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What's a boy to do?
"riverman" wrote in news:1162253876.219819.310220
@m7g2000cwm.googlegroups.com: Remember, we are looking at a conditional probability; dart B has already landed farther than dart A. So our list of outcomes is limited to: ABC ACB CAB Reducing this to a combinatorial problem is incorrect and misleading. Each dart has a distribution around some point, hopefully the center, with decreasing probability as you get further from the target. If darts A and B are three or four standard deviations out, dart C has a very high probability of being closer to the center. If A and B are a tenth of a standard deviation out, there is a very low probability of C being closer. We need to know the 2-D distribution of the dart C, and then we need to know where the first two darts landed. So "not enough information" is correct. -- Scott Reverse name to reply |
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Calif Bill wrote: You could have an almost infinite amount of darts the exact same distance from the center. The only limiting number is how big the circle is from the center and how big of diameter is the dart. There are an infinite number of points equidistant from the center point. And it depends on neither the precision or accuracy of the measurement. And in your measurement of the distance it would be more accuracy and not precision. Precision only gives more numbers after the decimal point. Bill: I'm not sure where to start, but there are a lot of little details in your assertations that are erroneous. There's some truth also, so don't lose hope :-) First of all, yes the definition of a circle states there are an infinite number of points in a plane that are equidistant from a given point, but the liklihood of getting even two darts to land on that circle is slim. (Just how slim is discussed in the second paragraph below.) We don't even have to agree on how slim for now, but the the more darts you want to have land on that circle, the less likely it is to happen, and it approaches zero as the number of darts gets larger and larger. Although the phrase 'almost infinite' is actually meaningless, I assume you mean we are looking at numbers that are growing huge beyond comprehension, so the liklihood of it happening is shrinking tiny beyond comprehension. Secondly, it IS a matter of precision, not accuracy. We don't care what the actual distance from the center is, what we do care about is whether or not two darts have the same measurement from the center, even if that measurement is wrong. If we use an inaccurate tool, then we might get a wrong amount (a broken ruler might show each dart to be 10.55 cm from the center, while they are both actually much less that that). That's 'inaccurate', but if the numbers match, then we can still assert that they are the same distance. If we use a ruler with really fat indicator lines, we might get both measuring 10.55 cm, however if we used a vernier caliper, calibrated or not, we might get one of them measuring 10.550000000000001 cm and the other measuring 10.550000000000002 cm. Those are measures of high PRECISION, and my assertation is that, no matter how the darts land, we can always use more precise measuring devices until we find where the numbers vary. And they always will, even if we have to go to electron microscope levels. Just as no two snowflakes are alike, no two darts can land the same distance from the center. Now, I appreciate that some people might have an ingrained prejudice against math because it doesn't always conform to their intuition (and this might be you, or it might not). But when faced with something that doesn't seem to 'fit' what we want to believe, there are two choices: find out the rules of math and learn to analyze things according to those rules, including learning the constraints and limitations and the meaning of those, or else continue to assert that what we believe is right because it 'feels right' to us, and use poorly structured arguments or misnomers to claim that nothing has any validity, so we can't possibly be wrong. That way lies madness. --riverman |
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Scott Seidman wrote: ...Each dart has a distribution around some point, hopefully the center, with decreasing probability as you get further from the target. I detect a troublesome ambiguity here. I see two ways to read this without the reader doing any violence to good sense: 1. Each dart has a distribution around some point, hopefully the center, with decreasing probability of being close to the center as the thrower gets further from the target. 2. Each dart has a distribution around some point, hopefully the center, with decreasing probability of landing at any point as distance from the center increases. I think no one will have much trouble with the first reading. The second is easier to defend than it may appear at a glance because it is.....partly.....true. I assume that a truly great dart thower can consistently place the darts very close to the center of the board (which, as we all know is not necessarily the object in every game, but is as good a spot as any other and does, at any rate, appear to be accepted for the purposes of this discussion) and I know that there are people whose skills are so abysmal that they rarely hit the board at all from the standard distance of what I believe to be 8 feet or so. Probabilities of distribution clearly vary widely (if not to say wildly) between these two extremes. What makes the second reading defensible is that for the best dart throwers the probability of landing at a given point DOES decrease with distance from the center.....um.....mostly. In fact, it also decreases as the point gets very very close to a precisely measured center. The same is also true for ALL dart throwers. The difference among them is that the diameter of the circle at which increasing or decreasing probabilities converge or diverge (depending on direction of travel toward or away from the circle) varies with the skill of the thrower, being very small for the very good and very large for the very bad. I haven't looked at this, or the larger discussion, closely enough to suppose that it will be a crushing blow to anyone's thesis but, on the other hand, I haven't seen anything yet that I think rules it out either. It may not even be relevant or interesting given the assumptions that have been (if only tacitly) agreed on. However, it does once again open the door to an examination of those assumptions. Given that no one showed any interest the first time I brought the matter up though, I guess I won't go into it in any depth here. I will simply confine myself to making a proposition open to anyone. Give me three darts and a prediction of where they will land relative to one another in terms of distance from the center of the target, and I will prove you wrong EVERY time. :) Wolfgang |
