In this post I won't go into that detail, I'm going to assume that you're now familiar with both versions and you know that the Monty version wins 2/3 of the time if you switch, while the Opponent version is still 50/50, even if you don't understand why. In the comments of that original post I described an alternate way to view the problem that, for me at least, makes this whole thing really obvious. I haven't seen it described that way elsewhere, so I want to elaborate in the hopes that this will help some people intuitively see why Monty Hall works, as well as why that isn't initially obvious.

I want to show you the smoke and mirrors at work.

I'm going to describe a number of scenarios. It will not be obvious at first, but every one of these scenarios will be equivalent to the standard Monty Hall problem. I will begin with a version in which it is (probably) super-obvious to you that switching is the right thing to do, and then move gently towards the standard Monty Hall description where things are less obvious but still identical when you look behind the curtain. In all versions, the problem is the same as the standard Monty Hall except in the ways described.

It is my hope that you will see that these problems are all the same, and that they only differ in how well they hide the advantage you receive by switching.

__Version 1. The super-obvious version:__

You pick Door 1, but it is not opened.

Monty gives you the option to instead choose to open both Doors 2 and 3, and if the prize is behind either one of them, you win.

Should you switch?

This one is obvious, or should be. You get to win if the prize is behind either Door 2 or Door 3, giving you a 2/3 chance to win if you switch. Critical concepts that will be repeated in all other versions are:

- This is a 2-for-1 guess, that is why you get a 2/3 chance to win if you switch.

- Another way to look at that is that you've been given the opportunity to make 1 choice that actually reveals a pair of doors.

All future versions will contain these same features; that is why they work.

__Version 2:__

You pick Door 1, but it is not opened.

Monty tells you that, if the prize is behind Door 2, he is going to move it to Door 3. If the prize is behind Door 3, he is going to leave it there.

You are then given the chance to open Door 3 instead of Door 1.

Should you switch?

Yes, absolutely. Door 3 now wins if it had the prize originally, or if the prize was originally behind Door 2. Only if the prize was originally behind Door 1 does switching lose. You have a 2/3 chance, and again we have been given a single choice that essentially reveals a pair of doors.

__Version 3__:

You pick Door 1, but it is not opened.

Monty tells you that you can pick Doors 2 and 3 as a pair, and if the prize is behind either one of them, he will open that door, and you will win.

Should you switch?

Yes. If the prize is behind either door you win, giving you a 2/3 chance by switching. Again the doors are treated as a pair.

__Version 4:__

You pick Door 1, but it is not opened.

Monty tells you that you can pick Doors 2 and 3 as a pair. If you do so and the prize is behind one of them, he will open the other one (the loser) and then let you actually open the winner.

Should you switch?

Yes. If the prize is behind either door you win, giving you a 2/3 chance by switching. Again the doors are treated as a pair. Notice how close we are to the description of the original problem.

__Version 5:__

You pick Door 1, but it is not opened.

Monty tells you that he is going to open a losing door from the pair of doors made by Doors 2 and 3. He will then give you the option to switch to the unopened door.

Should you switch?

Yes. Same as before, these doors are treated as a pair. 2/3 chance to win by switching. Notice how this is identical to the previous version.

__Version 6, the original problem re-framed to not specify which door Monty opened__:

You pick Door 1, but it is not opened.

Monty tells you that he is going open a losing door, but he will never open the door you've already picked. He will then give you the option to switch to the remaining door.

Should you switch?

Yes. 2/3 chance to win by switching. Though he hasn't said so explicitly, Monty is treating Doors 2 and 3 as a pair because he isn't going to open the door you already picked. His choice therefore is out of Doors 2 & 3, and by using his knowledge to eliminate a sure loser from that pair, he makes it so that you have 1 choice that wins if either of these doors has the prize. It is more obvious that they are a pair here, than in the standard Monty version, because we have not specified which of the two is a loser, and which remains.

__Version 7, the original problem:__

You pick Door 1, but it is not opened.

Monty tells you that he is going to open a losing door for you.

Monty opens Door 3, revealing a goat.

You are given the option to switch to Door 2.

Should you switch?

Yes. You have a 2/3 chance to win by switching.

**What makes this trick your mind is that you are likely to frame the problem as "Should I pick Door 1 or Door 2?" when the correct framing is "Should I pick Door 1 or the door that remains from the Door 2/Door 3 pair?"**(which in this particular case happens to be Door 2) As in all other versions, Doors 2 & 3 are treated as a pair. You get to win if the prize was behind either of them. You get a 2-for-1 guess, but if you're thinking of Door 2 as itself and not as representative of the pair, then your brain is likely to experience the disbelief that makes this a veridical paradox. Your choice is not between two doors, but between a single door and a pair of doors. When given the opportunity to switch to a pair of doors, your brain gets it. When given the opportunity to switch to a door that

*represents*a pair of doors, your brain can easily miss the hidden value there.

The Monty version (knowledge based choice) isn't the same as the Opponent version (random choice) because in the Opponent version the opponent is opening a door that might be a winner. If the prize is revealed, you lose. Monty never reveals the prize, he reveals information that gives you the option to make a 2-for-1 guess. In this way, Monty's action is guaranteed to help you, whereas the opponent will cause you to immediately lose 1/3 of the time (the times when he guesses correctly.) Nothing that the opponent does creates a situation where you can win if either of 2 doors has the prize. The opponent does not create a 2-for-1 guess for you.

Does this help make the problem intuitive for anyone else?