Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Pmb on 02/05/2013 04:54:28
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Hi everyone!
I'm trying to keep myself busy and am going to create some learning material for math and physics. I would like to ask everyone what the hardest parts of their learning in math and physics was and how we can help people overcome their difficulties.
Myself, I had a terrible time memorizing the multiplication tables. Even to this day I have trouble with it. I think its because I have a very poor memory. I learned that I sucked at arithmetic but was an ace in math because arithmetic is all about memorizing things and the algorithm to calculate numbers but math was easy for me because its abstract and visual and I'm good at those things.
When it comes to physics I had a terribly hard time with statistical mechanics. I think its because one has to have a solid grasp of combinatorics and I sucked at that.
This is the kind of thing I would like to ask you good folks to share even though I realize that it’s a hard thing to do. After all who likes to put out there what we suck at. I don’t mind because I’m no longer ashamed of it. I’ve come to understand and accept that part of myself. That way I can fight it.
If you have any suggestions on how I can overcome my difficulties in the above please share. Thanks!
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Makes me remember a good friend from long time ago Pete. He was gifted when it came to understanding the principles behind mathematics, and I was rather poor in it :) So he once said that mathematics could be understood as logical symbols, even things you could hold in your hands, that you by some defined rules could twist and put together in combinations and treat one way, or the other. I don't think he ever made something from it but it stuck in my mind as a good idea. Wonder if it would be possible? To make it a game that you could give to kids to play with and from that intuitively learn maybe not all rules, but some of the hidden logic regarding mathematics, whatever form it takes?
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It's been a while since I've thought about Maths and I think my main problem was not studying enough, but I can remember that whenever we learned a new formula, the teacher would take the effort to go through how it was derived from basic principles. I found this really helpful and especially useful in exams when I couldn't remember the formula, but could remember how to derive it.
The problem with education is that different ways work for different people. Some people might be better at learning a few basic principles and extrapolating them out for whatever purpose, other people might do better at memorising large amounts of facts to call upon when required.
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I have a very strong sense of pattern and spatial relationships. When I was quite young (about 5) my very wise father put up a chart on the back of the toilet door, where I would have plenty of time to contemplate it. It showed the relationships between numbers from 1 to 100 -- all of the numbers divisible by 5 were in 2 columns with an orange background shading. Those divisible by 8 were in blue squares, those divisible by 7 had a blue ring around them, divisible by 3 was a red font, by 9 a red font backed up with a red diagonal stripe, by 11 a black diagonal stripe running the other way.
This seemingly chaotic arrangement was capable of resolution into a number of patterns, especially as I was just learning to play chess at the time, and was fascinated by the knight's move. I found it expressed in the 8 times (10-2) and 7 times (20+1) tables. It started me out with a really good sense of number, based on an innate spatial sense.
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I recall doing Quantum Mechanics in my Physics degree. It took me 6 months to get my head around the mathematical concepts. In the end I passed but I really did not know what I was doing. It was no longer a Physics course; it became a Maths course.
One of the major problems that students have in doing the maths in the Physics course is that there must be certain prerequisites that must be achieved.
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He, hear what you're saying Lunar. You start with grasping some principles, then you meet obstacles confounding you, as advanced mathematics but without basic premises explained? And doing them you sort of lose sight of what you first thought of as being principles, just to come back to them after more time passed, as the mathematics suddenly starts to make sense again?
The wheel of life huh :)
And you must have had a good, and patient, teacher Bizerl. Myself I like finding out how people thought something up, it makes it so much easier to understand how they got their results.
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I really hate the Monty Hall problem. I just can't accept that changing doors makes a difference.
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from yamo:
I really hate the Monty Hall problem. I just can't accept that changing doors makes a difference.
What I cannot figure about the Monty Hall problem is why so many authorities say that the probability of success is only 1 in 2 if you change doors. The way I figure it is that you should try to pick the door that you are not expecting it to be behind, because if that fails (1 in 3) you will have a sure fire pointer to success (2 in 3).
