The Monty Hall problem

[Rasmus]

Provided he always opens a door and offers a switch, then it's always better to accept. Suppose there are 100 doors initially, then my initial choice has a 1% chance of being correct. Once Monty reveals a goat, the remaining 98 doors have a cumulative 99% probability of being correct, so the probability of any one of them being correct is 0.99/98, or about 1.01%. The general case is that the initial door chosen from n has a 1/n chance, whereas the switch after eliminating one door has a chance of (n-1)/{n(n-2)}. This is the initial probability multiplied by (n-1)/(n-2), which must always be greater than 1.

Dave

This I understand, of course. (But thanks for the maths).

Does anything change, though, if after the first switch is made Monty opens another door (that was previously not singled out in any way, say) and then offers me to switch? What if he then offers me to switch back to my original door, specifically?

The first situation is just like a new game I think. But what about the second? What if there are no more doors left than the ones I'Ve already picked? (We can force Monty to always open doors that I hasven't picked yet, right?)

 

[humber]

Wait, what...?
No, the best that can happen to the contestant is Monty opening all doors but the one he's chosen initially and one other door. (Okay, it would be better to just be given the car, but that would defeat the purpose of the puzzle.)

Yes. That is right.

Here is the standard Mantra.
"If the car is not behind the contestant's door, then 2/3 of the time it is behind one of the other two. Monty opens the door to reveal a goat, so the chance for the contestant is 2/3 if he swaps, or 1/3 if he sticks.

Yes, but that is simply a numerical coincidence. Not a change in probability.
If the car is not randomly placed, such that the probability of finding the car on the first guess is 1/3. Then you can see what is really at work.

Change it so that there is a car, but the chance of it being behind Door 1 is 1/2. If the contestant picks that door, he wins 1/2 if he sticks, and 1/2 if he doesn't. It must be behind Door 2 or Door 3. Monty always opens the door with the goat, no matter what.

The 100 door argument relates the chance to the number of doors, but it is actually the initial distribution that determines the result. Three doors works because that conveniently leaves Monty with only one door as an option.
The other 97 doors are totally irrelevant.

Try it for three doors. With the following distributions.
Door 1 p = 5/8
Door 2 p = 1/4
Door 3 p = 1/8
Total p = 1 ( there is a car)

Door 1
Dave sticks P win = 5/8
Dave swaps P win = 3/8

Door 2
Dave sticks P win = 1/4
Dave swaps P win = 3/4

Door 3
Dave sticks P win = 1/8
Dave swaps P win = 7/8

Woah. Seven times! And doubling causes a furor. What a lot of nonsense and analysis based on a simple error.

Take the extremes. If the car is never behind Door 1, and Dave picks it, obviously if he sticks, his chance is zero. But, the car will still be behind either Door 2 or Door 3 no matter what the chance is at Door 1.
Monty opens the remaining door with the goat, and if Dave swaps he wins 100% of the time.

If the car is always behind Door 1, and Dave picks it, obviously if he sticks he wins 100% of the time. The car will not be behind either Door 2 or Door 3. Monty opens the door with the goat, and if Dave swaps he loses 100% of the time.

The outcome "doubles" because it just so happens that 2/3 is twice 1/3.
But all the is at work is the initial probability as the sticking chance ( as you would expect) and the complement of that, is the swapping chance.

All because Monty always opens the door where the car isn't. No paradox, or anything at all interesting. A simple 100% predictable deterministic process ( Monty's logic) driven by variable input. The output follows the input.

Above, you are just rambling on and actually skipping over the important points.
Assume 100 doors. As soon as the contestant picks one and Monty opens another door to reveal a goat - what exactly do you think is now 1/99?

Yes.

The chances of the contestant having picked the car are *still* 1/100 assuming that Monty did not open a door randomly but was restricted to revealing a goat. If Monty then reveals a second goat the chances that the contestant picked the car are *still* 1/100.

Why? There are only 98 doors to hide the car. So it must be 1/98.

... continue until Monty has opened 98 doors to reveal 98 goats: The chances that the contestant has initially chosen the car are still at 1/100.

No. When there are 3 remaining doors the chance is 1/3.

Should he swap for the last other remaining door? Yes, yes he should! Only one time out of a hundred games will Monty have picked a door at random that he will leave closed to the very end. 99 times he will not have had a choice about which door to leave closed.

