It makes a difference how you got there, assuming in the second one that he reliably/intentionally never reveals the grand prize.
Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.
Marble 1 2
Bag
2R R R
BR B R
Out of the three possible equally likely worlds (given that you know you didn't roll into BR-1), two of them have you holding bag 2R, aka a 2/3rds chance.
If, on the other hand, you use a red-marble-picking robot, you get:
Marble 1 2
Bag
2R R R
BR R R
Now you have four possible worlds and are equally likely to be holding either bag, ie 1/2.
So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.
tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.
OK, so I pick a door, Monty randomly reveals a goat. Your claim is that this information should make me increase the odds that my initial pick was a car. That sounds right. In the normal Monty problem, Monty doesn't give me any information about my original pick, just about the remaining door.
Alright just working through your logic here. Let's say I pick the first door always. I'll write out all the possible configurationss with both possibilities for Monty.
CGG > Monty reveals mid (Goat)
CGG > Monty reveals right (Goat)
----
GCG > Monty reveals mid (Car)
GCG > Monty reveals right (Goat)
---
GGC > Monty reveals mid (Goat)
GGC > Monty reveals right (Car)
So there's 4 worlds where Monty reveals a goat. In two of them, I picked a car initially. In the other two, I get a car by switching.
By the time you have the decision, yeah, that's true.
There are other situations where the game is over before you get a decision.
Both GCG and GGC are eliminated half the time before you get to make a decision. But CGG is never eliminated before you get to make a decision.
Once you're making a decision and the game isn't already over, you can use this information -- that the game didn't end -- to assign a 50% chance of being in CGG, and 25% of each of the other two.
Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.
Out of the three possible equally likely worlds (given that you know you didn't roll into BR-1), two of them have you holding bag 2R, aka a 2/3rds chance.If, on the other hand, you use a red-marble-picking robot, you get:
Now you have four possible worlds and are equally likely to be holding either bag, ie 1/2.So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.
tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.
EDIT: formatting.