The Monty Hall Problem is a popular Probability brain teaser. It’s also one where when I first heard the answer, I just couldn’t wrap my head around it. Have you ever had something explained to you and it sort of makes sense to you rationally, and yet your intuition keeps shouting, "This cannot be!" Well, that’s how I felt when I first learned the solution.
Since we are currently sheltering in place in San Francisco (due to the ongoing pandemic), I have a lot of time to think, read, and write. So today I decided I will finally fully understand the Monty Hall Problem.
The Problem
From Wikipedia:
"Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
The Solution
Spoiler alert: the answer is that, yes we do want to switch. But why, and how much does our probability of getting the car increase if we do make the switch? Let’s find out.
This problem is designed to tease the brain, even for people trained in probability. It wants you to get lost in the probabilities until you just give up and say, "Aren’t they both the same?"
One non-mathematical and relatively simple way to arrive at the correct answer is to use the following line of reasoning:
- At first, when we randomly choose one of the 3 doors, our probability of getting the car is 1/3.
- If after the host opens a door, we stay with our choice, our probability of getting the car is still 1/3. Nothing has changed even though it feels like it has when the door was opened; so our probabilities remain unchanged as well. Think about it this way – if we are committed to our original decision, then the game becomes equivalent to one where we only get to choose once and are subsequently locked into our choice. In this case, whether the host opens all doors at the same time or he opens them slowly one by one (which he effectively does in the Monty Hall Problem), the probability of getting the car is always going to be 1/3.
- When he opens a door that has a goat behind it and gives us the option to choose again, instead of thinking of it as a continuation of the original game, we should think of it as a new game. In this new game, we only have 2 doors to choose between. If we randomly choose one of these 2 doors (say by flipping a fair coin), our probability of getting the car is 1/2. But we only get this boost in probability if we truly make a random choice again, so make sure to properly flip that coin. Pretty trippy right? Even though our initial choice is one of the two remaining possible choices, if we steadfastly stick with our initial pick, our probability of getting the car is still 1/3. But if we make a new random choice amongst the 2 remaining doors (and we still might choose our original pick), we end up with a 50 50 shot at a new car.
So the correct answer is to definitely choose again. But the actual correct answer is to pick the other door – the door that is not the one that you initially chose. Then your probability of landing the car goes up to 2/3!
How Did That Happen?
I coded up a simulator in Python for the Monty Hall Problem (you can find my code here). When I ran it 900,000 times and then stared at the results really hard, I realized what’s going on.
Here are the results if you do not change your door selection after the host opens one of the doors with a goat behind it. It is as expected – in 1/3 of the simulations, you get lucky and choose the correct door.
Obviously if you have a 1/3 chance of choosing the right door, then it follows that you have a 2/3 chance of choosing the wrong door. Wouldn’t it be nice to flip the odds? But the only way to do that would be if we got to choose 2 doors right?
Well what if I told you that by switching your door selection after the host opens one, you are effectively choosing 2 doors? Sounds too good to be true right? But that’s exactly what’s happening. It’s easier to see why if we invert the situation.
Let’s think about how we could lose if we decide to flip our door pick. In this case, we lose if our initial pick was the correct one. That’s because by opening a wrong door (one with a goat), the host is effectively giving us a free guess. And then we can complete the reversal of our initial decision by flipping our original pick. Our initial pick will be correct 1/3 of the time – and since we’ve now taken the other side of our initial pick, we get the car 2/3 of the time.
A more intuitive way to visualize this is to recast the Monty Hall Problem as a slightly different game. Imagine instead that you are playing against 2 other players. All 3 of you are randomly assigned a different door and you get to keep what’s behind the door that you’re assigned.
The Key Twist
Now right before you open the door the host comes to you and says:
I will give you a choice. Either you can keep what is behind your door OR you can keep what’s behind the doors of your opponents. And yes both of them!
Would you take that deal? I definitely would. Your two opponents get to open one door each (a total of 2 doors), so there is a 2/3 chance that they will get the car. That’s better than your 1/3 chance if you stay with your current door and worst comes to worst, you’ll end up with 2 new pet goats.
By choosing to switch your door pick in the Monty Hall Problem, you are effectively choosing your opponent’s doors (one opened by the host and the other by you after you’ve decided to reverse yourself).
We can see this in the simulated results:
Let’s go through the first 2 rows of the table to make sure you know what is going on. In the first row, we picked door 1. Then the host opens either door 2 or 3 and reveals a goat. We choose whichever of the doors that he did not open and we get a goat as well (because our initial selection was correct and the car was behind door 1). Too bad!
In the second row, we again pick door 1. But this time, the car is behind door 2. Knowing this, the host can only open door 3, otherwise he would reveal the car to us! Finally, we switch to door 2, open it, and get a brand new car!
In our 900,000 simulations, we are now winning the car in 600,000 of them (if we switch), a probability of 2/3 exactly like we reasoned above.
Hope you enjoyed this and found it insightful. I plan to explore more probability and Statistics brain teasers in the days ahead. Cheers and stay healthy everyone!