No Monty knows where the car is, and IF he opens that one, it's game over, and as you said, defeats the point of the puzzle. He does not "avoid" other than that.

Interestingly enough, I think that if Monty offers to switch at any time it would be advisable to switch, since every revealed goat makes "all the other doors" look slightly better than "all the doors" you were initially choosing from. But I haven't done the math for that.
Now, it will get really interesting if Monty keeps offering you to switch your doors every time he reveals another goat. Especially once you reach the point where you can no longer chose a door that you haven't played with already! (Did I just come up with a new twist to the problem?)

[/quote]
It is always better to swap, but the more remaining doors, the smaller the chance of picking the car. Three maximizes that chance. Four means that at swap time, the contestant still has two choices. Two doors is not a game, so three is the default and best number of doors to pull off this silly gimmick.

 

[humber]

How did this get to be all about me?

It seems to me that, if there is a strictly deterministic process, in which the only choice that's ever available to Monty after I've chosen a door is which goat to reveal if I happen to have chosen the car, then switching is always a better strategy. If Monty has the freedom to choose a course of action after I've made my initial choice, then the problem becomes one of game theory rather than one of probability, and I need to understand Monty's aims in order to determine the best strategy. So what it all boils down to is the rather obvious conclusion that, in order to analyse a game, it's necessary first to know what game you're playing.

Dave

Variations are not interesting other than something to squabble about.
Game theory works as does Bayesian analysis and calculus, because they would...

It's not the rules that are the cause of the confusion, but the expression of the outcome as being a result of probability. It's not, because Monty is reliably deterministic, so the only variable is the input distribution.

Monty simply inverts the native 1/3 (p_win), 2/3 (lose) chance, to become
1/3 (lose), 2/3 (win). 1 - pwin

I think of all those simulations, where a programmer has written the code that is Monty, and then fed it with random data and expects the outcome not to follow that random input. Did a light not come on at any stage? An electronics engineer would laugh at that.

ETA:
Schematically it is like this. The contestant either accepts the initial state and sticks with it, or the complement upon swapping. So, if you are saying it a simple game, I could not agree more, but how did it blow up to such a fuss that books have been written about it?

http://www.internationalskeptics.com/forums/attachment.php?attachmentid=21869&stc=1&d=1303130667

 

[Jack by the hedge]

Here is the standard Mantra.
"If the car is not behind the contestant's door, then 2/3 of the time it is behind one of the other two. Monty opens the door to reveal a goat, so the chance for the contestant is 2/3 if he swaps, or 1/3 if he sticks.

Yes, but that is simply a numerical coincidence. Not a change in probability.

I don't know what you mean by a "numerical coincidence". Which numbers coincide? It is not a change in probability in that the probability of the original selection does not change, but it is a change from a selection with one probability to a selection with a different, greater probability.

If the car is not randomly placed...

Then the probabilities change. So what?

The outcome "doubles" because it just so happens that 2/3 is twice 1/3.
But all the is at work is the initial probability as the sticking chance ( as you would expect) and the complement of that, is the swapping chance.

Yes. Why do you imply anyone is fixated on the word "doubles"? That happens to be the ratio of the two probabilities in the 3 door case. We all know that. So what is your point?

All because Monty always opens the door where the car isn't. No paradox, or anything at all interesting. A simple 100% predictable deterministic process ( Monty's logic) driven by variable input. The output follows the input.

Yet curiously a large number of people - some of them very smart people - have refused to believe it at first. Hence the 100 door variant, which merely makes it more obvious for many people.

Assume 100 doors. As soon as the contestant picks one and Monty opens another door to reveal a goat - what exactly do you think is now 1/99?

Yes.

What?

The chances of the contestant having picked the car are *still* 1/100 assuming that Monty did not open a door randomly but was restricted to revealing a goat. If Monty then reveals a second goat the chances that the contestant picked the car are *still* 1/100.

Why? There are only 98 doors to hide the car. So it must be 1/98.

You do not understand probability. Monty can open all but one of the other doors without telling you anything you don't already know. You chose your door from the original 100, not from the 98. Your probability of being right has not changed.

... continue until Monty has opened 98 doors to reveal 98 goats: The chances that the contestant has initially chosen the car are still at 1/100.

No. When there are 3 remaining doors the chance is 1/3.

