What is the correct answer?

...

You forgot to specify: "The host is not opening the doors at random, but deliberately reveals a goat behind a door you haven't selected"

Yes this is actually an important part of the problem specification due to preservation of evidence.

(Or to state it slightly differently: If the host followed the rule I outlined up top, you should switch for an increase from 1:2 to 2:1 odds of getting the car. If the host is picking a door at random from all three doors, switching doesn't matter, as you are changing odds from 1:1 to 1:1)

Everyone assumes that the door the host opens is removed

If you make a program that randomly selects a door, and you open a door then it will still have a 1/3 chance

-1/12

No it's not the same. Because the host eliminates one choice so now whe have a chance of 50% and before we had 33%.

>Because the host eliminates one choice
Then you analize the new sample space and reset the probabilities.
So it's still 50%

Depends on what we know about the hosts selection method. It is only 50% if we assume the host is picking one of the three doors at random.

Never switch.

goats >> cars

the probabilities are not reset

you are choosing between one door, and not one door

thats the binary choice you are given

Then surely you switch to the open goat?

It's easier to understand if you suppose that once you picked a door Monty lets you switch to the other 2 if you accept to give up on one goat (you'll have always picked at least one goat between the two). This problem is equivalent to the original one since in both cases one door is a 'non choice': in the original case because Monty opens it and you can avoid it, in this case because you give up on it after having switched. The idea of being able to pick two doors helps getting it because the other two door's content is only influenced by your original choice, which on the other hand was completely naive (in the sense that your choice was free and unconditioned, not touching the probability of picking the car, 33%). With this in mind, the probability that the car is on one of the two remaining doors is as naive as your first choice, and the probability of picking a car accepting the switch is an unaltered 66%.

Its 2/3, and as mentioned above, the host knowing which door has the prize makes the difference. Always switch, the answer is 2/3, countless sims have been run, there is nothinf to debate, end this thread.

This should be in the sticky by now god damn