The only way to lose, if you switch after, is to pick the door with the prize, out of 3 doors this being a 1/3 chance
If you pick a goat door, a 2/3rd chance, and then you switch, you will always win (since Monty removes the other goat door, leaving just the car door to pick)
You mean you switched from following the Smartest Person Alive to following the other Smartest Person Alive? 
The difference in the two scenarios is not what Monty reveals, but the restrictions on Monty for his reveals.
Monty is required to reveal a goat from the unselected doors. In your new case, he has three unselected doors and two goats. Either way, there will then be two unselected doors, only one with a car. You get to pick - 50:50.
In the standard scenario, you pick first out of three doors, leaving only two unselected doors for Monty. Monty is thus constrained by your choosing first. He will show you a door with a goat, leaving one unselected door. Because he was constrained by your initial choice, your initial choice percentage remains at 1/3 of winning. Therefore, the remaining door has 2/3 chance of winning.
Which brings us full circle on the other issue on the original Monty Hall problem. The problem works as stated so long as we know Monty is so constrained. A malicious Monty Hall can elect not to open any doors, or to only open a door when he knows you have already picked the car - and thus encourage you to switch when you are already a winner. I think that often the intuitive difficulty people have with the main problem is in getting past the very human cynicism about Monty’s motives. If the problem is expressed as “Monty must always open a door with a goat” it is slightly easier to work the logic. But people being people need to grok that Monty is an automaton, and not a slimy game show host.
I’m curious about something. Let’s say that in the Monty Hall problem that you start out with 2 doors. The contestant selects one or the other. Now monty adds a third door that he reveals has a goat. Do you still do better by switching?
No.
So how is it different from the classic problem? That’s what I’m trying to understand.
Doesn’t the contestant know in advance that’s precisely what will happen?
Assuming he always adds a third door and it always has a goat - then no - he isn’t eliminating any of the doors. He is simply adding a goat. If there were two doors in the beginning - and one had a car - one a goat - and he doesn’t do anything with those doors - it is 50/50 - and not very fun if he does do something with the doors 
Right, because of the original 1/3 probability. For some reason I always think of this as a coin flipping situation, but it really isn’t. Thanks.
Yes. Your hypothetical is the same as saying the game begins with three doors, one of which you cannot choose and which Monty must always open to reveal a goat. The game as set up, on the other hand, forces Monty to open one and only one door (though you don’t know ahead of time which one he will have to open) except on the one-in-three chance that you picked the car in the first place.
I think the problem that many people have in grasping the solution is that they don’t see the overwhelming benefit of switching because there isn’t one. Simply a 2/3’s probability of winning. One can easily imagine losing that scenario so you don’t think of it as such of an advantage and tend to discount it. Likewise a 1/3 shot of winning isn’t such an impossibility that one casually tosses it aside.
The simplest way of looking at it is to discount what Monty Hall does as meaningless on one hand (because you already knew that one of the other doors had a goat) and instructive on the other (he has eliminated one possible choice). Then you decide if you original 1/3 shot is accurate. If you decide that it likely wasn’t (and less than 50% means it wasn’t) then you should switch to the only remaining choice which flips the odds and gives you a 2/3’s chance of winning.
The “100 doors” example is what finally convinced me.
The OP’s question seems to be similar to that of the Three Prisoners (scanned PDF of the problem with solution).
Briefly, one of three prisoners will be executed in the morning, but none of them know which. Prisoner A asks the guard to give a letter to one of the other guys who’ll be set free, just in case he doesn’t make it. Prisoner A then asks the guard to tell him which person he gave the letter to, thinking that it can’t make a difference, since he knows at least one of them must go free. But then he realizes that now that it’s down to him and the guy the guard didn’t name, and his odds have worsened from 1/3 to 1/2.
The terrible thing is even just contemplating the question seems to have done this, because no matter who the guard names, his odds of surviving apparently decrease.
In the classic problem, you choose one of three; in the new problem, it’s one of two.
This yields the following probability of initially choosing the door with the big prize: Classic problem: 1/3; new problem: 1/2
In both cases, switching causes you to lose if the first choice was right, and to win if it was wrong: Classic problem: 2/3; new problem: 1/2
What Monty does with doors & goats that are unavailable to you has the same effect on the above as do goats on Mars: none at all.
The Monty Hall problem has been much discussed on the SDMB and elsewhere. I’ve read a lot of those discussions yet I’ve never seen anyone answer the most important question of all:
If you pick a door with a goat behind it, do you get to keep the goat? Or is it only a gag? Do you have to pay taxes for the goat?
Assuming the execution will go off without a hitch, the prisoner’s chance of survival is either zero or one, as the condemned has already been selected. The only thing that’s changed is his beliefs.
(If it helps, think of the prisoner who got the letter. He knows he’s going to survive.)
Actually, the prisoner should assign a probability of 1/3 to his own guilt even after learning who got the letter. I had to use a slightly less sophisticated method to prove that to myself, but I think it makes the error clearer.
There’s a very subtle issue with the linked analysis of the three prisoners problem. Finding it is a good exercise, so I’m going to hold off on pointing out what it is for a bit in case anyone else wants to try to find it and suggest a fix.
Now I’m confused again. I understand why the original prisoner’s chance of being executed remains 1/3, but doesn’t that then mean that the chance of the other prisoner’s chances of being executed is now 2/3? That makes no sense. (Because the exercise could have been done in reverse and flipped the odds for each prisoner).
Prisoner A has a 1/3 chance of being executed. The guard tells him he gave the letter to one of the other two, say prisoner B. From the point of view of prisoner A, based on the information he has, prisoner C has a 2/3 chance of being executed. But assuming neither prisoner B or C know about this letter, they still (based on the information they each have), think they have a 1/3 chance of being executed.
All through this, the guard presumably knows which prisoner will be executed, so based on the information the guard has, the chances are either 1 or 0.