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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? :)
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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
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Could an analogy to superstates be used? Rather than measuring where the Big Deal is, collapsing the function, we are measuring where it is not.
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If a cat can be half dead, and half alive,
Does it still need to be fed?
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What I find difficult about maths is what do all the strange symbols represent I understand about half of them and have recently learnt about Bras and Kets but am always quite defeated by matrix's.
Arithmetic is no problem having grown up with the strange imperial units and £.S.D money which hones ones skills.
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On a good day, with a following wind, I can do the maths for calculating angles in 512 dimensions (and, with the aid of a computer, I used to do it a lot).
But I have no real idea what it means.
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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).
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heh :)
I seriously think we're all in the galactic teapot, best described by the hitchhikers guide to the galaxy.
We should present it to the UN. I'm sure they could use it..
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About the Monty Hall problem.
1. we have differently sized infinities defined mathematically. Would now this reasoning give me a better probability of finding something in a lesser infinity, as you can see that as open and then taking aways 'some room' :)
2. Imagine i have two doors to choose between. Somebody tells me that there was three doors before I got there, but one got taken away containing a goat, to the right side. Will this guarantee better odds for me if I too pick the right side? If it does, isn't this a hidden parameter? Information?
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About the Monty Hall problem.
1. we have differently sized infinities defined mathematically. Would now this reasoning give me a better probability of finding something in a lesser infinity, as you can see that as open and then taking aways 'some room' :)
Sorry, I don't understand what you're asking here.
2. Imagine i have two doors to choose between. Somebody tells me that there was three doors before I got there, but one got taken away containing a goat, to the right side. Will this guarantee better odds for me if I too pick the right side? If it does, isn't this a hidden parameter? Information?
No, the probability of you choosing the prize depends on the choices currently available to you. If you have two options and one has the prize, your chances are 1 in 2 (50:50). If there had been a hundred other options removed before you got there, it would make no difference, your chances with the two choices left are still 1in 2. But if you'd chosen from the options before any were removed, your chances would clearly be 1/number of options, and by switching after the 'dead' options were removed, you'd improve your chances to 1 in 2.
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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 :)
Then we do 2. again, this time me picking one of the other doors, before getting informed of what door he opened, before removing it, finding that goat. Would my chances now increase in a switch?
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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?
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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).
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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?
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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.
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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
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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 :)
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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.
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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.
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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.
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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.
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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 :)
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(Eh, the last is a small joke.)
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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?
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Yes, but only the alive part..
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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.
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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..
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"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 "=".
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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.
"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.
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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.
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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.
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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.
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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".
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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.
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.
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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?
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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.
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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.
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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?
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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.
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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...
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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.
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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]
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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.
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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.
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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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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.
That's not how the game works! If you choose the wrong door, the host is obliged to show you where the prize is by revealing the 98 doors where he knows the prize is not, giving you a sure pointer to the prize. So your chances of winning are 1% if you stand, but 99% if you swap.
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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.
Can you explain the reasoning here? you surely have the same chance whether you change your number each time or not.
As I understand it, the only criteria for selecting a lottery number is to avoid one that other people might be likely to pick too; it doesn't help your chances, but if you do win, you're less likely to be sharing the prize.
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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.
At what point from atom to cat do you draw the line? Objects visible to the eye (40 microns, ~about 10 trillion atoms) have apparently been put into quantum superposition (see quantum microphone (http://www.scientificamerican.com/article.cfm?id=quantum-microphone)). I don't doubt that decoherence would rule out any animate organic creature, but what about viruses or bacteria at cryogenic temps? or a cryptobiotic tardigrade?
From the POV of size alone, it would be interesting to discover the practical size limit for a measurable duration of superposition.
The comments about Schrodinger's Cat were just whimsy.