Again this is wrong. Test it yourself with the 52 card version. See if you can beat Jeeves 1 in 3 times. Remember that to do so you must draw the ace of spades 1 in 3 times.

 

[Jack by the hedge]

It's not the rules that are the cause of the confusion, but the expression of the outcome as being a result of probability.

I disagree. The outcome is quite properly expressed as a probability. People's initial confusion is caused by their misunderstanding that the probability of the initial choice being right does not change from 1/3 to 1/2 when Monty reveals a goat.

 

[Rasmus]

Yes. That is right.

Here is the standard Mantra.
"If the car is not behind the contestant's door, then 2/3 of the time it is behind one of the other two. Monty opens the door to reveal a goat, so the chance for the contestant is 2/3 if he swaps, or 1/3 if he sticks.

That is not a "mantra", it is a simple summary of the facts of the Monty Hall Problem.

Yes, but that is simply a numerical coincidence. Not a change in probability.
If the car is not randomly placed, such that the probability of finding the car on the first guess is 1/3. Then you can see what is really at work.

Was that supposed to go there?

If the car was not randomly placed I find it hard to see how your chances ever could be 1/3. Anyway, I m fairly certain I do see what is at work.Anyway, I m fairly certian I do see wha you is at work.Anyway, I m fairly certian I do see what is at work. I continue to be torn between thinking either don't or you do but are struggling with the language involved.I continue to be torn between thinking you either don't or you do but are struggling with the language involved.

Change it so that there is a car, but the chance of it being behind Door 1 is 1/2. If the contestant picks that door, he wins 1/2 if he sticks, and 1/2 if he doesn't. It must be behind Door 2 or Door 3. Monty always opens the door with the goat, no matter what.

Yes. If you accept the standard answer to the standard puztzle this seems rather trivial.

The 100 door argument relates the chance to the number of doors,

Yes. Which is true, assuming that the car is placed at random just like the puzzle stipulates. The more doors you have, the less likely you are to initially find a car placed behind one of them randomly.

(And that rather presupposes that "random" implies "equal distribution" which doesn't have to be the case at all.)

but it is actually the initial distribution that determines the result.

But with random placement of the car, the number of doors determines what the initial distribution looks like!

Three doors works because that conveniently leaves Monty with only one door as an option.
The other 97 doors are totally irrelevant.

No, they are not "irrelevant". They do not change the general principles involved nor the right answer to the puzzle - but they do change the numbers and chances quite significantly!

Try it for three doors. With the following distributions.
Door 1 p = 5/8
Door 2 p = 1/4
Door 3 p = 1/8
Total p = 1 ( there is a car)

Door 1
Dave sticks P win = 5/8
Dave swaps P win = 3/8

Door 2
Dave sticks P win = 1/4
Dave swaps P win = 3/4

Door 3
Dave sticks P win = 1/8
Dave swaps P win = 7/8

Woah. Seven times! And doubling causes a furor. What a lot of nonsense and analysis based on a simple error.

What error? This is quite obviously a puzzle much different than the original. Why should anyone be surprised that the answer is different?

Take the extremes. If the car is never behind Door 1, and Dave picks it, obviously if he sticks, his chance is zero. But, the car will still be behind either Door 2 or Door 3 no matter what the chance is at Door 1.
Monty opens the remaining door with the goat, and if Dave swaps he wins 100% of the time.

If the car is always behind Door 1, and Dave picks it, obviously if he sticks he wins 100% of the time. The car will not be behind either Door 2 or Door 3. Monty opens the door with the goat, and if Dave swaps he loses 100% of the time.

Yes. So?

The outcome "doubles" because it just so happens that 2/3 is twice 1/3.
But all the is at work is the initial probability as the sticking chance ( as you would expect) and the complement of that, is the swapping chance.

Nobody ever suggested that the chances would always double, independent of the number of doors used. (If they did, they were probably mistaken and most certainly wrong!)

All because Monty always opens the door where the car isn't. No paradox, or anything at all interesting. A simple 100% predictable deterministic process ( Monty's logic) driven by variable input. The output follows the input.

So?

Yes.

You do realize that this doesn't answer my question?

Why? There are only 98 doors to hide the car. So it must be 1/98.

No! The choice is made between 100 doors. We *know* that Monty will always open 98 doors to reveal a goat. We could play the game a million times and Monty would reveal 98 goats every single time. One times out of every hundret, the car will be behind the door chosen before Monty started his tedious task.

