Please Explain (actually explain) The Monty Hall Problem

[phyti]

Yes, it does make a difference. There is a difference between an accidental result of rare luck versus the intentional act of avoiding opening one door. Consider the example (again) of the game with 100 doors. In the first case of luck, the only inference that can be made from opening 98 doors without revealing the prize is that a rare, lucky event was just witnessed.

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.
The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

 

[PeroK]

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.
The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

Are you saying that if you stick with your original choice then you win 50% of the time?

 

[FactChecker]

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.

In the Monte Hall game, the player only has one choice that he does before Monte opens any door.

The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

So you don't think that Rule 2 and the fact that Monte did not open that remaining door gives any sort of hint that should be used to adjust the odds but that same implication does not apply to your door? As has been explained dozens of times on this thread alone, you are wrong.

 

[DaveC426913]

The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

Except that you can be objectively shown to be wrong.

The simulation in post 27 proves that switching dors will result in winning twice as often.
If your logic/intuition do not lead to that result, then it is demonstrably faulty.

 

[Dale]

The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin

Only if the two options are equal probability, which they are not. Simply having two options does not mean that the probability is 1/2

https://www.ac-psych.org/en/download-pdf/volume/10/issue/4/id/164