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Reading my electricity bill :)
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That one is weird too, there are two ways and I don't know which one, or if both, are true, but they are contradictory. If i flip a coin and get a hundred tails then that's a very unlikely result. I would call that a very low probability. But the coin is in a way reset each time you flip it, meaning that it shouldn't care for those other results, instead treating each flip as a new instance, the other results having no bearing on the outcome. So one does not lead to the other, but statistically you should have a 50/50 probability of tail and head, flipping it in a series long enough. Maybe the question should be if there is a possibility of defining when a improbable series become so overextended, only tails, that you could expect a probability of it to come up as head? It makes me think of feigenbaum's constant this one, wondering if there is some mathematical way to define a 'overextension'?
that as common sense tell me that one thousand flips, all coming out tail, should be a very seldom seen pattern. Against it you have the equally valid point that each new flip should be counted as starting anew? Looked at that way there is no reason to choose any numbers before any other, although you might want them to be those that people don't tend to pick. On the other hand, there should be some breaking point to those tails, when they become so improbable that ??
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Keep jumping over words, and, eh, spellings :)
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One can think of it this way maybe, the pattern of a thousand tails is no more uncommon that any other combinations of set patterns like tail - tail - head, if repeated over and over, for a thousand flips. Which then just should make one thousand tails, or heads, uncommon, because it is so easy to recognize for us. That will make the one defining it as being 'reset' with each new flip the one making most sense. What i mean is that if you count the way a pattern evolve over time, flipping a coin, that all patterns possible have a equal chance of evolve, singling out no pattern as more probable. And a pattern would then be whatever way you found head and tails arrange themselves over a thousand throws.
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If a coin were flipped and came up tails one thousand times in a row, I would be inclined to bet on it coming up tails the next time, because I would regard the previous one thousand times as statistical evidence in favour of the coin (or the toss) being biassed.
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Yeah, I know Damocles :) Can't help it going against my basic instincts though. Which is why I don't gamble, can't trust those instincts ::))
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Then again, thinking of it as patterns, the probability of it coming out 'any which way' should still be 50/50, shouldn't it, as no pattern is more probable than any other? (over 1000, and 1, flips) . They should all be equally probable as it seems to me, otherwise we find a bias? And I think that one can look at the constant 'reset', as well as the 'equally probable patterns', as a logical proof for that, the coin flipped enough times, also must give us a equal amount of heads and tails.
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There is one more thing though, calling a outcome randomly before the flip. Would that give you a statistically better chance to be correct, more than a 50% probability, than always calling a same, set, outcome? As calling 'tail' before each flip? And if that would be true (introducing a random call before the actual throw), why would that be?
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Seems to me that if it's a fair (random) coin, you'll average 50% success calling the same each time or calling randomly. Each call has a 50% chance of being right, regardless of previous calls.
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Yes, that's what I thought too, but it seem to have been a professor in statistics, meaning that you by introducing randomness on your side too, will get a better probability? I don't think it can be right myself, but I'm not sure?
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If you can consistently get better than 50%, the coin sequence isn't random. Your choice (prediction) can't affect the coin odds.
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I think people have a lot of trouble with statistics (e.g. the Monty Hall problem is a very classic example), but statistics in general seems to fry people's brains.
Relativity, again, people usually can't hack it.
QM, practically nobody really has much clue!
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That's not how the game works! If you choose the wrong door, the host is obliged to show you where the prize is by revealing the 98 doors where he knows the prize is not, giving you a sure pointer to the prize. So your chances of winning are 1% if you stand, but 99% if you swap.
Then what I said was caca?? ;)
I don't understand. Suppose that the prize is behind door number 2. If I choose door number 1 then he reveals that doors numbers 3-100 have nothing behind them? If so then it seems to me that you have to choose door number 2 after that and have 100% chance of winning.
I don't think I understand that game. In any case I'm not interested. It's getting off topic for me.
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Can you explain the reasoning here? you surely have the same chance whether you change your number each time or not.
Do two experiments using a single die.
Experiment Number 1: Roll the die 100 times. Every time the number 1 comes up give yourself a penny.
Experiment Number 2: Roll the die 100 times.
On the first roll if the number that comes up is a 1 give yourself a penny.
Roll the die again. If the number that comes up is a 2 give yourself a penny.
Roll the die again. If the number that comes up is a 3 give yourself a penny.
Roll the die again. If the number that comes up is a 4 give yourself a penny.