No. When there are 3 remaining doors the chance is 1/3.

Wrong!

Again: Sit down and play the game. Use matchboxes, a deck of cards or anything. Tel us what the actual numbers are!

No Monty knows where the car is, and IF he opens that one, it's game over, and as you said, defeats the point of the puzzle. He does not "avoid" other than that.

There is no "IF", since the puzzle states that he will not open that door: Monty does know where the car is, and his task is to open 98 doors that fullfill all the following conditions:

a) Door must not be the door picked by the player
b) Door must not have the car behind it
c) Door must have a goat behind it

(b and c are of course synonymous)

So if we play the game 100 times, then 99 times Monty will be forced to leave one specific door closed other than the door the player picked. Only in the rare case that the player picked the door with the car does Monty have a choice of which door to leave closed.

It is always better to swap, but the more remaining doors, the smaller the chance of picking the car. Three maximizes that chance. Four means that at swap time, the contestant still has two choices. Two doors is not a game, so three is the default and best number of doors to pull off this silly gimmick.

I can't parse what you're saying here.

What do you mean by "remaining doors"?

Is it "Doors that were in the game initially", or "doors not yet picked by you or opened by Monty"?

In the latter case, it is tricially true that the game is better for you if Monty opens as many doors as possible.

But *IF* he does that, then it is better for you to have as many doors as possible in the game initially. If Monty is going to bring the game down to 2 doors, you would want him to start with a million doors rather than three.

 

[humber]

I don't know what you mean by a "numerical coincidence". Which numbers coincide? It is not a change in probability in that the probability of the original selection does not change, but it is a change from a selection with one probability to a selection with a different, greater probability.
Then the probabilities change. So what?

They don't.
The probability of car being behind any door is assumed to be 1/3. That is 1/(Number of doors). That results in the 2/3 that is assumed to occur because there are only two remaining doors, but isn't.
That latter part of the argument is wrong, and so is the 100 door variant, because the ratio of success to failure is initial probability and its complement.

If there are 3 doors, but one has the car 1/10 of the time, then the other 2 doors will have the complement, so the car 9/10 of the time
So
p sticking = 1/10
p swapping = 9/10

Not related to the number of doors, so the 100 door argument is not relevant.
ETA:
If the situation is reversed, so that the initial probability is 9/10 (or greater than 1/2) then sticking is the best option. From this argument, that inverts my initial and complement probabilities. I can counter that by randomly choosing the choice of first door. However, the assumption of the question is that my initial choice will be less than the complement.

Yes. Why do you imply anyone is fixated on the word "doubles"? That happens to be the ratio of the two probabilities in the 3 door case. We all know that. So what is your point?

Now you say that, but it is not the ratio of the "3 door case".
But the given probability at any door. If they differ for each door, so will the probability. Neither is related to the number "3" unless forced to be so.

Yet curiously a large number of people - some of them very smart people - have refused to believe it at first. Hence the 100 door variant, which merely makes it more obvious for many people.

That is the same nonsense about "intuition". Lawyers do not often know of elementary logic gates, and neither do many other "smart people"

Read this
http://204.14.132.173/pubs/journals/features/com-124-1-1.pdf

You do not understand probability. Monty can open all but one of the other doors without telling you anything you don't already know. You chose your door from the original 100, not from the 98. Your probability of being right has not changed.

One car and 100 doors.
The random chance is 1/100 when there a 100 doors. 1/50 when there are 50 doors and so on. Without Monty's action, each door has equal chance.
When there a 50, it is just as likely to be behind any door. When there are 3, the chance is 1/3.
However, Monty's strategy means that when there are 3, one door has p=1/3
and the other p=2/3. The probability does not actually change, but the conditions do, and that is only applicable when there are 3 doors, otherwise the initial probability at the chosen door does change.

Four unopened doors means p = 1/4 for the chosen door, 3 means 1/3, but two also means 1/3. See? That is the point about not changing.

Again this is wrong. Test it yourself with the 52 card version. See if you can beat Jeeves 1 in 3 times. Remember that to do so you must draw the ace of spades 1 in 3 times.

Again, a wrong argument give in support of another one.

