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- The Monty Hall problem 10 answers
I have no doubt that the probabilities are $2/3$ by switching; my problem is trying to explain it to someone. His first point is the two choices are unrelated events. My argument is:
$1$) If at first you chose a door with a goat ($2$ cases of $3$), then for the second choice you will have a goat in your door and the car in the other. You win by switching.
$2$) If at first you chose the car door ($1$ case of $3$), then for the second choice you will have the car in your door and a goat in the other. You win by staying.
Textually, what he answered was: (He calls $E1$ to the first event, ie to the first selection, and $E3$ to the second selection)
"You could not have picked a Goat or Car in E1, it's impossible, the Game assigns no value to Your E1 pick, it's an unknown by Game definition. What is known is the Value of each Door in E3.
The difference in definition between E1 as a Constant or E1 as a Variable is the crux of the issue. All previous definitions of The Monty Hall define the Result of E1 as Variable. The Value of E1 is irrelevant. E1 as a Constant is why E3 is an independent event. With Odds calculated as Roulette, not as BlackJack."
The important thing about his argument, which I had not seen from other people, is that he thinks that the fact that you can not see what is behind the first door picked makes it completely irrelevant, and so the two choices are separated events.