Roll the die again. If the number that comes up is a 5 give yourself a penny.
Roll the die again. If the number that comes up is a 6 give yourself a penny.
Roll the die again. If the number that comes up is a 1 give yourself a penny.
Roll the die again. If the number that comes up is a 2 give yourself a penny.
(keep doing this until you've rolled the die 100 times)
The probability of you getting more money during experiment number one is greater than that of number two.
My expertiese in combinatorics is too rusty to calculate the excact probabilites. It's been over twenty years since I took that course. Blech! :)
As I understand it, the only criteria for selecting a lottery number is to avoid one that other people might be likely to pick too; it doesn't help your chances, but if you do win, you're less likely to be sharing the prize.
Never worry about that because its beyond your control and doesn't affect the probability of winning or the amount.
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The probability of you getting more money during experiment number one is greater than that of number two.
I'm pretty sure that's wrong.
The probability of the number you guessed coming up is 1 in 6 for every roll, regardless of whether you guess the same number each time, a different number each time following a pattern, or a different number each time chosen at random. The expectation value for the number of wins will then be 100/6 for either experiment.
Can you explain your reasoning why you expect differently?
By the way, I ran the experiments using the random number generator in Excel. I know that it's not a high quality random number generator, but it should be good enough to imitate a fair die. I "rolled" 1000 dice at a time and compared the number of wins under the two assumptions. While I didn't conduct the experiments sufficient times enough to give a conclusive statistical analysis, both methods gave results within a reasonable error range from 1000/6, and neither method showed a tendency to win more often than the other.
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Can you explain your reasoning why you expect differently?
As I explained above, my combinatorics is very rusty. Think of changing the number is trying to hit a moving target rather than a stationary one. But who knows. I could be wrong. You need to have a solid knowledge of combinatorics to determine this exactly and I haven't done that in decades. So sure, perhaps you're right and I'm wrong. You can always try it and see what happens.
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The probability of you getting more money during experiment number one is greater than that of number two.
I don't see how that's possible. The odds are one in six each time; your choice can't change that. OTOH, if your idea had legs, we could clean out the casinos ;D
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Is it this you're thinking of Pete?
"If every object in a collection of objects fails to have a certain property, then the probability that a random object chosen from the collection has that property is zero. Turning this around, if the probability that the random object has the property is greater than zero, then this proves the existence of at least one object in the collection that has the property. It doesn't matter if the probability is vanishingly small; any positive probability will do.
Similarly, showing that the probability is (strictly) less than 1 can be used to prove the existence of an object that does not satisfy the prescribed properties.
Another way to use the probabilistic method is by calculating the expected value of some random variable. If it can be shown that the random variable can take on a value less than the expected value, this proves that the random variable can also take on some value greater than the expected value."
"In probabilistic combinatorics, the questions are of the following type: what is the probability of a certain property for a random discrete object, such as a random graph? For instance, what is the average number of triangles in a random graph? Probabilistic methods are also used to determine the existence of combinatorial objects with certain prescribed properties (for which explicit examples might be difficult to find), simply by observing that the probability of randomly selecting an object with those properties is greater than 0. This approach (often referred to as the probabilistic method) proved highly effective in applications to extremal combinatorics and graph theory. A closely related area is the study of finite Markov chains, especially on combinatorial objects. Here again probabilistic tools are used to estimate the mixing time.
Often associated with Paul Erdős, who did the pioneer work on the subject, probabilistic combinatorics was traditionally viewed as a set of tools to study problems in other parts of combinatorics. However, with the growth of applications to analysis of algorithms in computer science, as well as classical probability, additive and probabilistic number theory, the area recently grew to become an independent field of combinatorics."
And this http://www.goldsim.com/Web/Introduction/Probabilistic/MonteCarlo/
"In order to compute the probability distribution of predicted performance, it is necessary to propagate (translate) the input uncertainties into uncertainties in the results. A variety of methods exist for propagating uncertainty. Monte Carlo simulation is perhaps the most common technique for propagating the uncertainty in the various aspects of a system to the predicted performance.