Other than realising that the doors and Monty are a simple logic circuit,
Game Theory is the best explanation. It takes care of all secon-guessing or intent. Using the accepted question and even distribution;

1) I only know that Monty Hall always opens a door revealing a goat. That is all, no probability or anything else needed, save that a car is there.

2) I didn’t know what strategy Monty is going to use, other than he wants to keep the car, and that I want it.

3) A strategy of choosing a door uniformly at random, guarantees a win, but I don't know what that is.

4) If Monty opens a remaining door, then I win with the competent of the initial chance, whatever that is. I do not need to know if he would open my door. If it's not open, he didn't.
I can conclude that if it's not behind my door, then is it behind one of the the other two doors. The closed door may contain the car. I know my initial guess to be the minimum random chance, so swap to where it will be the complement of that initial choice.

3) On the other hand, if Monty hides the car at random, and randomly chooses to open a door if there is a choice, then I cannot win the car with a better probability than the complement ( the assumed 2/3)

4) My strategy gives me no worse than the complement, while Monty's strategy prevents me from doing better, and that proves they both my and Monty's are minimax strategies.

5) If Monty opens my door when the car is there, he loses. If it's not there he informs me that it is in one of the other 2, so I have p = complement/2 with each.
Monty can do nothing to reduce that, and it will happen only p=initial of the time.

Always swap, because my worst case (given assumptions) is p=1/2 or p=2/3 when swapping, whereas sticking is p=1/3.

 

[Vorticity]

humber:

You're not explaining yourself well. I have no idea what you're trying to say.

I've read through this entire thread, and it's clear that no one else here has any idea what you're trying to say either.

You need to find some better way of expressing yourself.

Here's a simple question for you:

In the standard 100-door case, what is a player's probability of winning if they stick? What is their probability of winning if they switch?

 

[SumDood]

If there are 3 doors, but one has the car 1/10 of the time, then the other 2 doors will have the complement, so the car 9/10 of the time
So
p sticking = 1/10
p swapping = 9/10

If that is the case, the placement of the car is not random, and thus is a poor example for this problem.

Four unopened doors means p = 1/4 for the chosen door, 3 means 1/3, but two also means 1/3. See? That is the point about not changing.

(my bolding)

I definitely do not see. How is the probability for two doors not 1/2?

 

[humber]

That is not a "mantra", it is a simple summary of the facts of the Monty Hall Problem.

It is wrong.

Was that supposed to go there?

Yes.

If the car was not randomly placed I find it hard to see how your chances ever could be 1/3.

The car can be placed with any frequency at any door.

Anyway, I m fairly certain I do see what is at work.Anyway, I m fairly certian I do see wha you is at work.Anyway, I m fairly certian I do see what is at work. I continue to be torn between thinking either don't or you do but are struggling with the language involved.I continue to be torn between thinking you either don't or you do but are struggling with the language involved.

........

Yes. If you accept the standard answer to the standard puztzle this seems rather trivial.
Yes. Which is true, assuming that the car is placed at random just like the puzzle stipulates. The more doors you have, the less likely you are to initially find a car placed behind one of them randomly.

Right, but that is not what happens when Monty's logic is there.
Contestant simply opens doors (can be)
3 doors 1/3 (2/3 the other doors)
2 doors 1/2 ( 1/2 the other door)

With Monty (always is)
3 doors 1/3 ( 2/3 the other doors)
2 doors 1/3 ( 2/3 the other door)

(And that rather presupposes that "random" implies "equal distribution" which doesn't have to be the case at all.)
But with random placement of the car, the number of doors determines what the initial distribution looks like!

They can all be different as long as the sum is 1.

No! The choice is made between 100 doors. We *know* that Monty will always open 98 doors to reveal a goat. We could play the game a million times and Monty would reveal 98 goats every single time. One times out of every hundret, the car will be behind the door chosen before Monty started his tedious task.

No, the chance will be 1/3 not 1/100. He works to only open goat doors.
Which is to say, it says nothing about the 3 door case, that the 3 door case does not already say. A car is always there at each game.

a) Door must not be the door picked by the player

Need not be to be to benefit from swapping.

b) Door must not have the car behind it

He loses.

c) Door must have a goat behind it

So he doesn't lose.

So if we play the game 100 times, then 99 times Monty will be forced to leave one specific door closed other than the door the player picked. Only in the rare case that the player picked the door with the car does Monty have a choice of which door to leave closed.

