Here is a riddle I posted on Mastodon. If you haven’t seen it yet, you may want to think about it before reading on.

You’re in a game show. There are three closed chests, 1, 2, and 3. One of them contains a prize, the other two are empty. You can choose one chest and choose chest 1. “Are you sure?” the game host asks. “You can still switch if you want to!”

At that very moment, chest 2 malfunctions and accidentally opens. It is empty. “Huh,” says the host, genuinely surprised. “Well, my offer stands.”

What should you do?

(A) Switch to chest 3. (B) Stay with chest 1. (C) It doesn’t matter.

Of course this looks a lot like a more famous scenario, the Monty Hall Problem, which goes like this:

You’re in a game show. There are three closed doors, 1, 2, and 3. Behind one of them is a prize, the other two conceal nothing. You can choose one door. But before it is opened, the game show host will open one of the other doors, always one which does not conceal the prize. You are then given the choice of staying with your own door, or switching to the other unopened one.

In the event, you choose door 1. The game host opens door 2, which is empty.

What should you do?

(A) Switch to door 3. (B) Stay with door 1. (C) It doesn’t matter.

The correct answer to the Monty Hall problem is that you should switch. Staying with the door you had gives you a 1/3 chance of winning, switching gives you a 2/3 chance of winning. Famously, however, people refuse to accept this fact. And not just people who don’t know about math! Even professional mathematicians can have a hard time wrapping their head around the situation. This is in itself an interesting fact; interesting enough that there have been quite a lot of papers by psychologists who did empirical research to find out why people have so much trouble in accepting the correct answer.

There are several ways in which the Monty Hall problem can be changed that make it easier for people to see the right solution. One is to increase the number of doors:

You’re in a game show. There are one thousand closed doors, 1, 2, 3, … 1000. Behind one of them is a prize, the others conceal nothing. You can choose one door. But before it is opened, the game show host will open 998 of the other doors, always ones which do not conceal the prize. You are then given the choice of staying with your own door, or switching to the other unopened one.

In the event, you choose door 1. The game host opens all other doors, showing that they are empty, except door 619.

What should you do?

(A) Switch to door 619. (B) Stay with door 1. (C) It doesn’t matter.

Here, it is much easier for people to see that it is quite likely that door 1 and door 619 remain closed for different reasons. Very likely, door 1 remains closed because it was the one I chose, while door 619 remains closed because it is the one that conceals the prize. In fact, this will be the case unless I happened to choose the right door the first time, a 1 in 1000 chance. So switching has a 999 in 1000 win rate.

Another way to make things easier for people (this is due to Bruce D. Burns and Mareike Wieth) is by making it clearer that the unchosen options are in a kind of competition with each other. Here is a riddle about boxers that is structurally equivalent to the original Monty Hall problem:

You’re in a game show. There are three boxers, 1, 2, and 3. One of them is invincible and will always beat the others. The other two are equally good and have a 50/50 chance of winning against each other. At the start of the show, you can choose your champion. Then, the other two boxers will fight each other; and the winner of that bout will face your champion in the final. But before they fight the final, you are allowed to switch your allegiance.

In the event, you choose boxer 1. Boxers 2 and 3 fight each other, and boxer 3 wins. (You didn’t see the fight and don’t know whether it was hard-fought or a push-over.) Now your champion boxer 1 will fight boxer 3; but you if want, you can still switch to boxer 3. What should you do?

(A) Switch to boxer 3. (B) Stay with boxer 1. (C) It doesn’t matter.

Burns and Wieths show that people have an easier time getting the right answer in this case than in the original. Thinking about the elimination of boxer 2 as a fight against boxer 3, makes it easier to see that we now have reason to believe that boxer 3 is more likely to be the best one than boxer 1. (The two approaches can also be combined: imagine 129 boxers. You choose a champion; then the other 128 fight an elimination tournament; and the winner of that tournament will fight your champion in the final. Of course you should switch to the person who has already proven themselves by outcompeting 127 others!)

Finally, here is a version of my own devising that I suspect should also get most everyone to give the right answer.

You’re in a game show. There are three closed doors, 1, 2, and 3. Behind one of them is a prize, the other two conceal nothing. You can choose one door. But before it is opened, the game show host will do the following. (1) If you chose the correct door, the one with the prize behind it, then the game host will open the lowest-numbered door that you didn’t choose. (2) If you chose the wrong door, one without a prize, then the game host will — if it’s not already there — physically push the prize to the highest-numbered door that you didn’t choose. And then he’ll open the lowest-numbered door that you didn’t choose.

You are then given the choice of staying with your own door, or switching to the other unopened one.

In the event, you choose door 1. The game host disappears for a little while behind the doors — either doing nothing or pushing the prize from door 2 to door 3 — and then returns and opens door 2, which is empty.

What should you do?

(A) Switch to door 3. (B) Stay with door 1. (C) It doesn’t matter.

Here, it is clearly the case that you should switch. There’s a 1/3 chance that the prize was already behind door 3, and a 1/3 chance that the host pushed the prize there, so you’ll have a 2/3 chance of winning. This scenario turns the probability-raising act of opening the door into a concrete causal action, making it much easier to see what is going on.

