Monty Hall Problem Intuition I was thinking about the Monty Hall problem and I thought of a possible intuitive explanation: 


You choose a door.
Monty gives you the option of sticking with your original choice or instead choosing both of the other two doors.
If you decided to switch (which now becomes an obvious choice), Monty first opens the door with the goat behind it (say, to add to the excitement), and then opens the other door.


My question then is, is this reasoning flawed? Is this even the same problem as before? Because now, choosing to switch from one door to two doors becomes quite obvious, and so does the $2/3^{rd}$ chance of winning the car on switching.
I cannot find a flaw in your reasoning.

My own reasoning (if you are interested).

Someone who sticks to his original choice will win if his original choice was correct. 

Probability on that: $\frac13$.

Someone who switches will win if his original choice was wrong. 

Probability on that: $\frac23$.
I think there's a slight difference from the original problem here. 

In your version, neither the elimination nor the switching has any effect. In the original version, these processes matter. Here's how:

Let's say you choose Door 1. There's a 1/3rd chance you're right, and a 2/3rd chance you're wrong.

At this point, if you switch Doors (without Monty eliminating a choice for you) you're just moving to another door with the same 1/3rd chance of being right.

Now, Monty opens Door 2. You're then asked to switch or not.

At this point, if you switch, you will win if you were wrong at first (probability 2/3).

If you were right at first (probability 1/3) and you switch, you lose!

Essentially, when Monty removed that empty door, the question "What are the chances that you will win if you switch" became equivalent to "What are the chances that you were wrong at first", which is 2/3rd.