[damocles]

Yor_on this is not exactly how the Monty Hall problem works, and it does make a significant difference.
The point is that if you first choose a door and it is wrong, then his hand is forced -- he has only one choice of door to open, and if you change you will be choosing the correct one. So with three doors your chance goes from 1/3 if you stay to 2/3 if you change.
With 100 doors, you have only a 1% chance of choosing the correct door to a 99% chance if you swap (because the presenter has absolutely no choice of which doors to open if you are not correct).

[yor_on]

Not sure what you mean there Damocles? I want to do it as the original thought experiment, the only thing I change is to get informed instead of seeing him opening that door. To me it shouldn't matter for it if he remove that door, leaving two choices for me, before I get there, as long as my information of what he did is valid? Or, does that matter, and if so why?

[damocles]

It matters because if you choose an incorrect door before he makes his choice, his choice is restricted -- so much restricted that it is a "gimme" in fact.

[damocles]

Here it is spelled out:

There are three equally likely possibilities:

Case 1: Goats Goats Prize
Case 2: Goats Prize Goats
Case 3: Prize Goats Goats

Suppose that you pick the right hand box.
Case 1: if you hold you will get the prize, and by swapping you will lose it
Case 2: when you pick the right hand box, the presenter will be forced to reveal the left hand box, and by swapping you will win the prize
Case 3: when you pick the right hand box, the presenter is forced to reveal the centre box, and by swapping you will win the prize

[yor_on]

His choice of doors you mean? Assume him to be informed of my choice then. That leaves him the same choices as in the original experiment. Although that wasn't specified in the original that I saw? Otherwise it seems the exact same experiment to me. The door he choose will be 'gone' no matter what choice I make, and for the doors left, be they 99 or two, I already made a choice in both cases. The only thing that differs is him removing the one he opened before I arrive.

Although it seem simpler describing using three doors to me

[yor_on]

The point I'm trying for is that the mathematics should be the same, as I have information of what door he picked. If he need to know what door I choose before a switch I'm not sure, although if he won't know he might open that one, destroying the example. But the rest seems the exact same to me, only differing in that the door he picked no longer is there although we both have information about it being there before, and that what it held was a goat.

[dlorde]

Alternatively getting informed of which door he opened to then remove, leaving two choices for me in which I first choose one door of two, to then switch it. Would my chances improve?

We already covered this. If you know there isn't a prize behind a particular door, you know its not a possible choice, so you only choose between the doors that might conceal a prize. If there's two doors and one prize, your chances are one in two. If you choose one of the two doors and then switch, it doesn't change the odds.

[dlorde]

Why wouldn't it give the same odds?
the situation is the same as if I stood before those three doors, seeing him open the one to the right, finding a goat? Instead of being there I get informed of what door that was. Ahh, I think I see, I didn't make that choice before getting informed

Yes; the situation isn't the same. If you know a particular door doesn't have a prize, you're not going to choose it, it's no longer part of the game.

[yor_on]

You can't have it both ways

Either the way I describe is equivalent to the original experiment, only one door missing as I arrive to switch my original choice, or it isn't equivalent. To me it actually is equivalent.
=
Ahh "Yes; the situation isn't the same." Sorry, saw the 'yes' first

If one want to define the mathematics on what doors that really is existent at the time I arrive you're putting a lot of weight on what exist, less on the mathematics being equivalent. As the situation is the exact same, except that instead of me standing there, watching him choose a door, I'm on my way to the game. You could imagine me seeing him on a television, or someone informing me per telephone. Otherwise it should be the exact same as it seems to me, although he remove the door he opened before I arrive.

If you now assume that the odds change because of the removal of a door that we both know to be wrong, then it seems to me that you also have to assume that 'kismet' steps in, to rearrange what's behind the two doors that's left, somehow?

Which then, assuming two possibilities (doors) left, give my first choice the same weight as the switch would have given in the original experiment
=

(Eh, the last is a small joke.)

[CliffordK]

If Schrödinger's cat is both alive and dead, or perhaps half dead and half alive.

Does one still have to feed it?

[yor_on]

Yes, but only the alive part..

[dlorde]

You can't have it both ways

Can't have what both ways?

Either the way I describe is equivalent to the original experiment, only one door missing as I arrive to switch my original choice, or it isn't equivalent. To me it actually is equivalent.

You've lost me - what do you mean be 'one door missing as I arrive to switch my choice' ? You're there the whole time. You choose one of three unknowns, which means you have one chance in three of having chosen the prize, and then one of the other unknowns is shown not to have a prize. You then decide whether to switch. The remaining unknown has one chance in two of having the prize, which makes it worth switching to from your one in three choice.

