[jashmenn]

Here's what worked for me:

Imagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)?

People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is.

(I learned this idea from Better Explained: https://betterexplained.com/articles/understanding-the-monty...)

[filoeleven]

I was lucky enough to get this explanation from my high school physics teacher, who first presented the classic Monty Hall problem and then illustrated the changing of the odds by substituting all of the lockers in our school for the three doors. Switching gives a clear advantage.

The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn.

We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor.

The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.”

“Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.”

The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about.

“...I couldn’t really tell the difference,” said the student.

“Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”

[thehappypm]

I don’t think this helps me understand it.

In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.

[cestith]

Try thinking of it this way. It may or may not be any more useful. I've seen different people come to understand the problem from different examples.

    1. Observe that 3/3 = 1. Pedantic, yes, but good for frame of mind here.
    2. Pick one of three doors. (1/3 odds)
    3. Gain information that one of the three doors is a loser.
    4. Note your odds on choosing the original door correctly are still 1/3.
    5. Note that if you change doors, there are still 2/3 doors there to choose.
    5. Note you're not going to switch to the known loser door, so if you change doors you know 100% which of the other 2/3 of doors to choose.
  

The intuition usually is that you're down to two doors after the loser door is opened, but that's not the case. There are still three doors. The host has just told you that if you trade doors, you know which door to trade for. So trade for it.

Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.

[thehappypm]

Step 4, kind of a mystery. Why are my odds still the same even though the situation is different?

[inetknght]

You choosing a door in the first place gave you 1-in-3 chance. But your 1-in-3 choice is deducted from the 3-of-3 choices that Monty could have had, so Monty only had a 2-in-3 choice: there's a 2/3 chance that the car is in the pool of doors from which Monty could select. Monty has a 100% chance of choosing a door without a car. Therefore, you inherit the 2-in-3 chances if you change your selection.

The first door will have the car 1/3 of the time. The second door's chances had been expanded to the remaining 2/3 percent thanks to Monty always choosing the last 1/3 door which is guaranteed to not have a car.

[Swenrekcah]

Because it isn't different really. It is always a goat door that is opened, so you don't gain any information about your door by the opening of the goat door.

I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.

[SamBam]

Because it really centers on the initial premise: Monty will always open a goat door after your choice, no matter what.

So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2?

Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?

[cestith]

What you knew about the door you initially picked hasn't changed at all. What you now know about the other doors has. By giving you information about 1/2 of the other 2/3 doors, Monty has given you an extra 1/3 chance if you pick among those.

[bartc]

Assume there are seven billion people on the planet. One of them knows the location of a specific hidden treasure. I know who it is and I ask you to guess who it is, but you have no possible way of knowing or even getting a hint about it.

You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in.

At this point, there are only two possibilities:

1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret.

2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is

[ragnese]

You don't need to know the exact odds to understand that it's higher. I think that's the main takeaway of making it a 1,000 door problem. It makes it intuitive that the correct solution is to switch. The exact probability doesn't matter.

[delecti]

IMO what helps is to imagine slightly changing the order.

First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.

[klmadfejno]

probability of winning if you switch is 1 - (1 / n) aka (n-1)/n.

probability of winning if you don't is 1 / n.

By switching, you are simply betting that your original guess of 1/n was wrong.

[Closi]

Try watching this video below, which explains the 1/100 analogy with a real world example:

https://youtu.be/GPoPSNxV1D4?t=365

I've timestamped the relevant bit - but you should watch the full thing from the start, it's very entertaining :)

[notsuoh]

Wow. I "understood" the reasoning behind the original and knew that was the right answer, but your anecdote just made it totally click. Of course if I choose one random locker there is a 1/1000 chance of getting a prize, and if Monty opens 998 other empty lockers, of course I should switch. That would make switching be the correct choice in 999/1000 times.

[lordnacho]

For someone who doesn't get it, the problem is then whether the correct extension is all the doors being opened. With 3 doors it's the same.

For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.

[NE2z2T9qi]

This is the most intuitive explanation by far. I'd even say 1 million doors to really drive the point home.

You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.

[Green_man]

When I first heard about the problem, I struggled with the seemingly 50/50 chances. either you picked the right door and now switch and lose, or you picked a wrong door and switch and win. Switching seemed a zero sum game.

The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.

[snapetom]

That's the explanation that helped me, and led me to realize something about this problem. The problem asks what is the best strategy. It doesn't ask what should you do at that moment. At that moment implies you should process the information available to you right then and there. The information available to you at that time - two doors, ignores prior information, which I think is the counterintuitive aspect that trips people up.

[kmm]

That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem.

A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".

Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.

[tzs]

You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous.

In the case of the clumsy Monty of your example, it goes like this:

1. There is a 1/1000 chance door 429 has the car.

2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000.

2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds.

3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch.

3a. Your door has the car, which happens 1/1000 games.

3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games.

In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching.

In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this:

1. There is a 1/1000 chance your door, 429, has the car.

2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours.

2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car.

3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door.

3a. You get there via 2a in 1/1000 games.

3b. You get there via 2b in 999/1000 games.

If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.

[Sohcahtoa82]

Your proposal is changing the scenario, though.

In the original problem, Monty always opens a door to reveal a goat. This strongly implies that he knows which doors have goats.

Your alternative is different. If the doors opened are truly selected randomly, then the odds are high that he would reveal a car, especially in the 1000-door version. What are the odds that Monty could choose 998 doors randomly out of 1000 and NOT pick the one with the car?

If Monty doesn't know where the car is, then you're arguing with a completely different version of the game that could result in him opening the door with the car, so the strategy is going to be different.

[kmm]

Changing the scenario is the exactly the point of why I am unsure increasing the number of doors gives true intuition. In the original problem with the original scenario, people intuitively think switching doesn't matter, when it provably does. With 1000 doors and the Monty Fall scenario, again the intuition is wrong. So are we gaining true intuition for the Monty Hall scenario by expanding the setup, or is it just some subconscious Bayesian thinking accidentally steering us toward the right answer.