He does not want to open it, or he loses.

I can't parse what you're saying here.
What do you mean by "remaining doors"?
Is it "Doors that were in the game initially", or "doors not yet picked by you or opened by Monty"?

Door not picked by contestant.

In the latter case, it is tricially true that the game is better for you if Monty opens as many doors as possible.

It is always better for Monty to open any door. If it's the car, I win, if not, my chances rise.

But *IF* he does that, then it is better for you to have as many doors as possible in the game initially. If Monty is going to bring the game down to 2 doors, you would want him to start with a million doors rather than three.

But he only gives me the maximum information when there are 3 doors.
So yes, 3.

 

[humber]

If that is the case, the placement of the car is not random, and thus is a poor example for this problem.

No it means that the conclusion of the given example is arithmetically correct, but not the reason.

(my bolding)
I definitely do not see. How is the probability for two doors not 1/2?

[/quote]

Because of Monty. He opens only goat doors. It is more a matter of the distribution of each door. Without Monty, each door has equal chance.

But, when there are 3, and he opens the last goat door, one has the initial chance of 1/3 and the other 2/3. It his "avoidance" of the car that does that. Agree? But that is not strictly correct.

If there is initially a chance of 1/10 for Door 1, then the last unopened door will be 9/10, not 2/3.

The ratio of chance to number of doors is assumed, right, OK.

But if you ignore that, and try other distributions, you see that its the initial probability for one door and it's complement for the other.
When you think about it that way, whenever is that not the case?

What is the fuss? Monty assures that happens, that is all.

 

[Jack by the hedge]

It's disconcerting when you explain a mistake to humber and receive an exhaustive restating of your own point as if it were what he had said before.

Humber, please, just so we can stop going round in pointless circles, is it your view of the 100 doors case that:
If the car is placed randomly, and if you choose a door randomly, then the probability of the car being behind your door is 1/100 and, further, if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

 

[Dr.Sid]

Simply read the article about the problem on wikipedia. If you don't understand it, the problem is on your side.

 

[humber]

I won't be the first person to fail at convincing anyone about this problem.
I think that the offered solution is wrongly expressed, and that causes the difficulty. Not just the definitions of the conditions, but the way the solution is expressed, and with the unnecessary call to intuition.

It seems that the difficulty and objections are dependent on occupation and skill. That has been tested in a trial held in Nijmegen in The Netherlands.

Mathematicians tend to prefer the Bayes. Lawyers are hard to convince, Game Theorists like the games approach, and many puzzle writers like intuition. How ever, it is none, but a simple state machine.

Richard Gill, a lecturer at Leiden wrote:

" It seems to me that adding into the question explicitly the remarks that the three doors are equally likely to hide the car, and that when the quiz-master has a choice he secretly tosses a fair coin to decide, convert this beautiful paradox into a dull probability puzzle, which any right-minded amateur wouldn’t want to solve."

What I am trying in to say in addition, that if that explicit condition of equal distribution is removed, then you see the more general case, where the initial chance is p (not matter what that may be) and the other door is 1 - p, the compliment.

How trivial is that?

I can get the same answer as published that way, as by any other more complex or convoluted means.

Why has that spawned a monster, and produced papers like this?
(the abstract is enough)
http://www.ncbi.nlm.nih.gov/pubmed/12656295

 

[humber]

Simply read the article about the problem on wikipedia. If you don't understand it, the problem is on your side.

I understand it well enough. Do you?
BTW. I can implement that state machine. The car can be represented by a weight placed on a pressure switch, the others are empty, while actuators open and close the doors. Using nothing more than a few relays, I can mimic Monty. You will not be able to tell that from the puzzle Monty, no matter what you throw at it.

 

[humber]

It's disconcerting when you explain a mistake to humber and receive an exhaustive restating of your own point as if it were what he had said before.

Humber, please, just so we can stop going round in pointless circles, is it your view of the 100 doors case that:

If the car is placed randomly, and if you choose a door randomly, then the probability of the car being behind your door is 1/100 and, further, if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

No. It becomes 1/3, 2/3 ( for the assumed distribution).
That only goes to Savant's support for her argument, as does the matter of if Monty knows or not. That last one is definitely false.
It works for 3 doors. Opening doors until you get there, says nothing other than to support the mistaken interpretation.