But is this scenario really the same as the original Monty Hall? Of course it is. It doesn’t matter whether there’s any actual pushing and shoving involved. The only thing that matters is that, if the prize is behind one of the unchosen doors, the quiz master opens the one that doesn’t contain the prize. Whether he does that by first pushing the prize and then opening a door, or just by opening a door, doesn’t make a whit of difference.

So, now it’s finally time to go back to the riddle I put on Mastodon. At the moment I am writing this, 308 people have given their answer. 60% chooses to switch to chest 3. 10% chooses to stay with chest 1. And 30% says that it does not matter. The last answers is correct. It does not matter. Switching or staying both give you a 50% chance to win.

I don’t think this answer will just be accepted by the 70% who chose something else, so let’s delve into it a little more! One reason to not accept it would be the following argument:

“In the Monty Hall case and in your case, the exact same thing is happening. I choose chest 1, then chest 2 is opened and revealed to be empty, then I have to make my decision. The only difference is that on the original version the chest is opened deliberately, and in your version it happens by accident. But surely the fact that something was done deliberately has no influence on where the prize is! So the answers in the two scenarios must be identical.”

It is true that the mental act of deliberation is irrelevant. But the two scenarios are nevertheless not identical. Let’s start by asking ourselves why the unopened chests remain closed. In the Monty Hall case, this is why:

• With probability 1/3 you have chosen the right door. In that case, the chosen door remain closed because you have chosen it and also because it contains the prize. The other door remains closed because the game host has randomly chosen to leave it closed.
• With probability 2/3 you have chosen the wrong door. In that case, the chosen door remains closed because you chose it, and the other door remains closed because the prize is behind it.

However, in my riddle, the reason that chests 1 and 3 remain closed has nothing do with which chest was chosen or which chest contains the prize. Chest 2 simply malfunctioned and opened. It would also have opened if I had chosen it. It would also have opened if it had contained the prize. Chest 1 and chest 3 remain unopened simply cause they did not malfunction — and that doesn’t generate an asymmetry between them.

Here’s another way to think about this. In the Monty Hall case, choosing a door has an influence on which door will be opened. In fact, 2/3 of the time, your choice forces the game host to open a particular door. If I choose door 1, and the prize is behind door 3, the host has to open door 2. If I choose door 1, and the prize is behind door 2, the host has to open door 3. Only if I choose door 1 and the prize is behind that very door, does the host have a choice. What this means is that the step of choosing a door cannot be skipped. If you don’t make a choice, and the host just beings the game by opening a door not containing the prize, no asymmetry is generated between the two remaining doors. It is because you choose a door and thereby protect it from being opened, that its being unopened at the end has a different status from the other — unprotected — door’s being unopened.

In my riddle, the choosing is completely irrelevant. Chest 2 malfunctions and opens. It could have done so twenty seconds earlier, before I made my choice. Obviously, I would then have no reason to prefer chest 1 to chest 3 or chest 3 to chest 1. And the mere fact that I made a choice before the opening doesn’t change that, because the choice had no effect on the opening. (Note that if I had chosen chest 2, then I would have had a very good reason to switch after it had accidentally opened and revealed itself to be empty! But chest 1 and chest 3 would still have been equally probable.)

Here’s yet another, perhaps more mathematical way to think about it. Let’s consider all the scenarios. For simplicity, assume that you choose door 1. Then there are six possible scenarios, because (1) there are three doors that could conceal the prize, and (2) the host could have decided to open either the lowest-number or the highest-number door that it is possible for him to open. In the Monty Hall case, the six scenarios play out as follows:

1. Prize behind door 1, host prefers to open lower door: Host opens door 2. You win by staying.
2. Prize behind door 1, host prefers to open higher door: Host opens door 3. You win by staying.
3. Prize behind door 2, host prefers to open lower door: Host opens door 3. You win by switching.
4. Prize behind door 2, host prefers to open higher door: Host opens door 3. You win by switching.
5. Prize behind door 3, host prefers to open lower door: Host opens door 2. You win by switching.
6. Prize behind door 3, host prefers to open higher door: Host opens door 2. You win by switching.

Once the host opens door 2, you know you are in one of the three equiprobable scenarios 2, 5, or 6. In two of the three you win by switching.

In my riddle, the six scenarios play out as follows:

1. Prize in chest 1, host prefers to open lower chest: Chest 2 malfunctions, is revealed to be empty. You win by staying.
2. Prize in chest 1, host prefers to open higher chest: Chest 2 malfunctions, is revealed to be empty. You win by staying.
3. Prize in chest 2, host prefers to open lower chest: Chest 2 malfunctions, is revealed to be full.
4. Prize in chest 2, host prefers to open higher chest: Chest 2 malfunctions, is revealed to be full.
5. Prize in chest 3, host prefers to open lower chest: Chest 2 malfunctions, is revealed to be empty. You win by switching.
6. Prize in chest 3, host prefers to open higher chest: Chest 2 malfunctions, is revealed to be empty. You win by switching.

