[dfxm12]

The Monty Hall problem is especially unintuitive if you've ever watched Let's Make a Deal, since the problem set up is oh so close to, but not exactly, the set up of the Big Deal in the show. It's too easy to conflate the rules of the show with the math problem, which will lead to confusion.

I think seeing the results of a simulation also elucidates the set up of the math problem vs reading a proof.

[cortesoft]

Yeah, I feel like the Monty Hall confusion goes away if you are explicit about the rules:

"Hall will always open one of the two non-chosen doors and will never reveal the prize"

I think most people who don't understand the problem miss that critical detail.

[bigstrat2003]

No, I don't think that detail makes it any easier. I know that but I still really can't accept the correctness of the Monty Hall strategy (I have to basically just take it on faith and stop trying to understand it). I was trying to put my finger on why, and I think it's this.

After Monty eliminates one of the three doors, then the prize is behind one of the two. If someone were to come in this point, with no prior knowledge whatsoever, their chance of picking the correct door at random is 1/2. And that is still true even if they pick the door which our contestant is being asked whether or not to switch from! This is a real mind fuck to try to accept, that the same state of what's behind each door leads to different odds of making a correct random choice, depending on when you make the choice.

I honestly don't think I'll ever be able to "get" the Monty Hall strategy. I think I get why it works (choosing to switch means you're going from a 1/3 probability to 1/2), but it makes no sense at all. It seems like even if you choose to stay on the same door, your probability is 1/2 (the same as if Joe came in off the street and chose the same door as you). Like I said, I just have to take it on faith.

[benchaney]

The best probability estimate you can make is constrained by the information you have available. The new person showing up has less information than the existing constant, so it makes sense that their best estimate would be less precise. Similarly, if someone with x-ray vision walked up in the middle of the game, they could pick the car 100% of the time, because they have access to more information than either of the existing contestants.

Your last paragraph isn't correct though, By switching you go from a 1/3 probability to a 2/3 probability. Based on the information the original contestant has, switching gets the car 2/3 of the time.

[bigstrat2003]

I don't see how a new contestant has less information, though? They know that one of the two doors contains the prize, which is all the previous contestant knows either.

[cortesoft]

The crucial bit of information that the new contestant doesn't have is that there was a door that was ineligible to be eliminated (the door chosen by the original contestant).

If the game had different rules, it would work like you are imagining. Specifically, if Monty randomly eliminated one of the two doors, meaning there was a chance for Monty to reveal the prize instead of a goat. If Monty has the chance to eliminate the prize before giving the contestant a chance to switch, then switching does not give you an advantage.

[benchaney]

When one door was opened it revealed information about the other two doors.

[bigstrat2003]

But it didn't. Before, we knew that one of those two doors could contain either a prize or a goat. After, we know the same exact thing. No information was gained there.

[dfxm12]

Maybe this will help understand it intuitively. You have a choice between doors 1 2 3. You pick door 1. You know the odds of the car being in door 1 is 1/3. The odds of the car being in door 2 or door 3 are 2/3.

Monty opens door 3, showing a zonk. You knew there was a 2/3 chance of the car being in door 2 or 3, but now you know there's a 2/3 chance of the car being in door 2 (since you know it is not in door 3).

All this didn't change anything you know about door 1. It has the same 1/3 chance it started with. Probability is all about what you know in the moment.

The math involves understanding the rules, that Monty will never open the door you picked and will never open the door with the car behind it. This is why one can't look above and say "well, there is a 1/2 chance of the car being behind door 1 after door 3 was opened and there wasn't a car there". This would only be true mathematically if the door Monty opened was random, but we know the door Monty picks isn't random. In fact, the pool of doors that could be opened depends on your initial pick. Monty was never going to open door 1 (the door that you picked), even if it was a zonk & Monty was never going to open the door with the car, therefore one can't make that assertion.

[cortesoft]

I have fun trying to explain this problem. Let me see if I can give an explanation that will help you.

So lets say you have just picked a door in the beginning. You know you have a 1/3 chance of being right.

If I then tell you, "I will give you two options... you can either bet you are right, or bet that you are wrong"

You would obviously choose to bet you are wrong, correct? Because you know you only have a 1/3 chance of being right with your guess, which means you have a 2/3 chance of being wrong. The smart bet is that your original guess was wrong.

This is actually what is happening in the game if you think about it. You pick a door and it has 1/3 chance of being the right one; since we know Monty is only going to ever reveal a goat and never the prize, we don't even NEED Monty to reveal the door at this point - we know he is going to reveal a goat, no matter what. We don't even have to wait to see which door he reveals, since that isn't going to give us more information (it is going to be a goat, no matter what). So when he asks you if you want to switch doors, he isn't asking you to switch to ONE of the other two doors, he is asking if you want to switch to having BOTH other doors as your choice. Whether he reveals the goat before or after you choose to switch doesn't matter, because you know it will always be a goat.

If that is still not clear, lets just write out all the options:

There are three doors, A B C. One has a prize, the other two have goats. Let see what happens with your two options (switch or dont switch).

In our first example, you pick door A and you are going to switch.

1/3 of the time the prize is behind door A. If the prize is behind door A, and you switch, you lose. This is 1/3 of the time, and you lose for switching.

1/3 of the time the prize is behind door B. You picked door A, so Monty reveals door C. You switch to the remaining door (B) and you win.

1/3 of the time the prize is behind door C. You picked door A, so Monty reveals door B. You switch to the remaining door (C) and you win.

Add up all those choices, and 2 out of the 3 times you win.

Now lets imagine that we DON'T switch.

1/3 of the time the prize is behind door A. Monty reveals one of the other doors, but you don't switch. You win.

1/3 of the time the prize is behind door B. Monty reveals door C, but you don't switch from A. You lose.

1/3 of the time the prize is behind door C. Monty reveals door B, but you don't switch. You lose.

So in this not switching world, you win 1/3 of the time.

In summary, switching wins 2/3rds, not switching wins 1/3.

Does that help at all?

[bigstrat2003]

That does help - if nothing else, adding up the outcomes manually helps to demonstrate it in a way that's not really easy to grasp on first glance. It still is kind of a spooky result which makes no sense, but at least I feel a little better about the correctness.