So in the simple version where the host doesn’t do anything, there’s 3 possibilities for the car. It’s either behind door number 1, 2 or 3. So when you pick one, there’s a 1 in 3 chance you pick the door with the car.
Now as for why the host’s choice matters:
The host always opens a door that you didn’t choose and that also does not have the car. This information seems like the useful part, but it’s actually not. No matter which door you pick, one of the other two will always have a goat. He’s just showing it to you for the show, but that doesn’t change anything, because you don’t get to pick which of the two other doors, only “switch,” that’s the important part. This recontextualizes the choice you made, effectively your choice is now between the door you picked, or the other two doors combined (the one he showed you which is wrong, and the other one).
When you first picked the first door, the odds of the car being behind it were 1 in 3, and those odds still haven’t changed. There’s still a 1 in 3 chance that your door has the car behind it. And when you first picked, there were 2 in 3 odds that the car was behind one of the two doors that you didn’t pick. Having the incorrect door revealed after the fact doesn’t change that fact about both doors. So now the door you chose initially has 1 in 3 odds and the other choice - switch to both doors - has 2 in 3 odds. That’s twice the odds, so you should definitely switch. You’ll be wrong 1/3 of the time, but it’s the best pick.
At least that’s how I think about it: by staying, it’s like you only got to open one door, and by switching it’s like you get to open both of those two doors.
If the host showed you the wrong door ahead of time, you’d be right though. Your choice would be between two doors and be 50/50, but by showing it after you’ve already picked one, the likelihood of the car being one of the doors you didn’t pick stays the same, and since the revealed door now has a 0 in 3 chance the last door must have that remaining 2 in 3.
If the host showed you the wrong door ahead of time, you’d be right though. Your choice would be between two doors and be 50/50, but by showing it after you’ve already picked one, the likelihood of the car being one of the doors you didn’t pick stays the same, and since the revealed door now has a 0 in 3 chance the last door must have that remaining 2 in 3.
This is also where the intuitive comprehension breaks down, and why I feel like an idiot when people explain this “there’s still 2/3rds probability” answer. That seems like we are selectively carrying-forward new information - maintaining the 2/3rds probability information but omitting the information that we exist in a revised 2-door universe after the host’s choice.
Similar to the questions I initially posed, we would typically assume probability changes when an action occurs. Set aside the door that the host ultimately chooses (which I know is probably changing the hypothetical, but work with me). There are two doors and in this example the “choice” of one of two doors is a 50/50 chance to contain a car or a goat. Now add the third door, which is always going to be a goat. Since it will always be a goat, because the host will never pick the door with the car, it will always be irrelevant. This doesn’t seem like an “action” in that it doesn’t give true new information.
To explain differently: presumably we would agree that that two-door hypothetical is always going to be a 50/50 choice, whether we switch or not. And isn’t that exactly the same as the three-door situation, if the host will always open a door with a goat? No matter the order, there are always going to be three doors: {Player-chosen} {Not-chosen} {Host-chosen}, where the {Host-chosen} is always a goat. Opening it seems a like a predestined, 100% deterministic act that can be removed from the equation.
The question is, why does that change when we add a third door that is always going to be irrelevant to that choice? Why is the pre-ordained act of the host always opening a third door with a goat not just some ritual - a superficial act that changes nothing to the two-door problem?
I know I’m wrong, I just can’t square that hypothetical.
The probability that you guessed right when you picked the initial door cannot change. Revealing additional information doesn’t retroactively change the odds that your guess was correct.
You had a 1 in 3 chance when you guessed, so the odds that your guess was wrong was 2 in 3. Those odds don’t change just because you gained more information later on. That’s why those 2/3rds chances that you’re wrong apply to the remaining door being correct.
So in the simple version where the host doesn’t do anything, there’s 3 possibilities for the car. It’s either behind door number 1, 2 or 3. So when you pick one, there’s a 1 in 3 chance you pick the door with the car.
Now as for why the host’s choice matters: The host always opens a door that you didn’t choose and that also does not have the car. This information seems like the useful part, but it’s actually not. No matter which door you pick, one of the other two will always have a goat. He’s just showing it to you for the show, but that doesn’t change anything, because you don’t get to pick which of the two other doors, only “switch,” that’s the important part. This recontextualizes the choice you made, effectively your choice is now between the door you picked, or the other two doors combined (the one he showed you which is wrong, and the other one).
When you first picked the first door, the odds of the car being behind it were 1 in 3, and those odds still haven’t changed. There’s still a 1 in 3 chance that your door has the car behind it. And when you first picked, there were 2 in 3 odds that the car was behind one of the two doors that you didn’t pick. Having the incorrect door revealed after the fact doesn’t change that fact about both doors. So now the door you chose initially has 1 in 3 odds and the other choice - switch to both doors - has 2 in 3 odds. That’s twice the odds, so you should definitely switch. You’ll be wrong 1/3 of the time, but it’s the best pick.
At least that’s how I think about it: by staying, it’s like you only got to open one door, and by switching it’s like you get to open both of those two doors.
If the host showed you the wrong door ahead of time, you’d be right though. Your choice would be between two doors and be 50/50, but by showing it after you’ve already picked one, the likelihood of the car being one of the doors you didn’t pick stays the same, and since the revealed door now has a 0 in 3 chance the last door must have that remaining 2 in 3.
This is also where the intuitive comprehension breaks down, and why I feel like an idiot when people explain this “there’s still 2/3rds probability” answer. That seems like we are selectively carrying-forward new information - maintaining the 2/3rds probability information but omitting the information that we exist in a revised 2-door universe after the host’s choice.
Similar to the questions I initially posed, we would typically assume probability changes when an action occurs. Set aside the door that the host ultimately chooses (which I know is probably changing the hypothetical, but work with me). There are two doors and in this example the “choice” of one of two doors is a 50/50 chance to contain a car or a goat. Now add the third door, which is always going to be a goat. Since it will always be a goat, because the host will never pick the door with the car, it will always be irrelevant. This doesn’t seem like an “action” in that it doesn’t give true new information.
To explain differently: presumably we would agree that that two-door hypothetical is always going to be a 50/50 choice, whether we switch or not. And isn’t that exactly the same as the three-door situation, if the host will always open a door with a goat? No matter the order, there are always going to be three doors: {Player-chosen} {Not-chosen} {Host-chosen}, where the {Host-chosen} is always a goat. Opening it seems a like a predestined, 100% deterministic act that can be removed from the equation.
The question is, why does that change when we add a third door that is always going to be irrelevant to that choice? Why is the pre-ordained act of the host always opening a third door with a goat not just some ritual - a superficial act that changes nothing to the two-door problem?
I know I’m wrong, I just can’t square that hypothetical.
The probability that you guessed right when you picked the initial door cannot change. Revealing additional information doesn’t retroactively change the odds that your guess was correct.
You had a 1 in 3 chance when you guessed, so the odds that your guess was wrong was 2 in 3. Those odds don’t change just because you gained more information later on. That’s why those 2/3rds chances that you’re wrong apply to the remaining door being correct.