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Jonathan Cook wrote: riverman wrote: solution. To answer the darts question, merely rely on the definition of probablilty: the number of ways to achieve your objective, divided by the number of possible outcomes. List all the possible arrangements of how the darts could land, and count how many fit our scenario. I saw your response to my post, but I still disagree. You are making all sorts of assumptions by reducing it to this simple model, not the least of which is some sort of uniform probability distribution, or worse yet, a 50-50 likelihood of landing inside or outside of dart A. And those assumptions _heavily_ depend on how closely dart A is to the target, among other things (like if I'm throwing, who has very little dart-throwing experience). Dart throwing is _nothing_ like dice rolling... First, list all the ways to throw three darts, A B and C. ABC ACB BAC BCA CAB CBA Sometime too simple _is_ too simple... But, Wolfgang's original post has got me thinking, and I'd be interested to see your response to a post I'm about to make from his original question (and Wolfgang's and any others' responses, too.) Jon. Hi Jon: Actually, I'm not assuming a uniform distribution, just an unchanging one. I think the only crucial assumption here (beyond the obvious ones: gravity is constant, the target is planar, the darts are self-similar, the wind doesn't change, etc) is that the skill of the thrower does not improve or deteriorate demonstrably between throws, and that the thrower is aiming for the same spot with each throw. Some people are NOT assuming these things, and I agree that it skews the results, and I'll acquiesce and say that I should have stated those things and not assumed them. The nonsense about two darts being able to be equidistant is just that: nonsense. But if we assume the conditions are identical for all three darts, then each throw is identical without prejudice, and the six outcomes are equally likely, and the answer is 2/3. Take this problem across the hall to the Stats professors and see what they have to say about it. Your other assertations about probabilities not being meaningful in the context of singular events sounds bizarre. I'll re-read what you wrote, mostly because I see you have some education in math and statistics, but it sounds like you are getting twisted around somewhere. Possibly within the definition of 'probability'. --riverman |
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riverman wrote: ....If we use a ruler with really fat indicator lines, we might get both measuring 10.55 cm, however if we used a vernier caliper, calibrated or not, we might get one of them measuring 10.550000000000001 cm and the other measuring 10.550000000000002 cm. Those are measures of high PRECISION, and my assertation is that, no matter how the darts land, we can always use more precise measuring devices until we find where the numbers vary.... If you can REALLY measure things on a scale several orders of magnitude short of an angstrom, I REALLY wanna come play in your shop! :) For the benefit of those who don't know what a vernier caliper is (a friend of mine....who should have known better.....once thought a micrometer I handed him was some sort of special application c-clamp and proceeded to see how tight he could crank it.....I nearly smacked his ****in' hea......well, never mind about that), suffice it to say that the numbers displayed above were used for effect but Myron's point is indisputable. Precision measuring instruments these days are such that finding two darts whose distance from a given point is so close that the difference can't be measured is vanishingly small. The REAL difficulty (whether in a pub or a laboratory) would be in getting agreement on exactly where to measure from. Wolfgang |
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"riverman" wrote in message oups.com... Calif Bill wrote: You could have an almost infinite amount of darts the exact same distance from the center. The only limiting number is how big the circle is from the center and how big of diameter is the dart. There are an infinite number of points equidistant from the center point. And it depends on neither the precision or accuracy of the measurement. And in your measurement of the distance it would be more accuracy and not precision. Precision only gives more numbers after the decimal point. Bill: I'm not sure where to start, but there are a lot of little details in your assertations that are erroneous. There's some truth also, so don't lose hope :-) First of all, yes the definition of a circle states there are an infinite number of points in a plane that are equidistant from a given point, but the liklihood of getting even two darts to land on that circle is slim. (Just how slim is discussed in the second paragraph below.) We don't even have to agree on how slim for now, but the the more darts you want to have land on that circle, the less likely it is to happen, and it approaches zero as the number of darts gets larger and larger. Although the phrase 'almost infinite' is actually meaningless, I assume you mean we are looking at numbers that are growing huge beyond comprehension, so the liklihood of it happening is shrinking tiny beyond comprehension. Secondly, it IS a matter of precision, not accuracy. We don't care what the actual distance from the center is, what we