Suppose after you make your initial choice, instead of opening a door, Monty simply asks if you'd like to switch to BOTH the other two doors, such that you win if the prize is behind EITHER of them.
That switch is intuitively a great deal, giving you 2/3 odds. The only way you can lose is in the 1/3rd case where you already picked a winner. The original scenario is equivalent to this, since by revealing a goat Monty is allowing you to pick "the best of" the two doors.
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Often people who don't like this problem complain that actually Monty, with his knowledge of the winning door, may be trying to trick you into losing. Maybe he offers this trade conditionally based on whether you've already selected a winner. Of course, this wouldn't support the most common intuitive answer of "it's 50/50", either, making this a bad excuse. If we accept this behavior from Monty -- not stated in the problem -- it becomes a very uninteresting problem based on whether we are dealing with Evil Monty who only offers the trade when it's a bad one (stay wins 100% of the time), Friendly Monty who only offers the trade when it's a good one (switch wins 100% of the time), or somewhere in the middle. It has no analytical answer. So for this to be a solvable problem at all we must assume we are dealing with Fair Monty who offers the trade all the time, or at least, offers it at random times not based on the contestant's initial pick.
It's amazing how even seeing the probabilities written out, or running simulations, doesn't really make it easier to truly understand the result.
If Monty always opens a door, and uses his knowledge of which door has the prize to ensure that the door he opens is always empty, then you should switch, because he's providing you with extra information about which door has the prize.
However, if Monty changes it up and sometimes doesn't open a door, or opens the door with the prize, the whole problem changes. If he only opens a door and offers you the chance to switch when you've picked the door with the prize, you should never switch of course. If he opens a random door and might reveal the prize (after which he obviously won't let you switch anymore), switching doesn't change your odds. I think a lot of people see it as one of those two situations.
In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two doors, and need to pick the one you think hides a car. So intuitively, your odds of choosing which of two doors has the car are 1/2.
It's hard to really understand why it matters that the previous rounds occurred at all — why it matters how you narrowed the selection to these two doors. I don't think the 1000 door version makes that easier to understand, because the same thing is true there — if you come into a game, see two doors, and are asked to choose between them, it's very hard to understand why it matters whether in a previous round there were 2, 3, 10, or 1000 doors — there are only two now, when you make your choice.
It's not hard at all. Humans are not perfectly rational beings. It's not purely about odds, it's about emotions and psychology. In a version where there's no previous selection and it's 50:50, than I pick one and live with it. In the canonical example, the previous selection means that the contestant has already staked their claim, and changing it to the loosing door would have a different emotional response than a simple 50:50 shot with no previous selection. There's a reason why Roulette shows you the last X spins and whether they're odd/even, red/black... because humans make the assumption that the last disconnected data point somehow impacts the current one.
You're also glossing over the fact that there was a 1:3 chance you picked the right door to begin with, and 2:3 chance the "other" door was right. The last round isn't 50:50, as you so claim, because there was prior information
This is the same reason why the 1000 door example helps explain things; because the math is fundamentally the same, yet significantly more imbalanced. We can also think about how we'd feel about our selection in the 1000 door version, which is likely significantly less confidence, and therefore more likelihood of switching. Whether it's 3 doors, or 1000, the math still say switching is optimal, and our psychology and emotions deal with the choice of switching different in each case.
Had to vs happened to makes a big difference.
Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.
Does it really change anything?
You had a 1/1000 chance of picking the right door at once.
Monty had a 1/1000 chance of opening only empty doors.
He had a 998/1000 chance of opening a door to a car, but that didn't happen.
So it's either of the other two cases, with equal probabilities.
If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases.
If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.
From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.
Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).
The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.
There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
I never liked the 10^x other doors version either. My thought was, what say me and a friend play a the same time, and pick the same initial door? If I NEVER change, and he ALWAYS does... I'm still going to win 1/3 of the time, and if Monty always shows the goat, ONE OF US has to win, so if I'm 1/3 by not switching, he HAS to be 2/3 by switching.
The problem stated with 3 doors is hard, with 5 is still not quite there for many. 100, 100, 10_000_000... becomes almost absurd.
In the tree it becomes blindingly obvious that the probabilities are.
That still leads to the wrong answer, because he is retiring one of the losers, leaving a 50:50 chance of getting a prize.
The act of retiring leads to the wrong answer, so it does not aid in understanding.
I came up with another way to visualize it that made more sense than increasing the number of doors, but it requires that one accept the fact that your chances of winning are 1/3 if you don't switch, and you know Monty always shows a losing door.
You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer.
Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.
... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the closed door that definitely still had a goat behind it).