Are you serious? Are you buying this? Ok - let me make this easy: There NEVER WAS a 33% chance. Ever. The 1-in-3 choice is a ruse. No matter what door you choose, Monty has at least one door with a goat behind it, and he opens it. At that point, you are presented with a 1-in-2 choice. The prior choice is completely irrelevant at this point! You have a 50% chance of being right, just as you would expect. Your first choice did absolutely nothing to influence the outcome! This argument reminds me of the time I bet $100 on black at a roulette table because it had come up red for like 20 consecutive times, and of course it came up red again and I lost my $$. A guy at the table said to me "you really think the little ball remembers what it previously did and avoids the red slots??". Don't focus on the first choice, just look at the second - there's two doors and you have to choose one (the one you already picked, or the other one). You got a 50% chance. (by the way - sorry if I posted this twice?? Or in the wrong place?)
Why would you need more than plain English to intuitively grasp Monty-Hall-type problems?
Take the original Monty Hall 'Dilemma'. Just imagine there are two candidates, A and B. A and B both choose the same door. After the moderator picked one door A always stays with his first choice, B always changes his choice to the remaining third door. Now imagine you run this experiment 999 times. What will happen? Because A always stays with his initial choice, he will win 333 cars. But where are the remaining 666 cars? Of course B won them!
Or conduct the experiment with 100 doors. Now let’s say the candidate picks door 8. By rule of the game the moderator now has to open 98 of the remaining 99 doors behind which there is no car. Afterwards there is only one door left besides door 8 that the candidate has chosen. Obviously you would change your decision now! The same should be the case with only 3 doors!
There really is no problem here. You don’t need to simulate this. Your chance of picking the car first time is 1/3 but your chance of choosing a door with a goat behind it, at the beginning, is 2/3. Thus on average, 2/3 of times that you are playing this game you’ll pick a goat at first go. That also means that 2/3 of times that you are playing this game, and by definition pick a goat, the moderator will have to pick the only remaining goat. Because given the laws of the game the moderator knows where the car is and is only allowed to open a door with a goat in it. What does that mean? That on average, at first go, you pick a goat 2/3 of the time and hence the moderator is forced to pick the remaining goat 2/3 of the time. That means 2/3 of the time there is no goat left, only the car is left behind the remaining door. Therefore 2/3 of the time the remaining door has the car.
I don't need fancy visuals or even formulas for this. Do you really?
Are you serious? Are you buying this? Ok - let me make this easy: There NEVER WAS a 33% chance. Ever. The 1-in-3 choice is a ruse. No matter what door you choose, Monty has at least one door with a goat behind it, and he opens it. At that point, you are presented with a 1-in-2 choice. The prior choice is completely irrelevant at this point! You have a 50% chance of being right, just as you would expect. Your first choice did absolutely nothing to influence the outcome! This argument reminds me of the time I bet $100 on black at a roulette table because it had come up red for like 20 consecutive times, and of course it came up red again and I lost my $$. A guy at the table said to me "you really think the little ball remembers what it previously did and avoids the red slots??". Don't focus on the first choice, just look at the second - there's two doors and you have to choose one (the one you already picked, or the other one). You got a 50% chance.
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Your analogy doesn't hold, because each spin of the roulette wheel is a separate trial, while choosing a door and then having the option to choose another are causally linked.
If you've really thought about XiXiDu's analogies and they haven't helped, here's another; this is the one that made it obvious to me.
Omega transmutes a single grain of sand in a sandbag into a diamond, then pours the sand equally into three buckets. You choose one bucket for yourself. Omega then pours the sand from one of his two buckets into the other one, throws away the empty bucket, and offers to let you trade buckets.
Each bucket analogizes to a door that you may choose; the sand analogizes to probability mass. Seen this way, it's clear that what you want is to get as much sand (probability mass) as possible, and Omega's bucket has more sand in it. Monty's unopened door doesn't inherit anything tangible from the opened door, but it does inherit the opened door's probability mass.
"Your analogy doesn't hold, because each spin of the roulette wheel is a separate trial, while choosing a door and then having the option to choose another are causally linked."
No, they are not causally linked. It does not matter what door you choose, you don't influence the outcome in any way at all. Ultimately, you have to choose between two doors. In fact, you don't "choose" a door at first at all. Because there is always at least one goat behind a door you didn't choose, you cannot influence the next action, which is for Monty to open a door with a goat. At that point it's a choice between two doors.