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Author Topic: What physics and math topics do people find hardest to grasp?  (Read 18122 times)

Offline yor_on

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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.
 

Offline dlorde

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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.
« Last Edit: 12/05/2013 17:35:59 by dlorde »
 

Offline dlorde

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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.
 

Offline yor_on

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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.)
« Last Edit: 12/05/2013 18:05:37 by yor_on »
 

Offline CliffordK

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

Does one still have to feed it?
 

Offline yor_on

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Yes, but only the alive part..
 

Offline dlorde

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You can't have it both ways :)
Can't have what both ways?

Quote
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.

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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.

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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.

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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.
« Last Edit: 12/05/2013 18:30:36 by dlorde »
 

Offline yor_on

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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 "=".
« Last Edit: 12/05/2013 18:45:36 by yor_on »
 

Offline dlorde

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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.


Quote
"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.
 

Offline yor_on

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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.
 

Offline yor_on

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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.
 

Offline damocles

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His choice of doors you mean?
Yes
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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.
 

Offline dlorde

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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".
 

Offline dlorde

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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.

Quote
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.
 

Offline yor_on

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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?
 

Offline yor_on

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There is no law deciding what door you choose. But as soon as you've done, in this example, you made a system out of it. The game leader opening one of the 'two' doors that's left, according to you, is in reality opening one door of three. The definition builds on the information you get from him opening that door, and the way you split it into two systems. If the odds get better by a split, and 'a later shift' then I want to see why. I can do the math, but I can't see how to describe it, other than this way.
 

Offline damocles

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Umm...

When I first joined this debate about the Monty Hall problem, I said that the main problem that I had with it was why so many authorities were saying that the odds increased from 1 in 3 to 1 in 2 if you swapped. In fact it increases from 1 in 3 to 2 in 3.
Wikipedia is one source that has the right answer:
http://en.wikipedia.org/wiki/Monty_Hall_problem
My reply #23 on this thread clearly points out (to my way of thinking) why this is the case. I simply cannot see why others cannot see it this way.
 

Offline yor_on

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so how would you define my thought example, it being the exact same amount of information, with one difference, the game leader instead of leaving the door (he opened), removing it all together. Would the switch then become meaningless Damocles?
« Last Edit: 13/05/2013 14:48:42 by yor_on »
 

Offline yor_on

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"The correct answer, that players who swap have a 2/3 chance of winning the car and players who stick have a 1/3 chance of winning the car, is based on the premise the host knows which door hides the car and will always reveal a goat but never the car. If the player initially selected the door that hides the car (a 1-in-3 chance only), then both remaining doors hide goats, the host may choose either door, and switching doors loses. On the other hand, if the player initially selected a door that hides a goat (a 2-in-3 chance), then the host has no choice but to show the other goat, and switching door wins for sure."

This reasoning tells me that I don't need to be there, and that my thought example is sound. Your link Damocles.
 

Offline dlorde

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Umm...

When I first joined this debate about the Monty Hall problem, I said that the main problem that I had with it was why so many authorities were saying that the odds increased from 1 in 3 to 1 in 2 if you swapped. In fact it increases from 1 in 3 to 2 in 3.
Wikipedia is one source that has the right answer:
http://en.wikipedia.org/wiki/Monty_Hall_problem
My reply #23 on this thread clearly points out (to my way of thinking) why this is the case. I simply cannot see why others cannot see it this way.
You're right of course; if your pick has one chance in three, the remaining option must have two chances in three. I must have been distracted with odd considerations of two door examples...
 

Offline yor_on

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Heh, I now know more of this weird Monty Hall problem than I ever wanted to know :) But it was very weird, and treated as information you might state that it had a hidden parameter, which to me then would be the game leader never opening the door with the car, knowing which door it was. And naturally we have to assume that the guy choosing a door in the beginning must inform the game master about which one, as it otherwise could be one containing a goat that the game master also might open. Which in that case should mean that the guy would stand before a 50/50 chance of getting it right. If now I got it right :) weird stuff.
 

Offline bizerl

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THat's all very well, but it doesn't tell me how I can win the $200 000 on "Deal or No Deal".

 ;D
 

Offline Pmb

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That's a weird one alright. Saw someone explain it with a hundred doors instead, suddenly making it make sense. You have a hundred doors to pick from, you pick one of them. Then the game leader opens 98 of the other doors showing you nothing in them but goats. Now the question becomes one of keeping your original one that you picked randomly out of a hundred, not opened, or use the last door out of 99 that the game leader opened? It's a question of odds, and you picked one randomly from a hundred closed doors, but the 'other side' of it is the one where 98 doors was opened to find nothing, one left. It's like two games, the one you had from the beginning being the hardest to guess, wheres the one the game leader had being the 'foolproof' one. In reality it can't be foolproof as it could be your door too, but imagining it as two separate games makes it easier to see the reasoning.

and it is weird as you could imagine yourself not choosing any of those doors, waiting until the game leader opened 98 of them, then having two doors left to choose between. In that case you would have a 50/50 % probability of getting the right one, as I see it. Statistics as magic? :)
It's simple, really. As time progresses you know more about the odds of winning. There is never a reason to change the doors.
 

Offline Pmb

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Thankyou Yor_on! your example just highlights my reasoning! Because you only had a 1 in 100 chance of getting the first door right, then you are 99% sure if you change doors that you will be right after the game leader has opened 98 of them
That's not how probability works. Take a guess out of the 100 doors. Your probability of guessing right is 1/100. A door is opened and its empty. Regardles of whether you keep or change doors the probabiligy will be 1/99 of choosing the right one, and so on. This is different if you were playing the lottery. When playing the lottery always play the same number since its your goal to win in your lifetime, not merely today even if the chances of the new number you pick has the same probability of winning as any other number. Each problem is specific and needs to be addressed in each case. In the Montey Hall problem the winning door is never changed whereas in the lottery problem the number is always changed.
« Last Edit: 14/05/2013 13:44:23 by Pmb »
 

Offline Pmb

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If a cat can be half dead, and half alive,

Does it still need to be fed?
If it's Schrodinger's cat, it's both dead and alive, so you only need to feed the living version. You bury the dead version.

I once learned how to mentally calculate the day of the week of any given Georgian calendar date. It was too complicated and wasn't useful enough, even as a party trick, to remember once the novelty wore off (on a Thursday).
I disagree with these interpretations of quantum mechanics. A cat is a macroscopic animal whereas an atom is not. A cat is either alive or dead and not in a superposition of both.

Einstein was pointing this out when he asked "Is the moon there when nobody is looking?"  The answer is "Yes." Just like we know that the sun was there before life was here.
 

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