[4star3star]

The way the problem makes sense to me is this.

Doors: Goat Goat Car

You pick a door. Monty shows you a Goat. You switch or stay.

Monty will never show you the Car before offering a switch. He always shows you a Goat. It doesn't matter which Goat he shows you - it's just "not the Car".

If your first choice is a Goat, switching will win you the Car. If your first choice is a Car, switching will win you a Goat. You have a 2/3 chance of picking a Goat, so, effectively, you want to pick a Goat so that you switch to the Car.

[tacitusarc]

For sufficiently analytical folks that works, but for lay people it tends to still be confusing.

The best way I’ve heard it explained to help people get it through intuition is by changing the number of doors and goats. Say there are 100 doors, and they all have goats except one, which has a car. You pick door 1. Monty then proceeds to open doors 2 through 48, skips door 49, and then opens the remaining doors. After all that, he stops and asks you, would you like to switch?

[thaumasiotes]

There's a better way to think about it.

1. You pick a door.

2. You get the offer "Do you want to keep that door, or choose both [all] of the other doors? In either case, you'll keep anything that isn't a goat."

3. Nobody opens any doors.

Should you keep your one door, or switch to the two doors?

[tomrod]

This is the right way to think about it. Very clear and concise, thank you.

[s1artibartfast]

I always feel like there is something fundamental missing from the examination of Monty Hall problems.

I think it has to do with the difference between "probable outcome in reality" and "probably outcome based on personally known information".

Lets say when you get down to doors #1 and #49, Monty brings in someone new, with no information and says pick a door. For that new person, standing right next to you, doors #1 and #49 have a 50-50% chance, while for you they are a 2% vs 98% chance.

How can door #1 simultaneously have a 2% chance for you and a 50% chance for Bob? The answer is that the chance is not a single fixed property of the door itself- which is hard to wrap ones head around.

And for that matter, Monty Hall himself knows one of the doors is 100% and the other is 0%.

[Natsu]

There is something missing: regular stats don't differentiate between doing things and observing things and these two are not at all the same. If I have a digital thermometer and I observe it to show a high temperature, then I will note an association between that and feeling warm. But if I merely set the thermometer gauge to a high value artificially, it's not going to make me feel any warmer.

This ambiguity is resolved by something called do calculus - https://arxiv.org/pdf/1305.5506.pdf

[s1artibartfast]

I think it is more fundamental than that, and not even mathematical. I think the issue is that people conflate or blur the difference between reality and their models of reality.

Your personal, information limited calculation of the chance a car is behind door #1 has no impact on if there is a car behind door #1. Reality is binary and constant. There was always a car there, or there always wasn't.

Most people correctly intuit that of course the real probability that the car is behind door #1 cant change with reveled information. It isn't a quantum car. They just get caught up on the fact that predictive chance is a attribute of the model, not the real door.

[Natsu]

I agree that the map is not the territory, but there is a better model here that captures the difference.

The car isn't moving, as you say, but that intervention by the host lets us trade one door for both of the other doors.

[evouga]

The situation is now counterintuitive in the other direction: if Monty Hall had opened those 48 doors at random and they just happened to not contain the car, then there is no advantage to switching, though many people would insist otherwise.

[Sohcahtoa82]

> if Monty Hall had opened those 48 doors at random

The fact that Monty Hall opens the doors deterministically (not randomly) is KEY.

In the original problem, Monty ALWAYS opens a door with a goat. In using 50 doors, Monty would ALWAYS open doors containing goats, and not the car. It's not random.

Knowing it's not random, it should be very intuitive.

[joshuamorton]

But the doors weren't opened at random. You know he won't open the car, because that's part of the rules of the game.

Let's demonstrate with a slightly different construction: You're no longer playing with monty, but with a demon. This demon wants you to lose, but also picked a very bad game for themselves. You pick a door, then the demon opens all-but-one of the remaining doors. Then, you can pick any door, open or closed, and you get what's in it.

If the demon opens doors at random, nearly all the time (with 100 doors) you'll see the car and be able to pick it directly. In this situation, switching between the closed doors doesn't really matter, but you'll usually know exactly which door to pick, because you can see the car.

So instead, the demon only opens doors that don't have a vehicle behind them. You only ever see goats. At this point, he's not opening doors at random. If he were, you'd see the car 98% of the time, but you never do. At this point, since he's using additional information, it is in your best interest to switch.

[matsemann]

I've never been happy with that explanation. I don't get why the host would not just open a single door, that's what the host does in the other scenario to me.

[cstrahan]

> that's what the host does in the other scenario to me.

Is it, though? It seems apparent that, after the first guess, the host opens all but the last two doors, which just so happens to be 1 door.

To check the math:

Start with the $NUM_DOORS open doors. Now open all but the last two. So that’s $NUM_DOORS-2, which is 3-2, which equals 1 open door.

[matsemann]

My interpretation of the host opening a single door and asking if you want to switch is equally valid when you expand it to 100 doors.

[joshuamorton]

Sure, it could be that the host only opens one door, or it could be that he opens all but one door. In every case, however, it is better to switch. The all-but-one example is hyperbolic but still follows precisely the same mathematical rules. Your interpretation is valid, but so is the all-but-one example, and they all lead to the same result, it's just more obvious when you open nearly all the doors.

[aplusbi]

The best way that I've thought about it is like this:

You pick a door, then Monty let's you switch to the two remaining doors and if the car is behind either of them you win.

Obviously choosing the two remaining doors is better.

The trick is to realize that Monty showing you the contents of one door and letting you choose the other one is identical to Monty letting you choose both the remaining doors.

[LoganDark]

This is the best explanation I've seen for the problem so far. Thank you

[lukasb]

Even simpler - you pick, knowing nothing, so there's a 2/3 chance you're wrong.

If you're wrong, Monty points to toward the right door.

So you should switch.