But I don’t know why the parent comment doesn’t fail in this exact way, unless it’s sneaking in a discussion of conditional probabilities.
Because, of course, the door to which you’re offered the switch can’t have a 1⁄2 chance of being a car, given that there’s only one of it.
Consider it in terms of scenarios. There are two possible ones:
Scenario 1 (happens 1/3rd of the time): You picked the car. You are now being offered the chance to switch to a door which definitely has a goat.
Scenario 2 (happens 2/3rds of the time): You picked a goat. You are now being offered the chance to switch to a door which definitely has the car.
The only problem is that you don’t know which scenario you find yourself in (indexical uncertainty), so you have to average the expected utility of each action (switch vs. no-switch) over the prior probability of finding yourself in each of the scenarios. This gives you 2/3rds of a car for the “switch” action and 1/3rd of a car for the “no-switch” action.
Edit: A previous version of this comment ended with the sentence “No conditional probabilities are required.” This, of course, is not strictly accurate, since my treatment is isomorphic to the conditional-probabilities treatment—it has to be, or else it couldn’t be correct! What I should’ve said, and what I do maintain, is simply that no explicit discussion (or computation) of conditional probabilities is required (which makes the problem easier to reason about).
Replying to the edit:
Because, of course, the door to which you’re offered the switch can’t have a 1⁄2 chance of being a car, given that there’s only one of it.
Consider it in terms of scenarios. There are two possible ones:
Scenario 1 (happens 1/3rd of the time): You picked the car. You are now being offered the chance to switch to a door which definitely has a goat.
Scenario 2 (happens 2/3rds of the time): You picked a goat. You are now being offered the chance to switch to a door which definitely has the car.
The only problem is that you don’t know which scenario you find yourself in (indexical uncertainty), so you have to average the expected utility of each action (switch vs. no-switch) over the prior probability of finding yourself in each of the scenarios. This gives you 2/3rds of a car for the “switch” action and 1/3rd of a car for the “no-switch” action.
Edit: A previous version of this comment ended with the sentence “No conditional probabilities are required.” This, of course, is not strictly accurate, since my treatment is isomorphic to the conditional-probabilities treatment—it has to be, or else it couldn’t be correct! What I should’ve said, and what I do maintain, is simply that no explicit discussion (or computation) of conditional probabilities is required (which makes the problem easier to reason about).