If one wrong door is always opened, your chance was never 1/3 to begin with, so you are thinking about this problem with the wrong premise, making it hard to grasp. You were just assuming it was 1/3 because you didn’t know one door would be taken away.
As soon as the wrong door is opened, your odds are never 1/3 nor 2/3. It’s 1/2 because there’s only two doors. What did you think the number after / stood for?
EDIT: Now I’ve tried to look through the examples in the article, and it honestly just makes it worse.
The example about picking a door at 1/1000, and then Monty removing 998 of the doors, leaving two doors, therefore making it more likely you should pick the one Monty left open, is also stupid - because it’s not comparable.
The above example is true. The likelihood of Monty being right is much higher.
But your pick is never 1/1000 when there’s only 3 doors, making the example not compatible with the other. The 1000 door example is not wrong - you just can’t compare them.
And now to explain why it’s different:
In the 3 door example, your “pick power” is 1. Means you can pick 1 door. Montys “pick power” is also 1, making you both equally strong.
This means that you picking a door gives as much intel as Monty picking a door does. No matter what, you will always be left with 1 door not being picked.
Now you look at the 2 doors. The one you picked, and the one nobody did. Now this problem suggests that Monty has given you new information because he removed a door, but he didn’t give you that, and here’s why:
The problem suggests that Monty gives you intel by removing a door in a 1/3 scenario. But he doesn’t. That’s an illusion.
From Montys perspective, he only has 2 doors to pick from, because he can NEVER remove yours, no matter what you picked.
Now Monty has made his choice, and this is where we turn the game around making it clear it was a 1/2 choice all along.
Because the thing you are picking between is not the doors anymore. It was never about the doors.
You are picking between if Monty is bluffing or not.
Let’s say you always pick door 1 as your first option. Monty will always remove 2 or 3. Either Monty removes door 2 or 3 because he helps you, or he’s doing it because he’s bluffing.
If you didn’t get any more help, this WOULD’VE been a 1/3. You’d have to choose between if Monty bluffed at door 2 or he bluffed at door 3, or he bluffed at both, because it was your door.
But then Monty goes ahead and removes a door, let’s say 3 (or 2 if you want, it doesn’t matter). He tells you it’s not that one. Now you have to choose if he’s bluffing at door 2 or he’s bluffing at your door.
To calculate the odds of getting a passing grade (at least one question correct) or a perfect score (both questions correct) through brute-force guessing, let’s break down the scenario:
Each question has 4 possible answers, and only one is correct.
You have two attempts to answer each question.
You remember your previous answers and do not repeat them.
First, we’ll calculate the probability of guessing at least one question correctly. There are two scenarios where you pass:
Scenario 1: You get one question right and one wrong.
Scenario 2: You get both questions right.
For each question:
Probability of getting it right in one of the two tries = ( 1 - ) (Probability of getting it wrong twice)
Probability of getting it wrong in one try = ( \frac{3}{4} ) (since there are 3 wrong answers out of 4)
Probability of getting it wrong twice = ( \left( \frac{3}{4} \right)^2 )
So, the probability of getting at least one question right in two attempts is ( 1 - \left( \frac{3}{4} \right)^2 ).
For two questions:
Scenario 1 (One Right, One Wrong):
Probability of getting one question right (as calculated above) multiplied by the probability of getting the other question wrong twice.
Scenario 2 (Both Right):
Probability of getting each question right (as calculated above) and multiplying these probabilities together.
The overall probability of passing (getting at least one question right) is the sum of the probabilities of these two scenarios.