EVE Search - Statistics brainteasers

All Channels

Out of Pod Experience

Statistics brainteasers

» Click here to find additional results for this topic using Google

Monitor this thread via RSS [?]

Pages: 1 [2] 3 4 5 6 7 8 :: one page
Author	Thread Statistics \| Show CCP posts - 0 post(s)
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 05:57:00 - [31] Edited by: Akita T on 16/04/2011 06:19:01 Originally by: Whitehound Who cares? If it is a trap question then you have confirmed the BS. What else is there to know? There's no "BS" whatsoever here, just math. The "trap" is a logic trap of your own creation, you tend to generalize where you shouldn't - just because 1D is easily solved does not mean the rest are. Again, I repeat the offer of sending you clear, mathematics-based proof that you were wrong (and how), under the condition that you do not disclose the proof. _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 06:28:00 - [32] Edited by: Whitehound on 16/04/2011 06:36:07 Originally by: Akita T Originally by: Whitehound Who cares? If it is a trap question then you have confirmed the BS. What else is there to know? No one gives a crap about your BS, Akita. It's a "trap" in the sense that the result is counter-intuitive for a novice mathematician (and not just them), most people are inclined to make the same type of mistake at first. There's no actual "BS" here, just math. And NOT just basic math. No, Akita. You still are wrong and I will explain it to you. Take the 1-dimensional problem. You say the probability of moving to the left and the right is the same. Saying that having a probability of a movement is however not the same as having a movement. All you have got is a probability, but not an actual movement. As a consequence, if you have no movement at all will the probability of moving to the left and the right still be the same. It only is not moving at all and this is where you were BS'ed. In other words, having a probability of a movement includes having no movement as well as having a movement, whereas having a movement excludes having no movement. When you then ask what the probability is to get to the origin do you not ask for the probability, but imply that it has got a movement. So as long as there is movement will you get to the origin just like you will get to any point. And if it has got no movement will you be sitting on the origin for all of eternity, not reaching any point, but you will "reach" the origin. (Of course, "reaching the origin" means again that there is a movement, because "sitting on it" and "reaching it" is not the same either.) --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 06:36:00 - [33] Edited by: Akita T on 16/04/2011 06:38:07 Originally by: Whitehound No, Akita. You still are wrong and I will explain it to you. You ALWAYS make a 1-pixel move in each iteration, the direction of the move is randomly chosen. The probability of picking any of the directions is equal (1/2 for 1D, 1/4 for 2D, 1/6 for 3D). In step 1, you will have absolutely, certainly and unquestionably moved away from the origin. Step 2 is the earliest you can land back on the origin. You can only land back on the origin in even steps : step 2, step 4, step 6, ... , step 2x, ... Absolutely no verbal trickery anywhere. _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 06:44:00 - [34] Edited by: Whitehound on 16/04/2011 06:46:25 Originally by: Akita T You can only land back on the origin in even steps : step 2, step 4, step 6, ... , step 2x, ... Absolutely no verbal trickery anywhere. It does not matter if it takes even steps, because it does not matter if it takes an infinite amount of time or twice as much. Multiply infinite with a finite number and it stays infinite. Cantor has then shown that one can take a 2-dimensional plane and string it into a 1-dimensional line if the elements are only countable infinite. As a consequence of Cantor's proof is anything you say about 1 dimension just as true as for 2 dimensions, or 3 and 4 and n-dimensions. Again, as long as the elements are countable infinite. I am really curious how you are going to make a difference. You have already indicated that you will be making one. --
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 06:49:00 - [35] Edited by: Hieronimus Rex on 16/04/2011 07:07:04 Originally by: Akita T Edited by: Akita T on 16/04/2011 06:40:52 Originally by: Whitehound No, Akita. You still are wrong and I will explain it to you. You ALWAYS make a 1-pixel move in each iteration, the direction of the move is randomly chosen. The probability of picking any of the directions is equal (1/2 for 1D, 1/4 for 2D, 1/6 for 3D). In step 1, you will have absolutely, certainly and unquestionably moved away from the origin. Step 2 is the earliest you can land back on the origin. You can only land back on the origin in even steps : step 2, step 4, step 6, ... , step 2x, ... Absolutely no verbal trickery anywhere. The correct answer is NOT "100% for each of the three cases". I don't know if this is the "lazy logical generalization" you were talking about...but is it just.. In case 1, half of the time you'll return in 2 moves, 1/4 of the time you'll return in 4 moves, etc So 1/2+1/4+... = 100%, as others have said. In case 2, 25% of the time you'll return in 2 moves (otherwise you go in one of the other three directions), 1/16 of the time you'll return in 4 moves (you basically need to return to your starting point two moves ago, but twice). so 1/4+1/16+..... = 1/3 for Case 2 In case 3, 1/6 of the time you'll return in 2 moves, and applying the same logic as below.... 1/6+1/36+..... = 1/5 for case 3 DO I WIN A PRIZE?
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 06:57:00 - [36] Edited by: Whitehound on 16/04/2011 06:58:34 Originally by: Hieronimus Rex DO I WIN A PRIZE? You will get half of a prize. Or perhaps a third of a prize. Or just any fraction of it. It still is a full win, isn't it? --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 07:09:00 - [37] Edited by: Akita T on 16/04/2011 07:16:32 Originally by: Whitehound Cantor has then shown that one can take a 2-dimensional plane and string it into a 1-dimensional line if the elements are only countable infinite. As a consequence of Cantor's proof is anything you say about 1 dimension just as true as for 2 dimensions, or 3 and 4 and n-dimensions. Again, as long as the elements are countable infinite. Discounting the controversy over whether Cantor's proof is actually accurate or not to begin with, and assuming for now that it actually is, and also assuming you are using it correctly in this particular case (of which I am not convinced), the generalization to more than 2 dimensions would require further proof, you can't just claim it's a consequence. I am not aware of any such proof, but feel free to correct me. In fact, if you are indeed accurate in your inference that Cantor's proof would necessarily imply that the result for 3D in this particular problem should be 100%, you just successfully shown that Cantor's proof is wrong. Originally by: Hieronimus Rex I don't know if this is the "lazy logical generalization" you were talking about...but is it just.. In case 1, half of the time you'll return in 2 moves, 1/4 of the time you'll return in 4 moves, etc So 1/2+1/4+... = 100%, as others have said. Well, at least the RESULT is correct, even if the proof is not. See the hints on previous page for more details. Quote: In case 2, 25% of the time you'll return in 2 moves (otherwise you go in one of the other three directions), 1/16 of the time you'll return in 4 moves (you basically need to return to your starting point two moves ago, but twice). so 1/4+1/16+..... = 1/3 for Case 2 Not even close. Quote: In case 3, 1/6 of the time you'll return in 2 moves, and applying the same logic as below.... 1/6+1/36+..... = 1/5 for case 3 Still no. Quote: DO I WIN A PRIZE? Sorry, no _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 08:09:00 - [38] Edited by: Whitehound on 16/04/2011 08:10:37 Originally by: Akita T Discounting the controversy over whether Cantor's proof is actually accurate or not to begin with, and assuming for now that it actually is, and also assuming you are using it correctly in this particular case (of which I am not convinced), the generalization to more than 2 dimensions would require further proof, you can't just claim it's a consequence. I am not aware of any such proof, but feel free to correct me. In fact, if you are indeed accurate in your inference that Cantor's proof would necessarily imply that the result for 3D in this particular problem should be 100%, you just successfully shown that Cantor's proof is wrong. You probably already know how to string a 2-dimensional plane into a 1-dimensional line: a-b c-d e-.. \| \| \| \| f-g h i j .. \| \| \| \| k-l-m n o .. \| \| p-q-r-s t .. => a-b-g-f-k-l-m-h-c-d-i-n-s-r-q-p-u-v-w-x-y-t-o-j-e-... \| \| u-v-w-x-y .. : : : : : A 2-dimensional plane of pixels is stringed into a line of pixels by moving through the rows and columns in a zick-zack pattern, like in the above example. Cantor did this by using diagonals, which is why it is called the "diagonal proof". You do the same with a 3-dimensional cube of pixels by including movements into the nearest upper and lower planes to string an ever increasing cube into a line. So I do not see how the 1-dimensional problem differs from the 2- and 3-dimensional problem in its outcome. You will not reach every point in a finite amount of time and therefore will be able to give a probability, but you will reach every point as soon as you have infinite time. In infinite time is it possible, depending on how you move to the next pixel, that some pixels are reached more often than others, but as long as one can consider the movement to be free will every point be reached. --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 08:19:00 - [39] Edited by: Akita T on 16/04/2011 08:21:07 Originally by: Whitehound You do the same with a 3-dimensional cube of pixels by including movements into the nearest upper and lower planes to string an ever increasing cube into a line. I'm not quite sure that's accurate. I tried to picture it and I either end up with holes in the cube grid or locked in. Proof is needed. _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 08:30:00 - [40] Originally by: Akita T Originally by: Hieronimus Rex I don't know if this is the "lazy logical generalization" you were talking about...but is it just.. In case 1, half of the time you'll return in 2 moves, 1/4 of the time you'll return in 4 moves, etc So 1/2+1/4+... = 100%, as others have said. Well, at least the RESULT is correct, even if the proof is not. Assume "0" means "left" and "1" means "right. Step #2 has 2 out of 4 possible combos that lead back to the origin, indeed : 01 and 10 (while 00 or 11 do not). But step #4 has actually 6 out of 16 possible combos that end up on the origin (37.5%) : 0011, 0101, 0110, 1001, 1010, 1100 (out of which 4 have actually reached the origin before in step 2 too, so only 2/16, or 1/8 extra) ...also 4 (25%) which do not land on the origin in step 4 have reached the origin in step 2 already : 0100, 0111, 1000, 1011 ...and 6 out of 16 which never landed on the origin (37.5%) : 0000, 0001, 0010, 1101, 1110, 1111 ...so 10 out of 16 possible combos (62.5%) have stopped on the origin at least once, NOT 75% (i.e. 1/2+1/4, or 12 out of 16 combos) as you seem to imply by your attempted proof. Whoops I double counted....
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 08:41:00 - [41] Originally by: Akita T I'm not quite sure that's accurate. I tried to picture it and I either end up with holes in the cube grid or locked in. Proof is needed. Whenever you get close to locking yourself in do you escape into the next, nearest plane for a new point. You stay as close to the origin as possible. You will eventually get further and further away from the origin and end up with a line of pixels which runs through all dimensions. If it helps, think of a ball of wool. A ball of wool is a 3-dimensional object, consisting out of a single string. --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 08:43:00 - [42] Edited by: Akita T on 16/04/2011 08:44:21 And that's how you end up with holes. Seriously, the answer for 3D is not 100%, like for 1D and 2D. So the construct you are trying to describe must be impossible anyway. _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 08:46:00 - [43] Originally by: Akita T The answer is NOT 100%. The answer is 1, Akita. The origin can be reached and therefore will be reached in infinite time. There is nothing stopping the movement from getting back to its origin, just as there is nothing stopping the movement from getting away from it. --
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 08:49:00 - [44] Originally by: Akita T Edited by: Akita T on 16/04/2011 08:44:21 And that's how you end up with holes. Seriously, the answer for 3D is not 100%, like for 1D and 2D. So the construct you are trying to describe must be impossible anyway. No. You do not get holes. If you skipped a pixel, left it out, then you are doing it wrong. Nothing is stopping you from creating an endless string of pixels in a 3-dimensional space. Only your imagination does. Where you see a hole do I see a pixel that goes next onto the string. --
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 08:52:00 - [45] Edited by: Hieronimus Rex on 16/04/2011 08:56:59 The internet delivers SPOILER ALERT: This is why cases 1 and 2 are 100% This is why case 3 is ~34.054%
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 08:56:00 - [46] Hey, no Google cheating in this classroom ! Student, hand in those papers and destroy your cheat sheet _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Netheranthem Eve Engineering Finance Eve Engineering	Posted - 2011.04.16 08:59:00 - [47] Damnit, it looks like the gambler problem I had at my oral exam, and I don't remember nao, I'll try scratching my head harder :( Also, I could do a real proof and all, but I prefer to keep it like that so it's readable and I don't have to pull out fantastic induction tricks. For 1D, you will move either "left" or "right" with 50% probability for each (1/2). Probability you'll hit 0 from 1 pixel away: p1 = 1/2 * p0 + 1/2 * p2 (p0 = 1) Probability you'll hit 0 from 2 pixels away: p2 = 1/2 * p1 + 1/2 * p3 And etc with pn = 1/2 + 1/2 * pn+1 Let's try to substitute in p1: p1 = 1/2 + 1/2 * ( 1/2 * p1 + 1/2 * p3). 3/4 * p1 = 1/2 + 1/4 * p3 p1 = 2/3 + 1/3 * p3 Substituting again: p1 = 2/3 + 1/3 ( 1/2 p1 + 1/2 * p4) 5/6 * p1 = 2/3 + 1/6 * p4 p1 = 4/5 + 1/5 * p4 Or, more generally p1 = a + bpn Where a -> 1 and b -> 0 (And a+b = 1). For 2D, you will move either "left", "right", "up" or "down" with 25% probability for each (1/4). For this, let's consider \|\|.\|\|1 as the norm we'll use to see if we're further from 0 or not. For each point at the same norm, the probability to move back to 0 is the same, as the same amount of steps split them from 0. thus, p1 = 1/4 p0 + 3/4 * p2. then, p2 = 1/4 * p1 + 3/4 * p3 If it's in straight line to 0. otherwise p2* = 1/2 * p1 + 1/2 * p3. Will probably have to think a bit about it, but I'm starting to see it won't net you 1 as answer.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 09:05:00 - [48] Originally by: Hieronimus Rex The internet delivers: This is why cases 1 and 2 are 100% This is why case 3 is ~34.054% It is wrong. It is a proof over uncountable infinite dimensions, not however over finite countable dimensions. Ask yourself, what is the point of saying that the origin can be reached to 34%? If you take a car, do you then arrive with only 34% of your car? The answer can only be "yes" or "no", and not 34%. Only the probability to arrive at the same point for a large number of movements is 34%. If the movement occurs over an uncountable infinite dimensions in infinite time will it be 34%, and only if you allow for one movement per second in order to make time itself countable infinite. If you however allow for infinite movements between fractions of a second will you again come back to 1 (or 100%). Therefore is Akita still wrong. --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 09:08:00 - [49] Originally by: Whitehound Originally by: Hieronimus Rex The internet delivers: SPOILER ALERT EXPLANATION FOR 3D CASE It is wrong. /facepalm _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 09:10:00 - [50] Originally by: Akita T /facepalm Just go and cover your face. The shame is all yours. --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 09:15:00 - [51] Wait, now we have to add "failure to properly recognize/use internet memes" to the list of things you're not quite so good at as you might think ? Also, see alternate explanation in edit above, a slightly more accessible version, with far more explanations as to why that is actually the case. _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Frecator Dementa Caldari Perkone	Posted - 2011.04.16 09:17:00 - [52] Originally by: Whitehound Originally by: Akita T The answer is NOT 100%. The answer is 1, Akita. The origin can be reached and therefore will be reached in infinite time. There is nothing stopping the movement from getting back to its origin, just as there is nothing stopping the movement from getting away from it. At first that's what I thought, but then I remembered the Achilles&Tortoise paradox, I had fallen into the same logical fallacy: Assuming that a sum of infinite number of positive, non-zero numbers must be infinity Infinite series can be both divergent and convergent The number of times that a random walk in 1D or 2D will reach the origin can be only expressed as a divergent series (ie: infinite number of times; ie: probability to pass through origin = 100%) The number of times that a random walk in 3++ dimensions will reach the origin however can only be expressed as a convergent series (ie: the random walk will pass through the origin a finite number of times, therefore there must be a nonzero chance that the random walk will never reach the origin => Probability of reaching origin is less than 100% ---- <sig goes here>
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 09:32:00 - [53] Originally by: Whitehound Edited by: Whitehound on 16/04/2011 09:12:31 Originally by: Hieronimus Rex The internet delivers: This is why cases 1 and 2 are 100% This is why case 3 is ~34.054% It is wrong. It is a proof over uncountable infinite dimensions, not however over finite countable dimensions. Ask yourself, what is the point of saying that the origin can be reached to 34%? If you take a car, do you then arrive with only 34% of your car? I think it means in 34% of all infinitely long trials, you reach the origin at least once. In 66% of them, you don't.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 09:39:00 - [54] Originally by: Frecator Dementa At first that's what I thought, but then I remembered the Achilles&Tortoise paradox, I had fallen into the same logical fallacy: Assuming that a sum of infinite number of positive, non-zero numbers must be infinity Infinite series can be both divergent and convergent This has got little to do with it. You can also say that, for the 1-dimensional problem, your first two movements are to the right and from there on alternate between left and right. The movement will never get back to its origin while at the same time the requirement for the probability of the movement (being equal for left and right) is still being fulfilled, given infinite time. Only for finite time will this example be invalid, because the probability requirement for left and right is not fulfilled. However, for an infinite amount of movements will the extra jumps to the right at the start become negligibly small, the probability for a movement to the right and left becomes equal, and therefore does this example become valid for infinite time. So how can the movement come back to its origin when it alternates on the spot somewhere away from the origin? --
Akita T Caldari Navy Volunteer Task Force	Posted - 2011.04.16 09:46:00 - [55] So you just don't want to read (or refuse to comprehend) the linked proofs, do you ? Maybe you actually think you know better than professional mathematicians ? Or is it just too painful to admit you were wrong ? _ CCP LEADERSHIP MENTALITY NEEDS TO CHANGE FAST ! "New junky features sell, old polished content doesn't" ? KILL IT WITH FIRE.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 09:51:00 - [56] Edited by: Whitehound on 16/04/2011 09:54:28 Originally by: Hieronimus Rex I think it means in 34% of all infinitely long trials, you reach the origin at least once. In 66% of them, you don't. No. It means it is 34% for finite trails. How else could you ever get to a result if a trail never ends? --
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 09:53:00 - [57] Originally by: Akita T So you just don't want to read (or refuse to comprehend) the linked proofs, do you ? Maybe you actually think you know better than professional mathematicians ? Or is it just too painful to admit you were wrong ? I read into it and got bored pretty quickly, seeing how loosely the proof was stitched together. So no, I do not know if the proof applies to your example. What I do know is that I am a mathematician. --
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 10:02:00 - [58] Edited by: Hieronimus Rex on 16/04/2011 10:04:42 Originally by: Whitehound Originally by: Hieronimus Rex I think it means in 34% of all infinitely long trials, you reach the origin at least once. In 66% of them, you don't. No. It means for it is 34% for finite trails. How else could you ever get to a result if a trail never ends? You don't actually run the trials to get the result. EDIT, also: Originally by: Whitehound Edited by: Whitehound on 16/04/2011 09:12:31 Originally by: Hieronimus Rex The internet delivers: This is why cases 1 and 2 are 100% This is why case 3 is ~34.054% It is wrong. It is a proof over uncountable infinite dimensions, not however over finite countable dimensions. It is a proof for a random walk on a lattice, which is countable infinite dimensions.
Whitehound The Whitehound Corporation Frontline Assembly Point	Posted - 2011.04.16 10:12:00 - [59] Originally by: Hieronimus Rex You don't actually run the trials to get the result. Which makes it an excuse for a result that does not have anything to do with the problem. --
Hieronimus Rex Minmatar Infinitus Sapientia New Eden Research.	Posted - 2011.04.16 10:23:00 - [60] Edited by: Hieronimus Rex on 16/04/2011 10:23:34 Originally by: Whitehound Originally by: Hieronimus Rex You don't actually run the trials to get the result. Which makes it an excuse for a result that does not have anything to do with the problem. This is the correct interpretation (of the 34% result): "On a three-dimensional lattice, a random walk has less than unity probability of reaching any point (including the starting point) as the number of steps approaches infinity. The probability of reaching the starting point again is 0.3405373296.... This is one of P=lya's random walk constants." Were you really just nitpicking about the "as steps approach infinity" vs. "infinite steps" distinction?

Pages: 1 [2] 3 4 5 6 7 8 :: one page
First page \| Previous page \| Next page \| Last page

COPYRIGHT NOTICE
EVE Online, the EVE logo, EVE and all associated logos and designs are the intellectual property of CCP hf. All artwork, screenshots, characters, vehicles, storylines, world facts or other recognizable features of the intellectual property relating to these trademarks are likewise the intellectual property of CCP hf. EVE Online and the EVE logo are the registered trademarks of CCP hf. All rights are reserved worldwide. All other trademarks are the property of their respective owners. CCP hf. has granted permission to EVE-Search.com to use EVE Online and all associated logos and designs for promotional and information purposes on its website but does not endorse, and is not in any way affiliated with, EVE-Search.com. CCP is in no way responsible for the content on or functioning of this website, nor can it be liable for any damage arising from the use of this website.