3. Is the Monty Hall problem the same even if the door opened by the host is chosen...

Is the Monty Hall problem the same even if the door opened by the host is chosen at random?

When the host chooses one of the two doors you didn't pick at random and reveals what's behind it, and it happens to be a goat, it doesn't matter if you switch or not

I'm also not seeing it.

I ran a simulation in my head.

Scenario: Contestant D plays a game called Goatse. Contestant D has randomly selected a door with no knowledge other than behind 1 is a car, the other two are unfavorable (goats).

What's behind this door is unknown, and at this point has a 1/3 chance of being a car.

Gamemaster opens a door other than Contestants selection. It it is a goat.

It is always favorable for "D" at this point to change doors, and it does not matter what Entity revealed the goat and it does not matter the reason.

Why?

Consider this:

You, yourself are playing a new game, Goatse1. Goatse1 is always the same. You have before you three doors. Behind one is a goat, behind one is a car or goat, and the other a car and a goat. However, the last two doors are not the same.

Why?

Because they carry information with them from Contestant D's game of Goatse.

When the host chooses one of the two doors you didn't pick at random and reveals what's behind it, and it happens to be a goat, it doesn't matter if you switch or not

You've proposed a game of Goatse1. Goatse1 is always the same to you! Of course it is! Because you only get to play Goatse1 if "D" has played Goatse and given you this setup! Your supposition

When the host chooses one of the two doors you didn't pick at random and reveals what's behind it, and it happens to be a goat, it doesn't matter if you switch or not

is Goatse1!!!!!

So:

We're playing Goatse1 right? Goatse1 is always the same, except, it isnt. It has 3 different possibilities, 2 of which are winners if you switch, and 1 is not.

  1. D has chosen a car. But, you always switch when you play Goatse1 no matter what. So, you have 1 revealed goat. You have 1 door that is the D's choice, and you never ever choose that door. So you chose the door with the other goat. The untouched door. You loooose!

  2. New game of Goatse1. D has chosen a goat. Before you are the Chosen door, the door you never choose. Good for you, cuz this time it's a goat, the other door that's open is a goat, and the one you ALWAYS switch to is a car!

  3. New game of Goatse1. D has chosien the other goat behind that door you never choose! You don't know this, all you know is that you ALWAYS switch, and there's a goat on the open door, so where's the car? The door you ALWAYS switch to!

There are only 3 possible setups to Goatse1. And you only play the game one way. Thus, 2/3 times you win car when you ALWAYS switch, and 1/3 you CANNOT win.

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