Once the chest is revealed to be empty, you know you are in one of the four equiprobable scenarios 1, 2, 5, or 6. It doesn’t matter whether you stay or switch.

We can also get more clarity by looking what the other version of Monty Hall look like if you add the fact of accidentality. I think boxers is especially clear.

You’re in a game show. There are three boxers, 1, 2, and 3. One of them is invincible and will always beat the others. The other two are equally good and have a 50/50 chance of winning against each other. At the start of the show, you can choose your champion. Then, the other two boxers will fight each other; and the winner of that bout will face your champion in the final. But before they fight the final, you are allowed to switch your allegiance.

In the event, you choose boxer 1. The boxer 2 and 3 are about to fight, but then boxer 2 trips over the steps to the stage and breaks a leg. The game host shrugs. “Oh well, he was not the invincible one anyway. Let’s move on to the finals.” Now your champion boxer 1 will fight boxer 3; but you if want, you can still switch to boxer 3. What should you do?

(A) Switch to boxer 3. (B) Stay with boxer 1. (C) It doesn’t matter.

If boxer 3 had defeated boxer 2 in combat, it would have given us a defeasible reason to believe he might be the invincible one; a reason that we don’t have for boxer 1. But in this case, when boxer 2 tripped and no fight took place, we have no such reason. The situation is symmetrical between boxers 1 and 3, and it doesn’t matter whether you switch or not. (I’m assuming that non-invincible boxers do not, in general, trip and break legs. This was really a freak accident.)

On Mastodon, I’m seeing some people actually confusing themselves with the increased number of doors version, failing to see that the increased number of chests version of my riddle is very different from the increased number of doors version of the original Monty Hall. Here’s what my riddle would look like:

You’re in a game show. There are one thousand closed chests, 1, 2, 3, …. One of them contains a prize, the others are empty. You can choose one chest and choose chest 1. “Are you sure?” the game host asks. “You can still switch if you want to!”

At that very moment, 998 chests malfunction and accidentally open. They are empty. Only chest 1 and chest 617 are still closed. “Huh,” says the host, genuinely surprised. “Well, my offer stands.”

What should you do?

(A) Switch to chest 617. (B) Stay with chest 1. (C) It doesn’t matter.

What we should realise is that something extremely unlikely has just happened. In the Monty Hall case, it is 100% certain that all of the 998 opened boxes will be empty. That’s how the rules of the game work. In the present scenario, however, there was only a 0.2% chance of all the 998 opened boxes being empty. It was vastly more likely that the prize would have been revealed. So there’s clearly no equivalence between the many doors and the many chests example.

Furthermore, in the Monty Hall version, there is a 99.9% chance that chest 617 is closed because it contains the prize. Conversely, there is a 99.9% chance that chest 1 is closed because I chose it. That’s why you should switch to chest 617. But in the current version, that’s not true. Both chests are closed because they happened not to malfunction. Since by hypothesis this has nothing to do with being chosen and nothing to do with containing the prize, no anti-correlation between being chosen and containing the prize has been established. There is perfect symmetry, and switching doesn’t matter.

It may be helpful to illustrate this in a somewhat different guise. Suppose I have one thousand people. Suppose furthermore that one of these thousand people was born on February 29, and that one of these thousand people can play the Nocturnes of Chopin. And suppose, finally, that these two traits are statistically unrelated.

Scenario 1: You are told to select two people from these one thousand, making sure that you select both the February 29 person and the Nocturnes person. “On the off chance that they happen to be the same person,” you are told, “just select a random second person.” Now you show these two people to me, and tell me that the one on the left can play the Nocturnes. What should I believe about their birthdays? Obviously, it’s almost certain that the one on the right is the one who was born on February 29. It is very likely that you chose the one on the right because they have that birthday.

Scenario 2: You are told to select two people from these one thousand, purely at random. After having done so, you find out that one of them is born on February 29. You also find out that one of them can play the Nocturnes. You tell me all of this. “Who can play the Nocturnes?” I ask. You point at the person on the left. What should I believe about their birthdays? Well, there’s no correlation between being able to play the Nocturnes and being born on the 29th of February, so learning that the left person can play the Nocturnes gives me no information about their birthday. Things remain at 50/50.

If that seems strange, think of it this way. If you randomly select two numbers from 00 to 99, and you then find out that exactly one of them ends in a 2 and exactly one of them begins with a 2, what is the chance that you have selected the number 22, which combines these properties? The answer is: exactly 50%. There are 81 ways to select 22 and a number with no 2 in it. There are also 81 ways to select one number that begins but does not end with a 2 and one number that ends but does not begin with a 2. Learning that your random selection of two numbers contains a number ending with a 2 and a number beginning with a 2 makes it quite likely that you have selected a number combining these properties. Learning that your random selection of two people contains a person with one rare property and a person with another rare property makes it quite likely that one person has both these properties.

Given the history of the Monty Hall problem, perhaps this blog post will not convince people. But if it doesn’t, it’s not because it wasn’t long enough! *wipes sweat from his forehead and keyboard*