In Monte Carlo simulation, the entire system is simulated a large number (e.g., 1000) of times. Each simulation is equally likely, referred to as a realization of the system. For each realization, all of the uncertain parameters are sampled (i.e., a single random value is selected from the specified distribution describing each parameter). The system is then simulated through time (given the particular set of input parameters) such that the performance of the system can be computed. This results is a large number of separate and independent results, each representing a possible “future” for the system (i.e., one possible path the system may follow through time). The results of the independent system realizations are assembled into probability distributions of possible outcomes. As a result, the outputs are not single values, but probability distributions."
That one sound close to what I called 'patterns' to me. And the number fits too :)
Anyone that have a simple example of it, maybe?
How using uncertainty to make a guess more certain, without hidden parameters?
Or do I need to assume hidden parameters for it to work? As the game master 'knowing' which door that contained the car, and so never opening that one. (As well as he can't open my first choice of door, as that destroys my later choice, well, as it seems to me?)
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You don't have to do 100 times to check it. There's nothing fundamentally different between the way the dice behave after 100 rolls or 2 rolls, and you can substitute 2 sided dice (a coin with numbered faces) for 6 sided dice without changing the fundamentals of the problem. Your possible outcomes of two flips are:
1,1
1,2
2,1
2,2
and they're all equally likely.
If you choose method 1 (predicting 1,1 as the outcome), your earnings are:
2 cents 25% of the time
1 cent 50% of the time
0 cents 25% of the time
If you choose method 2 (predict 1,2 as the outcome), your earnings are:
2 cents 25% of the time
1 cent 50% of the time
0 cents 25% of the time
And its easy to verify that any guess will have the same odds of winning, since each roll is independent of the others and you have a 50% chance of winning. This extends straightforwardly to more rolls and 6 sided dice. Things do change if you just want to guess at the numbers rolled, independent of ordering. If someone just asks what the two values in 2 flips of this die are, you're best off guessing 1 and 2, since that combination shows up half the time.
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That's pure probability, as I read you JP :)
But using uncertainty to lower a uncertainty of the other side, aka, not knowing the dices outcome before they show it? Is there really a way to do that? And a example of it please :)
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Or better expressed, can you fight the dice or lottery's randomness by introducing your own randomness. It's not the exact same as what Pete suggested but the idea caught my imagination. Is it possible? And in what ways/situations?
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In Monte Carlo simulation, the entire system is simulated a large number (e.g., 1000) of times. Each simulation is equally likely, referred to as a realization of the system. For each realization, all of the uncertain parameters are sampled (i.e., a single random value is selected from the specified distribution describing each parameter). The system is then simulated through time (given the particular set of input parameters) such that the performance of the system can be computed. This results is a large number of separate and independent results, each representing a possible “future” for the system (i.e., one possible path the system may follow through time). The results of the independent system realizations are assembled into probability distributions of possible outcomes. As a result, the outputs are not single values, but probability distributions."
That one sound close to what I called 'patterns' to me. And the number fits too :)
Anyone that have a simple example of it, maybe?
Weather forecasting. They run numerous projections with a number of models, varying the initial parameters. This gives them a spectrum of possible futures. If the weather is in a reasonably non-chaotic state there will be groups of similar patterns in the result spectrum. The sizes of the groups can be used to give a probability estimate for each predicted weather pattern.
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Nice example dlorde, I reminds me of a 'fractal approach' too, as you look for larger patterns in the patterns visible. It also reminds me of assigning 'weights' in neural networks, as those 'clumps' of patterns closest to each other might be said to be 'weighted up' by probability.
Then we just have randomness left it seems, and Pete's suggestion going the other way defining a ordered approach. Both involves decision making though, even if random in the first case. And those two are the ones I find most difficult to imagine, but rather intriguing.
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Then we just have randomness left it seems, and Pete's suggestion going the other way defining a ordered approach.
There is another major option - chaos. For example, when the weather is in a chaotic state, the simulations come out very different regardless how close the initial parameters are set. It's not random, it's entirely deterministic; but it's totally unpredictable... non-linear dynamics, the Butterfly Effect; it was all the rage n the '80's.
