I’m not an expert, but I don’t think that the scenario you’ve proposed is comparable to the original. If you had amnesia, you’d lose the information you had previously, so you’d have a 50/50 chance of choosing correctly. But that wouldn’t change the fact that there’s still a one-third chance of the prize being behind the door you originally chose and a two-thirds chance of it being behind the other one, which you would know if you didn’t have amnesia.

When you picked the first door, you had a one-third chance of guessing correctly. Nothing the host does to any of the other doors changes this, since the prize is never moved – the host opening a different door with no prize does not make it any more likely that the prize is behind your door.

A different way to think about it is a situation where instead of the host opening a door with no prize, they combined both doors that you didn’t pick into one, so it’s now one door with any prizes from either of the other two behind it. You would want to switch in this case – there are still two doors, but the one you didn’t pick (which was previously two) has two doors worth of chances to win a prize.

The original Monty Hall problem presents the exact same scenario, but instead of combining two doors, they just eliminate an empty one. This is functionally the same. Your options for the final decision aren’t actually just door A or door B, they are; a) your original door (1/3 chance), or b) not your original door (2/3 chance). Obviously one of the two doors you didn’t originally choose would still have no prize behind it no matter where the prize is, since it can’t be behind two of them – this would be obvious to you before you made your decision. It doesn’t actually matter which it is, you haven’t learned anything new. What the host is doing by opening a door with no prize is essentially the same as combining the two doors, since you’d now be getting the prize if it was originally behind either of them.

The situation would be different if the host opened a door with no prize behind it completely at random, including your door in their options. But the host won’t ever open your door, regardless of where the prize is. So all they’re doing is distilling both other options into one option, which has no impact on your door’s probability at all.

Since you would know this, you have a two-thirds chance of winning if you switched. If you had amnesia, you’d have a 50/50 chance of winning regardless of what you picked, but there’s still a two-thirds chance that it’s behind the door you originally picked – you just don’t know that any more, so your choice is random. Even if there’s a 99% chance of the prize being behind one door and a 1% chance of it being behind the other, or if the game was rigged and it was guaranteed to be behind one door, you’d still have a 50% chance of being correct if you didn’t know this and chose at random.

The mere act of making the new choice is, agreeably, a 50/50 choice. Why does that require a switch?

Imagine, for example, you choose door #2. The host opens door #1 to reveal a goat. Suddenly, a loose light falls on your head and you get knocked out. When you wake up, you have short term amnesia and can’t remember which door you chose in the first round (besides the obvious fact that it was not door #1).

That is to say, you forget the initial choice entirely and now, in choosing between door #2 and door #3, you happen to choose door #2. It’s not a choice to “stay” because both options are on the table; it’s making an entirely new choice (seemingly to you with amnesia) and choosing door #2.

With that logic in mind, what if you didn’t get amnesia? What if you merely set aside the previous decision, and create a whole new choice between door #2 and 3, and you choose #2 yet again? How is that different? Switching is unnecessary.

The point of a conditional probability problem, which was ignored by both Marilyn vos Savant and those who criticised her, is to assign initial probabilities to all of the possibilities, and removing what is inconsistent with what is observed. In this problem, that means (1A) 1/6, (1B) 1/6, (2) 1/3, and (3) 1/3. When the host opens a door, two of these case are eliminated, leaving either (1A) and (3), or (1B) and (2).

Either way, there was a 1/6 chance of reaching the current state with the car behind #1, and a 1/3 chance if a goat is behind #1. You multiply these by 2 to make the sum of all possibilities are up to 1, making it a 2/3 chance that switching wins. And if the split between (1A) and (1B) is not even, the answer is not 2/3. What the above solution calculates, is the chance of winning if you must decide to switch before the host opens a door. And that’s a different problem.

Marilyn vos Savant has never provided an actual solution. Her first column was just a different way to use intuition (“If there were 1,000 doors, and the host opened 998, you’d switch pretty quick”). I won’t defend the answer provided by many (but not all, or as many as she implied) with PhDs, but they were right to call her out on this.