If you don't choose until after one of the unknowns has been shown not to have a prize, you then have a choice of two unknowns, one of which has a prize. Your chance of the prize is one in two, and doesn't change if you decide to switch.

If one want to define the mathematics on what doors that really is existent at the time I arrive you're putting a lot of weight on what exist, less on the mathematics being equivalent.

I don't know quite what maths you're referring to, but if the maths doesn't match what exists, i.e. reality, you've probably made a mathematical error.

As the situation is the exact same, except that instead of me standing there, watching him choose a door, I'm on my way to the game. You could imagine me seeing him on a television, or someone informing me per telephone. Otherwise it should be the exact same as it seems to me, although he remove the door he opened before I arrive.

If he removes a door that doesn't have a prize, there are two doors left, one of which has a prize. When you choose one of the two doors, you have a one in two chance of the prize.

If you now assume that the odds change because of the removal of a door that we both know to be wrong, then it seems to me that you also have to assume that 'kismet' steps in, to rearrange what's behind the two doors that's left, somehow?

The odds change because there are fewer choices. That's the point of the 100 door explanation. If there is one prize and a hundred doors, and you choose one door, you have one chance in a hundred of the prize. If all the other doors except one are shown not to have the prize, that one has a 99 in 100 chance of having the prize. If you switch, you're choosing a 99 in 100 chance over your original 1 in 100 chance.

I'm not quite sure where your difficulty lies, but a surprising number of people do find it confusing.

[yor_on]

Well, the mathematics won't care where you, or the door, are. As long as you're informed about the game as I see it. That simple..
=

"You can't have it both ways" referred to both your posts before, misread you there, and commented on that in the post, take a second look under the "=".

[dlorde]

Well, the mathematics won't care where you, or the door, are. As long as you're informed about the game as I see it. That simple..

Post the maths so I can see what you mean. The way I see it, the maths you use when you have three choices is the same maths you use when you have two choices, but the parameters change, so the results are different.

"You can't have it both ways" referred to both your posts before, misread you there, and commented on that in the post, take a second look under the "=".

OK.

[yor_on]

But the parameters didn't change, you know them just as good as if you had been standing in front of three doors the whole time, and that's my point.

[yor_on]

And no, I think you're reading me wrong assuming that I don't get the example. What I'm wanting to discuss is whether you can assume the odds to still be there after that 'one door' is gone too. And I presume that you should be able to, assuming that you have the same information as if standing in front of those doors the whole time. Otherwise it becomes a example of a mathematics based not on 'information', instead based on? Tactile reality? As you then should need all of those doors existing, to be able to 'switch' door for getting those better odds, in the end.

[damocles]

His choice of doors you mean?

Yes

Assume him to be informed of my choice then. That leaves him the same choices as in the original experiment.

The whole point is that it does not! He may no longer choose the box that you have chosen, and if you are wrong in your (original) choice, that is forcing him to reveal the location of the prize.

[dlorde]

But the parameters didn't change, you know them just as good as if you had been standing in front of three doors the whole time, and that's my point.

The parameters of the number of choices you have for a chance of the prize changes from 3 in one case to 2 in the other. Surely that's obvious?

I think we may be talking at cross-purposes. If you post up the maths you have in mind, or explain precisely the situations you're excerpting in your comments, it might help. As it is, I'm trying to make sense of ambiguous snippets such as "you know them just as good as if you had been standing in front of three doors the whole time".

[dlorde]

What I'm wanting to discuss is whether you can assume the odds to still be there after that 'one door' is gone too.

Your question is opaque. Please clarify what you mean - which odds are still where? which 'one door' is gone too? You seem to be thinking aloud but not communicating clearly.

And I presume that you should be able to, assuming that you have the same information as if standing in front of those doors the whole time. Otherwise it becomes a example of a mathematics based not on 'information', instead based on? Tactile reality? As you then should need all of those doors existing, to be able to 'switch' door for getting those better odds, in the end.

I can't make sense of that. The maths is quite simple. When you choose one of three doors blind, you have a 1 in 3 chance of the prize. When a non-prize door is then revealed or removed, the remaining door (that you didn't choose) has a 1 in 2 chance of having the prize. Therefore you're better off switching to it.

Why would assuming anything about the odds make a difference? The odds are fixed.

[yor_on]

It's the way I think of it dlorde. As information. And I see your point, but I'm trying to see why we would get those extra odds, and to me that is about information. That's also why I made the example.  The one I presented with a hundred doors is what I call two 'systems' in where you artificially split it in two, treating it as probabilities for each system to contain the prize. And you doing so is dealing in information, is that very hard to understand?