The matter of "remaining the same" is about Erdos' objection. I have been unable to find out what he actually meant. Maybe it's subtle, but whatever is was, it has been drowned out by accusations that he is a stubborn bonehead. But, it looks to me that the simulations that are said to have silenced him, are not what they appear to be. That is, he was quite likely to have been given an explanation, that itself was a product of the programme.

What I think he meant, is that if two doors remain, the probability of the remaining closed will not change. He is right, because it doesn't.
It does not switch from being 1/3 when all doors are closed, to 2/3 when the goat is revealed. I can see how he sees that as violating some fundamental ideas of probability.
But, if you say the probability of the chosen door is p, and the remaining (unchosen) closed door is 1- p, there is nothing to discuss, because it always was that way, and the reason can be easily inferred. No question, of course!
Monty is deterministic, so it is not really a matter or probability, but a game in the "game theory" sense.
Conditional probability is IF then, and not WILL then.
Try again
Wiki
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the (conditional) probability of A, given B" or "the probability of A under the condition B". When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B.

It can't be said that the probability that something is certain to happen is anything but 1.
Writing it as 1 -p makes that clear.

 

[Rasmus]

No. It becomes 1/3, 2/3 ( for the assumed distribution).

Now it is clear: You don't get it.

 

[humber]

Now it is clear: You don't get it.

If you say not, then you are contradicting the usually accepted answer. I don't say that is not the case, but only the why. I was caught up in explaining something to you, that I think JBTH does not say is the case.

 

[Vorticity]

No. It becomes 1/3, 2/3 ( for the assumed distribution).

So, let's make sure I'm absolutely, 100% clear what you're saying here.

1) All 100 doors initially equally likely to have the car.

2) I pick a door at random.

3) Monty opens 98 of the remaining doors, revealing goats.

You say that my probability of winning if I stick with my original door is 1/3, and my probability of winning if I switch is 2/3?

If so, would this work for any number N > 2 of doors? A billion doors, say? I'd still have a 1/3 chance of winning if I stick?

Or have I misunderstood you?

Opening doors until you get there, says nothing other than to support the mistaken interpretation.

What is the mistaken interpretation?

The matter of "remaining the same" is about Erdos' objection. I have been unable to find out what he actually meant. Maybe it's subtle, but whatever is was, it has been drowned out by accusations that he is a stubborn bonehead. But, it looks to me that the simulations that are said to have silenced him, are not what they appear to be. That is, he was quite likely to have been given an explanation, that itself was a product of the programme.

What I think he meant, is that if two doors remain, the probability of the remaining closed will not change. He is right, because it doesn't.
It does not switch from being 1/3 when all doors are closed, to 2/3 when the goat is revealed. I can see how he sees that as violating some fundamental ideas of probability.
But, if you say the probability of the chosen door is p, and the remaining (unchosen) closed door is 1- p, there is nothing to discuss, because it always was that way, and the reason can be easily inferred. No question, of course!
Monty is deterministic, so it is not really a matter or probability, but a game in the "game theory" sense.

I literally cannot parse any of the above.

Speak slowly, using few words.

Read what you write before you post it. Please.

Conditional probability is IF then, and not WILL then.
Try again
Wiki
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the (conditional) probability of A, given B" or "the probability of A under the condition B". When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B.

It can't be said that the probability that something is certain to happen is anything but 1.
Writing it as 1 -p makes that clear.

The above is elementary probablity theory, which I learned in college. I don't see how it supports anything other than the "orthodox" view of the Monty Hall problem.

 

[humber]

So, let's make sure I'm absolutely, 100% clear what you're saying here.

1) All 100 doors initially equally likely to have the car.

2) I pick a door at random.

3) Monty opens 98 of the remaining doors, revealing goats.

You say that my probability of winning if I stick with my original door is 1/3, and my probability of winning if I switch is 2/3?

Oh, I am not going to discuss that. No, that is done and dusted.

"Suppose you’re on a game show, and you’re given the choice of
three doors: Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what’s
behind the doors, opens another door, say No. 3, which has a
goat. He then says to you, “Do you want to pick door No. 2?” Is
it to your advantage to switch your choice?"

The accepted answer is swap, because the probability is 2/3 for that door.
Thousands of references you can consult.
The questioner offers the 100 door example as support.
And "parse" can also mean "don't understand"
I am not interested in starting again with the MHP. Just the one point.