> my mind says that if I pick a door at random out of two I have the same probability of getting a goat.

In the proper setup of the problem, at no point do you pick a door at random out of two.

You start with three doors.

* You choose one, and you have a 1/3 chance that you have picked the car.

* There is a 2/3 chance that the door is behind one of the other doors.

Now, and this is a key point, Monty knows where the goat is.

* Of the other two doors, he can choose one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

* Of the other two doors, he does choose one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

* Of the other two doors, he opens one that hides a goat. That doesn't affect your chance of being right, which remains at 1/3.

Now there is one other unopened door.

* The chance that the door you chose holds the car remains at 1/3;

* The chance the car is behind one of the other two doors is 2/3;

* One of those two doors has been opened!

So the other closed door has a 2/3 chance of holding the car.

Yeah, my simplest explanation is the first guess you have a 1/3 chance. If you switch, you are actually just betting that you were wrong on the first guess… and there is a 2/3 chance you were wrong, so that is a good bet.

Indeed. Looking at this from an information point of view, the fact that the host can open a door tells you nothing as to whether your first choice was right. So the chance you picked the right door is always 1/3.

The information you get is which door to open if you are allowed to switch.

The confusing part is that him opening the door doesn't reveal any new information. In most cases knowing that an event happened changes other probabilities. Or in other words, probability of A is equal to probability of A given B in this case, where A is that your initial pick is correct and B is that he opened a door with a goat. For example, lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? It is 100% of course, since it is the only door left, him opening two doors with goats revealed new information about your door.

So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. It is true in this case, but not true in the general case of problems like this since probabilities change as new information is revealed.

> The confusing part is that him opening the door doesn't reveal any new information.

It does reveal new information, just not about the door you chose.

> lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct?

But you can't guarantee that he can do that. If you do guarantee that he can do that then you are giving me more information.

It's always the way, if you change the problem you change the analysis.

> So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply.

Just because it doesn't normally apply, that doesn't mean it's wrong in this case. It does apply in this case (as you say) and therefore it's right.

Knowing when it's right is hard, I agree.

If it helps any, you're supposed to be misled by the setup.

Monty could have asked the question as "Do you want to stick with the door you chose or do you think it's in one of the other two?" and skipped even opening one of them but that wouldn't have been as fun to watch.