But you have been given information about the door he didn't open, because he didn't open it. That's why it's possible for the odds on that door of holding the prize can change.
And yes, we do now have a choice between two doors, one with a goat and one without. The error is in believing that these are equal choices.
I roll a die, and you can choose "1" or "not 1". You have two choices, but they have unequal chances of being correct. Similarly with the doors. Just because there are two choices, they may have different odds.
In the Monty Hall problem, they do.
When or where are any odds changing?
Then the host opens one. The probability of the door opened holding the prize goes to 0. But the probability on the set of two is still 2/3, and so the probability of the door that is both unchosen and unopened goes to 2/3.
But the Monty Hall Problem as stated is about the probabilities, not on the psychology. Computing the probabilities is simple math, once you understand the situation. My explanation was to help the reader understand why the two choices given don't have equal probability.
If I chose a door before, then something happens that leads to only two doors being left, both those doors have the same probability so I could just choose the same door again.
aw3c2> If I chose a door before, then something happens that leads to only two doors being left, both those doors have the same probability so I could just choose the same door again.
The original said this:
CW> The information you have gained is not about the door you've chosen. It doesn't matter what door you choose, the host can always open a door to reveal a goat. So there is no information given about the door you've chosen, so the chances of that door containing the prize remain at 1/3.
CW> But you have been given information about the door he didn't open, because he didn't open it. That's why it's possible for the odds on that door of holding the prize can change.
So let's recap what's going on. There are three doors. For the sake of concreteness let's call them A, B, and C. You choose one of them. For the sake of concreteness let's suppose you choose A.
So now there are two doors, B and C, remaining unchosen by you. Currently those two doors, B and C, each have probability 1/3 of having the prize. The door you chose, door A, has probability 1/3 of holding the prize.
Now the host opens a door, taking care to open a door that does not hold the prize. So the pair {B,C} still has total probability of holding the prize, but you are being shown that one of them certainly does not. This doesn't affect the probability that your chosen door, door A, holds the prize -- the probability that the prize is behind door A is still 1/3.
The pair {B,C} still has total probability 2/3 of holding the prize. You're now given the choice of staying with A, or switching.
Quoting you again, you said:
aw3c2> something happens that leads to only two doors being left, both those doors have the same probability ...
That turns out not to be the case. Just because there are two doors they don't have to have equal probability of holding the prize, and in this case they don't. The probability that your door holds the prize has not changed and is still 1/3. The probability that the door neither chosen by you nor opened by the host holds the prize is now 2/3.
Does that help?
Let's change the number of doors !
First, I'll rephrase the original problem ================================================
You have 3 doors, 1 hides a car, all others hide goats.
You pick one door, then __ALL doors you have NOT selected AND that hide goats are revealed, except one__
Here it's only one door because we have a total of 3. But it's one door that is indeed, ALL except one, for this particular situation.
Do the same, but with 100 doors ====================================================
You have 100 doors, 1 has a car, all others have goats.
You pick one, then __ALL doors you have NOT selected AND hide goats are revealed, except one__
Now you get:
- 1 selected, 99 not selected => the car is most probably among the 99 ones.
- Monty reveals 98 goats
- 1 selected, 98 goats, 1 not selected => the car is most probably in the later one.
The problem itself can be rephrased in an even simpler manner where Monty doesn't open any doors at all. Once you've picked a door, he then gives you the option of either staying with that door or changing to both of the other doors (where you get to keep the best prize behind them). Given that formulation, almost everybody would switch.
If you always switch to the door you didn't initially choose your odds of winning rise to 2/3.