The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.
I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.
To be clear, if the host picks randomly (whatever happens to the door he picks) your odds are the same whether you switch or not.
Will try and run a simulation tonight or this weekend.
Specifically, if you have a 50% chance of winning on swap, you have a 50% chance of losing by not swapping. So, yeah, by the time you get to the swap, no matter what, you are at a 50% chance of winning. Swap or not.
I usually find this problem annoying, not because it's all that difficult, in fact it's quite intuitive - when you're told the exact parameters defining the Monty Hall Problem and systematically work through them.
In my experience though, it's used more often as an exercise in diminution, a sick wet dream of probability teachers, where the learning party isn't aware of the problem, and usually either hasn't been explained, or doesn't quite grasp, the exact circumstances around whether the host's choice is random or decided.
There are lots of "it depends" moments that can be applied to incomplete descriptions of the problem, including (amazingly) whether the host offers a choice at all - this is the one that seems to trip up most people, as they might start to question the "motives" of the host (which are irrelevant in the actual statistical problem).
I think to enumerate the possibilities you'd have to see that if you picked the winning door, there are two ways the host could leave doors for you to swap to and lose.
If you picked a losing door, there is only a single way for the host to reveal a losing door.
So, at the point you are looking at a losing door and making a swap, there are 4 ways you could have gotten there. You picked the winning door, and the host showed either of the two losers. Or you picked either of the two loser doors and the host showed you the other loser. Four possibilities, two of them you win if you swap.
It finally jives with me that your odds at swap time are only 50%. Seems kind of obvious when you think of it as random events and you are at the end with it definitely behind one of two doors. Either door is clearly as likely. Now, your odds of getting to this point are vanishingly slim, the more doors there are. Which makes sense.
I think my intuitive block comes in in that your odds of winning the game are not increased in this scenario at all. Which, I knew. I think I even stated it at some point. Still a hard block to get around.
That's really interesting and counter-intuitive.
https://gist.github.com/ecdavis/da8f67258860e9f35620
EDIT: After thinking about it for a while it seems obvious and I feel fairly stupid.
1/3rd of the time your initial choice was correct. The random host always reveals a goat, you take the opportunity to switch and lose as a result.
1/3rd of the time the random host reveals a car, the game ends without an opportunity to switch.
1/3rd of the time the random host reveals a goat, you take the opportunity to switch and win as a result.
First, let's make it clear exactly what variation of the game we are playing.
1. Prize is assigned to a random door with each door being equally likely. Neither you nor Monty know which door.
2. You pick a door. Because the prize was assigned randomly and you don't know where it is, it is irrelevant how you pick your door. Without loss of generality (WLOG) we can assume you always pick door #1.
3. The host picks a door and opens it. Because the prize was assigned randomly and Monty does not know where it is, it is irrelevant how Monty picks a door. WLOG we can assume he always opens door #2.
4. If Monty revealed the prize when he opened his door, the game ends and you lose.
5. If Monty did not reveal the prize, you are given the opportunity to switch to the remaining door (door #3).
6. Your door is opened. You win if the prize is behind it. Otherwise you lose.
There are three equally likely cases to consider.
1. The prize is behind door #1. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is bad.
2. The prize is behind door #2. This occurs 1/3 of the time. Monty opens #2. The prize is there and the game ends. Note that in this case, YOU ARE NOT GIVEN THE OPPORTUNITY TO SWITCH.
3. The prize is behind door #3. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is good.
Note that in the cases where you are given the opportunity to switch (#1 and #3), switching wins in one of them and switching loses in the other. Each of these cases is equally likely (occurring in 1/3 of the games of played), and so in this version of the game switching makes no difference.
Here's another way to look at it. Since neither you nor Monty know where the prize is when you pick doors, we could change the game so that the prize is not placed until AFTER Monty opens a door, and this would not change any probabilities.
So, in this modified but equivalent game, we play like this:
1. You pick a door.
2. Monty picks a door and opens it. There is nothing behind it, because the prize has not yet been placed.
3. You are asked if you want to switch to the other unopened door.
4. The prize is placed randomly.
5. If the prize is placed behind the opened door, the game ends and you lose.
6. Otherwise, your door is opened and you win if the prize is behind it.
It should be clear that you have a 1/3 chance of winning the car in this game no matter how you pick your door or whether or not you switch. At the time the prize is placed, there is a door that is now your door, and you win if and only if the prize gets randomly placed behind that door.
That's part of the setup. If it's not, it's not being told correctly. It's usually in the form of "...and Monty opens a door that is always a goat..."