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Yes, I agree. Let me put it this way though, when I think of randomness then I do it from chaos. Maybe that's not correct but to me they become equivalent, although you might want to define randomness to superpositions microscopically, as a example of how I think :) versus chaos macroscopically. Even though you can call chaos deterministic, as in if we only knew all parameters we could describe it, I personally relate it to a randomness.
Maybe I could express it as I don't think there ever will a possibility of knowing the whole history, of anything. It seems to go through all physics that one, no matter what scales you look at it from?
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That's pure probability, as I read you JP :)
But using uncertainty to lower a uncertainty of the other side, aka, not knowing the dices outcome before they show it? Is there really a way to do that? And a example of it please :)
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Or better expressed, can you fight the dice or lottery's randomness by introducing your own randomness. It's not the exact same as what Pete suggested but the idea caught my imagination. Is it possible? And in what ways/situations?
No, you can't if the dice are actually random. You don't have to take my word for it, though. The 2 coin case is easy enough to think about--you can list out all the possibilities very easily and each flip has a 50/50 chance of producing either 1 or 2.
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Yes, I agree. Let me put it this way though, when I think of randomness then I do it from chaos. Maybe that's not correct but to me they become equivalent, although you might want to define randomness to superpositions microscopically, as a example of how I think :) versus chaos macroscopically. Even though you can call chaos deterministic, as in if we only knew all parameters we could describe it, I personally relate it to a randomness.
Well, 'deterministic' means 'non-random' by definition, so perhaps 'pseudo-random' is a better description for the results of chaotic functions.
Maybe I could express it as I don't think there ever will a possibility of knowing the whole history, of anything. It seems to go through all physics that one, no matter what scales you look at it from?
'Having no history' - that could almost pass as a definition of randomness that leaves the question of knowledge open - i.e. can one distinguish a chaotic sequence from a random sequence without knowing the prior history of the system? If I recall correctly, a characteristic of chaotic systems is that islands of order may appear at regular intervals among the apparent randomness, so it should be possible to distinguish randomness from chaos for long enough sequences. How long 'long enough' might be is an open question...
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A intriguing idea dlorde. And yes, assuming Feigenbaums constant one should be able to assume a fractal behavior to chaos giving you mystical attractors and some regularity in that randomness. But then the question becomes how to prove what is random, doesn't it? Haven't thought of that one :) but it seems to become a mathematical concept defined by what circumstances, degrees of freedom etc something have. In flipping the coin we then have three possibilities, the coin giving us head tail and its edge. Of those we ignore the edge defining it as head and tail, and define it as a 50/50 chance for each throw. The randomness referred to here is not something able to make the coin disappear, it's a mathematical definition of what 'choice' of two it may make landing.
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Now, as far as I remember, although Feigenbaums constant give you a regularity, somethings bifurcations (splits) can not be back-tracked. What does that make it, a random behavior? What is random here? Normally used I would refer to something random as something I can't foresee, meaning its exact outcome. Although I might expect myself able to back-track it after a outcome. But that isn't perfectly correct, is it :) thinking of that coin throw. Even though I couldn't foresee it, after its outcome I sincerely doubt myself to be able to define exactly why it gave me a 'tail', but, should I be able too? If knowing all parameters?
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This one is nice http://www.imho.com/grae/chaos/chaos.html
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Then again, assuming a time symmetry. if I can back-track it I also should be able foresee it, using the same presumption (knowing all parameters). So that one doesn't really say anything. Well, it do state something, if we define it as being of two choices.
Determinism relative indeterminism. Then what we have found so far is not determinism as I think, that as I see no real experimental proof for it. Historically we've used determinism though as a leading star, expecting a linear causality to define the universe, as some clock work. And working from an assumption of the universe wanting it to be as simple as possible, indeterminism must win :)
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So, either we define it as a 'random behavior' inherent in Chaos equations, or we call them deterministic. If they really are deterministic then we should be able to simplify a chaotic system to the degree where we can control all parameters, and so 'predict' its outcomes. But we can't do that, as far as I know. So that leaves us a randomness, but is it a same randomness as in the coin throws?
