"In the first scenario Alice knows she had a 2/3 chance of being wrong when she made her initial choice and the 2/3 probability doesn't go away once one of the other options is removed. That 2/3 probability gets moved to the remaining option."

I don't think flipping the order changes the results. If you do, and are able to explain why, it would help both of us to better understand where and why we disagree. That seems relevant to me, at least insofar as the point of this conversation is to better understand these problems.

I'm saying that before he dies Bob doesn't know whether he will. Before Monty opens a door, he know whatever door he opens, it'll be a goat, because he knows what's behind the doors.

"You don't get to reassess the probability just because an option was eliminated, just like you don't get a 50/50 chance in the Monty Hall problem because the question is whether you should switch from your original door."

"Alice and Bob survive the crash and see 3 bushes, 2 of which they know are poisonous. Alice thinks the yellow bush is safe. Bob eats from one of the non-yellow bushes and dies. What are Alice's chances of survival?"

"Scenario B 1/3 chance: she chose goat 2 therefore he will reveal goat 1 and she should switch"

"The difference is that in the scenario that the video is proposing Bob 'is certain to pick a bad bush'."

I get that we're only interested in situations where Bob dies, but that doesn't mean you can just ignore the possibility that he lives. If it's a real possibilities (and on my understanding of the puzzle, it is), then it needs to be taken into account in order to accurately calculate the probabilities. In the Monty…

"Bob is dead. End of story. Him being alive has been eliminated as an option."

As I've said elsewhere, there's a crucial difference between "does pick a bad bush" and "is certain to pick a bad bush." Bob picks a bad bush because that's the story we're telling, but the actual outcome isn't relevant to the underlying probabilities. Monty picks a goat because he knows where the car is. He always…

"Any other scenarios involve Bob living which would be the same as Monty revealing the car. Neither can happen, because Monty knows where the car is, and Bob is laying dead on the ground."

But there's difference between a scenario not being interesting (Bob picks the safe bush) and a scenario being impossible (Monty revealing the car).

Agreed. Agreed. Agreed. Disagreed.

I'm focusing on that fact because it's the crucial fact that makes the Monty Hall problem so counterintuitive.

(How does one type out a long exhale? I want to do that).

Since Alice learns nothing about the difference between bushes 1 and 3 from Bob's dying (which is not the case in the Monty Hall problem) the only thing Bob's death does is increase her chances of surviving from 1/3 to 1/2. But it doesn't matter what bush she chooses.

I completely agree that switching leads to a win 2/3 of the time for the Monty Hall problem and I completely agree that that is a proven, resolved issue. The disagreement is over whether the bushes scenario is analogous to the Monty Hall problem in the relevant ways such that the proof for the latter can be applied to…

"The other time that he picks edible berries, you will just eat from the same bush. The choice between 2 poison berries is eliminated, just like the choice between 2 goats."

"Wrong. Bob dies, meaning there is a 100% chance he picked a poison bush."

Okay, let's try with more math, looking just at the results of switching: