When I first came across this, I didn’t believe my own analysis and wrote two different simulators before I accepted it. Now it’s so baked into my worldview that I don’t understand why it was difficult back then. It was confounded by the common misstatement of the problem (that is, the implication that Monty might not offer a switch in some cases), but it was also a demonstration that I have a “system 1.5”, which is kind of a mix of intuition and reflection. 50% felt like I was rationally overriding my gut reaction that switching was wrong (because it’s a game show, built on misleading contestants), and it was hard for me to further override it by calculation.
The usual explanations of the Monty Hall problem are needlessly confusing. In fact there’s no reason to talk about conditional probabilities at all. Here’s a far simpler and more intuitive explanation:
You had a 1⁄3 chance of picking the car and a 2⁄3 chance of picking a goat.
If it so happens that you picked the car, you shouldn’t switch. But, if it so happens that you picked a goat, then you should switch.
Since you don’t know what you picked, you take the expectation: 1⁄3 × “shouldn’t switch”[1] + 2⁄3 × “should switch” = “should switch” wins.
[1] The formal/precise/rigorous phrasing speaks, of course, of the expected number of cars you get in each case, i.e. either 1 or 0, and calculates the expectation of # of cars for each action and selects argmax, i.e. this is a straightforward decision-theoretic explanation.
Huh. Wouldn’t this argument suggest that, before the games master opens one of the doors, were he to offer you iterated switches, you should each time accept a switch because it has higher expected utility?
Edit: Iterated switches = the games master keeps offering you to switch doors, and you accept because each time has a 2⁄3 probability. But I don’t think it is actually a problem. Just parse the question as ’do you want to accept your 1⁄3 probability, or do you want to switch to a 1⁄2 chance between two doors who in total have a 2⁄3 chance of containing the car, which shows that they’re equivalent. 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.
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).
What does “iterated switches” mean? In the 3-door game, there’s only one switch possible.
That does remind me of another argument I found compelling—imagine the 10,000 door version. One door has a car, 9,999 doors have nothing. Pick a door, then Monty will eliminate 9,998 losing options. Do you stay with your door, or switch to the remaining one?
When I first came across this, I didn’t believe my own analysis and wrote two different simulators before I accepted it. Now it’s so baked into my worldview that I don’t understand why it was difficult back then. It was confounded by the common misstatement of the problem (that is, the implication that Monty might not offer a switch in some cases), but it was also a demonstration that I have a “system 1.5”, which is kind of a mix of intuition and reflection. 50% felt like I was rationally overriding my gut reaction that switching was wrong (because it’s a game show, built on misleading contestants), and it was hard for me to further override it by calculation.
The usual explanations of the Monty Hall problem are needlessly confusing. In fact there’s no reason to talk about conditional probabilities at all. Here’s a far simpler and more intuitive explanation:
You had a 1⁄3 chance of picking the car and a 2⁄3 chance of picking a goat.
If it so happens that you picked the car, you shouldn’t switch. But, if it so happens that you picked a goat, then you should switch.
Since you don’t know what you picked, you take the expectation: 1⁄3 × “shouldn’t switch”[1] + 2⁄3 × “should switch” = “should switch” wins.
[1] The formal/precise/rigorous phrasing speaks, of course, of the expected number of cars you get in each case, i.e. either 1 or 0, and calculates the expectation of # of cars for each action and selects argmax, i.e. this is a straightforward decision-theoretic explanation.
Huh. Wouldn’t this argument suggest that, before the games master opens one of the doors, were he to offer you iterated switches, you should each time accept a switch because it has higher expected utility?
Edit: Iterated switches = the games master keeps offering you to switch doors, and you accept because each time has a 2⁄3 probability. But I don’t think it is actually a problem. Just parse the question as ’do you want to accept your 1⁄3 probability, or do you want to switch to a 1⁄2 chance between two doors who in total have a 2⁄3 chance of containing the car, which shows that they’re equivalent. 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.
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).
Iterated switches to what, exactly?
Edit: Regardless, the answer is no; there’s no reason to conclude this.
What does “iterated switches” mean? In the 3-door game, there’s only one switch possible.
That does remind me of another argument I found compelling—imagine the 10,000 door version. One door has a car, 9,999 doors have nothing. Pick a door, then Monty will eliminate 9,998 losing options. Do you stay with your door, or switch to the remaining one?