees success.
In the last scenario, keeping results in success, but that's just 1/3 of the possibilities.
It's clear now that switching gives you an advantage. The fact that Monty's move can be forced or not is what amplifies your chances. Monty has to give you a choice between a goat and a car, so he can't choose randomly, and that means it makes sense for your odds to no longer be so random... The probabilities have been manipulated!
If you'd like to see some other explanations of the Monty Hall problem, feel free to check out this video:
Note: This problem can actually be extended beyond the 3 door setup to any number of doors. It turns out that if there are n doors and Monty reveals one or more doors, your chances of winning always go up from 1/n if you switch. If Monty opens all but one, giving you a choice of two doors, your probability of winning with switching is (n-1)/n, which is totally in line with what we found above because 3 doors give (3-1)/3 = 2/3, exactly what we showed above. I won't go into all the detail but feel free to read about it here: https://math.stackexchange.com/questions/608957/monty-hall-problem-extended
These probability puzzles show us that we can't always trust our intuition in statistics problems. In the birthday problem, we saw that we often have trouble grasping probabilities that grow exponentially, and in the Monty Hall problem, we see that sometimes it's hard for us to detect when seemingly random probabilities have been manipulated behind our backs. So let me give you a word of warning: Don't bet on something you haven't calculated first!
I hope you enjoyed the post! Comment below with any questions you may have or any feedback you may have for me. What are some other cool problems you've seen around? Is there anything you'd like me to write about? Also, like this post if you want to see more like it!