The Monty Hall problem is notorious for screwing with the heads of mathematicians, amateur and famous alike. But once you get your head around the Monty Hall problem, you then need to tackle the far less famous but arguably far more confusing Monty Fall problem. In this post I will introduce you to both problems, and hopefully persuade you that both make complete sense when thought about in the right way.
⚠ Warning!
There's no alt text on the images in this post. This is mostly because I don't think I can write an alt text for them that accurately portrays the images. I have tried to include enough information in the text to convey what they are showing, however. But if you're better at writing alt text than I am and can write something that makes sense, please let me know.
I wasn't kidding when I said this problem is notorious for screwing with the heads of famous mathematicians, by the way. Paul Erdős himself supposedly said
No, that is impossible, it should make no difference.
when told the answer. And if a man known for ... well ... so many things that they have their own Wikipedia page can't comprehend the answer, then what hope do we mere mortals have?!
Well for one, we have the internet spraying us with so many explanations that we're bound to see one we understand.
But more importantly you're much smarter than Paul Erdős even the best mathematicians don't know everything.
In the Monty Hall problem, famous TV host Monty Hall is hosting a game show. In this game there are three doors, hidden behind one door is a car and behind the other two are goats. Monty gives the contestant a choice of what door they would like to choose. He then reveals the location of one of the goats[1] , and asks the contestant if they would like to switch which door they have chosen. Once the contestant has made their second choice the remaining doors are opened. If the contestant has settled on the door with the car, they win!
The crux of the problem, then, is whether the contestant should switch or not after the first door is revealed. If this is the first time you've seen this problem, I urge you to take a moment to consider the answer before moving on.
If you answered "surely, it shouldn't matter", then congratulations! You've joined the ranks of such esteemed mathematicians as Paul Erdős! Unfortunately ... you're wrong. If the contestant decides to switch, they actually have a \(\frac{2}{3}\) chance of winning the car.
Below is a tree showing the possible things that can happen during the game. In the final layer, I have colour-coded the choice of whether to switch doors or stay with the door initially picked. The left-hand of each pair of branches is coloured purple and represents the choice to stay with the door that was initially picked. The right-hand of each pair of branches is coloured orange and represents the choice to switch doors.
The first layer represents the moment after the contestant has a door, but before Monty has revealed a goat. At this point the contestant has a \(\frac{1}{3}\) chance of picking the door hiding the car. The second layer represents the moment after Monty has revealed a goat behind one of the other doors.
This is an important point, and I think is what leads most people to get confused. When first thinking about this problem, I think most people think of the revealing stage as random. But it isn't, there is a 100% chance that Monty will not reveal the car. This is what breaks the symmetry, and causes switching to favour the car.
Finally we have the third layer, showing the final results. If the contestant decided not to switch, they end up with the car in only one of the three outcomes. If they do switch, however, they will get the car in two of the three outcomes.
To put it another way, when the contestant first picks a door they have a \(\frac{2}{3}\) chance of getting a goat. Monty then manipulates the problem such that if they chose the door hiding a goat \(\left(\frac{2}{3}\right)\) then switching will always get them a car. That manipulation doesn't affect the initial probabilities, so switching must be the beneficial choice.
⚙ Note
If you're particularly observant, you might be wondering why there's only one branch coming from the node where the contestant picked the door with the car when first choosing even though at that point Monty could open either of the doors. The reason for this is subtle. When they pick a door that has a goat behind, Monty is forced to open a certain door, but if they pick the door with the car Monty could open either of the others. Since the door with the car is picked as often as the other two, if I included both branches then they would be taken half as often as the other branches in that layer. This breaks the implicit assumption that all the final states occur equally as often. Instead what I've done is sum up those two branches such that all the outcomes in all the layers have the same probability of occurring.
If that didn't convince you, we can try examining the one-hundred-billion-door Monty Hall problem. I personally never found this as intuitive and I can't think of a way to use it for the Monty Fall problem, but apparently it's a helpful way to think about the Monty Hall problem for a lot of other people.
In this version of the problem, there are 99,999,999,999 doors with goats behind and one door with a car behind. When the contestant first picks a door, they are going to pick a door with a goat behind. No ifs, no buts, they ain't getting that car on the first try baby[2] .
When Monty comes to reveal the goats, he will reveal every single one of the other 99,999,999,998 goats. This means that the contestant will have one door to switch to, and that door will be hiding the car.
Unfortunately, if neither of these explanations was helpful to you then you might just need to sit down and apply Bayes' theorem yourself. But make sure you do understand what's happening here, because now we're switching the channel[3] .
In the Monty Fall problem, infamous TV host Monty Fall is hosting a game show. The setup to this game is the same as that of the Monty Hall problem, but Mr Fall is clumsy. When Monty F goes to reveal a goat there's a 50% chance that he will trip, fly across the studio, and accidentally reveal a car. Of course not wanting to embarrass poor F-Dog the producers of the show will continue on as if this were planned, even though a loss is now guaranteed.
So the question is this: After getting the contestant to pick a door, Monty trips and by a stroke of good luck reveals a goat. Should the contestant still switch? Again I want you to really think about the answer here before reading on.
If you answered "surely, it shouldn't matter", then congratulations! You're right!
Wait...
If in both situations there was a goat revealed, why does it matter whether the goat was revealed intentionally or not? What gives?!
Let's go back to the start and look at all possible outcomes, not just ones in which a goat was revealed.
The first thing you might notice is that if the initially chosen door hides a car, I've now included both branches coming from that whereas before I hadn't. As I mentioned in my note, one of the implicit assumptions required for these tree diagrams to make sense is that in each layer all the outcomes must have the same probability of occurring. Now that Monty has an equal chance of revealing a car or a goat (when there is a car to reveal) we would no longer be hitting the two different goat branches (when there is no car to reveal) half as often and so they must both be included.
The most important thing to notice, however, is that now there are branches where switching or staying will both net the contestant a goat. This means that whether they switch or stay, the probability of getting a car is always \(\frac{1}{3}\).
But what if we know Monty's revealed a goat? The key insight here is to realise that this doesn't change any of the probabilities, it merely tells us that we are not in some of the branches[4] . Those branches are the ones where a car has been revealed, and I've coloured them red in the diagram below.
We can see then that, even once a goat has been revealed, the probability of getting a car does not depend on whether the contestant switches or stays. Although once a goat has been revealed, the chance of getting a car increases to \(\frac{1}{2}\).
The Monty Hall problem is complex and breaks the brains of most people when they first encounter it. The Monty Fall problem is even more complex and still breaks my brain, even though I understand it perfectly well. That is to say don't worry if you still don't get it, I hope that at the very least I have nudged you along on the path to getting it.
At some point I may do another blog post where I go through the maths and just chuck Bayes theorem at it, because at least then I can convince you that the maths makes sense. Until then, however, good luck!