Put another way, you're part of a chaotic system, population wise, environmental, Earth itself. As you throw that coin, the coin also must become a parameter in chaos.
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And what more, well, if one believe in a absolute time symmetry, how do one explain indeterminism? And 'randomness'. I define time symmetry to a logic. I can also use Feigenbaums constant to define a logic, but inside that logic we find randomness and, as I think, also a description of indeterminism. Time symmetry works as a logic because we need it, it gives us a past and and a present and a future. But it's no proof of a arrow able to go backwards, not as long as you use your local clock defining that experiment proving your concept.
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So, either we define it as a 'random behavior' inherent in Chaos equations, or we call them deterministic. If they really are deterministic then we should be able to simplify a chaotic system to the degree where we can control all parameters, and so 'predict' its outcomes. But we can't do that, as far as I know. So that leaves us a randomness, but is it a same randomness as in the coin throws?
As I previously said, chaos is deterministic, not random. It is unpredictable because of sensitive dependence on initial conditions. For example, this means that given the initial parameters, we can't predict on which wing of Lorentz's butterfly we'll find ourselves after n iterations. To discover this, we must iterate the function n times with those initial parameters. In the real world, we can't obtain the initial parameters for chaotic processes with sufficient precision to obtain useful results by iterating the function(s) even if we know it(them). Hence the Monte-carlo simulations for weather, where even for chaotic states there is some hope of identifying the major attractors (although this may mean predicting, say, a 40% chance of stormy, wet weather and a 60% chance of clear, sunny weather!).
The real world has both chaotic and random components. Chaotic because of non-linear dynamic processes, and random because of quantum indeterminacy. The contribution of quantum indeterminacy to chaotic processes in the real world introduces a clouding degree of randomness at small scales, but should be statistically averaged out for macro-scale processes.
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Are you telling me that we can't define a smallest chaotic system to experimentally define its 'determinism' :)
And therefore it becomes deterministic?
Ah well :)
I think I would say that any system unable to be deterministically defined have a very little probability to be proven deterministic, other than theoretically. After all, we should use experiments to define theory, not the other way around.
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On the other hand, theory do define experiments, to test a theory. and that's what I would like to see here :) Determinism being proven by carefully chosen initial parameters. Otherwise it seems to me that no matter how small those initial parameters are we still won't prove a systems determinism, although we can show a 'deterministic' constant repeating itself. So one might then be able to say that even though outcomes are unpredictable, its pattern is mathematically predictable.
Doesn't it remind you about the discussion of 'free will' too? I know it does to me, statistical trends relative individual choices..
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I'm very displeased that this thread has been taken so off topic that the actual topic is no longer being addressed.
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Yes, afraid you're right Pete. Maybe we should split it?
You better talk with a moderator, to agree on where, as you're the originator of the thread.
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I'm very displeased that this thread has been taken so off topic that the actual topic is no longer being addressed.
You may be right; but perhaps the direction of the thread is evidence that the difference between randomness and chaos (which involves both physics and maths) is a topic that's hard to grasp?
Personally, I'd rather have a live thread than a dead one - unless someone has some other hard-to-grasp topic to discuss?
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Wrote this, and thought it made sense at the time :)
"One can think of it this way maybe, the pattern of a thousand tails is no more uncommon that any other combinations of set patterns like tail - tail - head, if repeated over and over, for a thousand flips. Which then just should make one thousand tails, or heads, uncommon, because it is so easy to recognize for us. That will make the one defining it as being 'reset' with each new flip the one making most sense.
What i mean is that if you count the way a pattern evolve over time, flipping a coin, then all patterns possible should have a equal chance of evolving, singling out no pattern as the one, more probable. And a pattern would then be whatever way you found head and tails arrange themselves over a thousand throws."
Still think it makes sense, but read this..
"In contrast, you wouldn’t expect all of the dice to be 4 at the same time, or otherwise assume one particular pattern. That would be a very unlikely and low entropy outcome." http://www.askamathematician.com/2011/12/q-why-does-the-entropy-of-universe-always-increase-and-what-is-heat-death-of-the-universe/
So a unordered sequence is then more probable?
What makes it so?
Stupid question :)
You got a larger probability (number) of 'unordered sequences' than 'ordered'. Then again, we are the ones defining them as ordered or unordered, isn't that so? We mark them, and so define them.
In 'reality' no pattern should be more probable, as I think?
Maybe it's a stupid question, but there seems a slight difference to me? What we define as a probability relative a logic. Or maybe it's just me not thinking it through.
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(Sorry Pete, forgot, still, this is a rather good thread I think, 'organically growing' if you see how I read it.)
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I think there was a moderator discussing the numerical theory. Anyway, if more posts are on that subject, we can split, if not, I'll leave it alone. Certainly there would be benefits of discussing probability and numerical theory. And, I'm quite rusty on my factorials and "Choose" equations.
I saw this puzzle online.
What is next in this series? 1, 4, 10, 19, 31, _ (no need to post the answer here).
It reminds me of the old SAT questions (Scholastic Aptitude Test). Anyway, there may be benefits of discussing how to approach determining an unknown sequence or series.
Another, more well known series:
1, 1, 2, 3, 5, 8, 13, ...
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I'm very displeased that this thread has been taken so off topic that the actual topic is no longer being addressed.
That's a shame Pete, because I really wanted to talk about the Monty Hall problem again. I got all excited because after reading this thread, I finally get it. But I'll address the topic first and say that Maths never really bothered me much at the fairly low levels I studied it at.
What always baffled me was "particle" physics and what a "particle" actually consists of. I think that trying to simplify it diagrams of an atom made of billiard balls actually makes it harder to grasp on a deeper level of what is actually happening.
Now, Monty Hall. I find it interesting that you can think of the situation of 100 doors, where you are asked to choose a door, but then you are asked to decide whether the door has the prize, or a goat. Probability at that stage would say that you've probably chosen a goat, which is why when you are given the opportunity to change in the original scenario, you would.
What interests me is that if at the stage of having 98 goats staring out of opened doors and 2 closed doors, one of which has a prize, someone new enters the studio and is asked to choose a door, would he (or she) be given any advantage by asking what the previous person chose?
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From bizerl:
What interests me is that if at the stage of having 98 goats staring out of opened doors and 2 closed doors, one of which has a prize, someone new enters the studio and is asked to choose a door, would he (or she) be given any advantage by asking what the previous person chose?
The answer to that is yes -- it is extra information that modifies the odds from 50:50 to 99:1.
From pmb:
I'm very displeased that this thread has been taken so off topic that the actual topic is no longer being addressed.
I apologize Pete for my part in this, although I think you can see that Probability and Combinatorics is a part of Maths that both fascinates and confuses many people.
One part of advanced Maths/Physics that confuses me is how to relate the character tables of say trigonal groups like C3v or D3 to that of the underlying C3 group (there are other examples of the same sort of thing where a degenerate representation might consist of a pair of complex conjugates or a conventional E type representation with a character of 2 for the identity.)
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I just discovered the Khan Academy, https://www.khanacademy.org/
This covers a number of areas of Maths & Science in a progressive and clear manner (I needed a statistics refresher).
...although I am puzzled about how & why they combine "Science & Economics" into a single category??
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I just discovered the Khan Academy, https://www.khanacademy.org/
This covers a number of areas of Maths & Science in a progressive and clear manner (I needed a statistics refresher).
...although I am puzzled about how & why they combine "Science & Economics" into a single category??
The reason for this is because physics and econometrics are the major areas of application of advanced mathematics, and because (at least in Canberra) there were a huge number of postdocs in physics at ANU who could not find employment in physics when their contracts ran out, but they were easily able to move across and develop the field of econometrics with jobs in the public service (perhaps explaining why econometrics has not turned out to be particularly successful). I do not know about the situation in other countries, but can imagine it was much the same.