do care about is whether or not two darts have the same measurement from the center, even if that measurement is wrong. If we use an inaccurate tool, then we might get a wrong amount (a broken ruler might show each dart to be 10.55 cm from the center, while they are both actually much less that that). That's 'inaccurate', but if the numbers match, then we can still assert that they are the same distance. If we use a ruler with really fat indicator lines, we might get both measuring 10.55 cm, however if we used a vernier caliper, calibrated or not, we might get one of them measuring 10.550000000000001 cm and the other measuring 10.550000000000002 cm. Those are measures of high PRECISION, and my assertation is that, no matter how the darts land, we can always use more precise measuring devices until we find where the numbers vary. And they always will, even if we have to go to electron microscope levels. Just as no two snowflakes are alike, no two darts can land the same distance from the center. Now, I appreciate that some people might have an ingrained prejudice against math because it doesn't always conform to their intuition (and this might be you, or it might not). But when faced with something that doesn't seem to 'fit' what we want to believe, there are two choices: find out the rules of math and learn to analyze things according to those rules, including learning the constraints and limitations and the meaning of those, or else continue to assert that what we believe is right because it 'feels right' to us, and use poorly structured arguments or misnomers to claim that nothing has any validity, so we can't possibly be wrong. That way lies madness. --riverman Your statement was no two darts could land equidistant from the center. As I state, totally wrong. There is an infinite number of points on the circle where the first dart landed from the center. You change the statement to say the more darts that are thrown, the likelihood that two are equidistant from the center diminishes. The more darts tossed the greater the likelihood two are equidistant. Not all equidistant, but two can be. Sure, your vernier calipers can have more precision than a wooden school rule, but it is still the accuracy of the measurement. Your statement is akin to saying you have increased precision when you multiply 1.02 times 2.04 and get 2.0808. You still have only 2 decimal points of precision. As to using an accurate tool, does not need to be. The actual measurement may be wrong, but as long as the measurement is the same is all we need to know for the stated question. As long as the precision and accuracy is great enough in the measuring instrument to have repeatability, the actual distance matters not. Maybe you should go back and review the rules of math. And review your arguments. |
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Scott Seidman wrote: "riverman" wrote in news:1162253876.219819.310220 @m7g2000cwm.googlegroups.com: Remember, we are looking at a conditional probability; dart B has already landed farther than dart A. So our list of outcomes is limited to: ABC ACB CAB Reducing this to a combinatorial problem is incorrect and misleading. Each dart has a distribution around some point, hopefully the center, with decreasing probability as you get further from the target. If darts A and B are three or four standard deviations out, dart C has a very high probability of being closer to the center. If A and B are a tenth of a standard deviation out, there is a very low probability of C being closer. We need to know the 2-D distribution of the dart C, and then we need to know where the first two darts landed. So "not enough information" is correct. But consider that outcome of throwing each dart is independant. IOW, there will always be one dart that is closest, one that is farthest, one that is in between. The combinatoric arrangement merely gives all the arrangements. Yes, if Dart A is very very close, then the probability of dart B being closer gets smaller. But that same distribution gets repeated again with dart B being closer than A if we throw dart B first. So, unless the ability of the thrower changes between the two darts, AB is identical to BA. --riverman |
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What's a boy to do?
"Kevin Vang" wrote in message t... In article t, says... Your statement was no two darts could land equidistant from the center. As I state, totally wrong. Two darts could conceivable land equidistant from the center; however, the probability of that happening is 0. Explaining why will require a bit of less than elementary probability theory, with integral calculus as a prerequisite. We can go there, if you are up to it... Kevin Explain why could not happen. |
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Calif Bill wrote: "Kevin Vang" wrote in message t... In article t, says... Your statement was no two darts could land equidistant from the center. As I state, totally wrong. Two darts could conceivable land equidistant from the center; however, the probability of that happening is 0. Explaining why will require a bit of less than elementary probability theory, with integral calculus as a prerequisite. We can go there, if you are up to it... Kevin Explain why could not happen. I don't know if you have taken Integral Calculus, so I won't use that to explain it. The simplest 'geometrically appealing' explanation is to say that the probability of hitting some target is directly related to the size of the target. The bigger the spot you are trying to hit, the more likely it is to hit it. Now, if you are aiming for a curved line....specifically the line that describes a circle, you have to consider the width of that line. Since circles are a collection of points, and points have no width, then you are essentially aiming for something that has a width of zero. Which means the possibility of hitting it is zero. You can hit it ONCE, because its position is unknown until the first dart actually lands. Once that dart establishes the position of the circle, it becomes a target of width zero which another dart cannot possibly hit. You can reduce the size of your miss (lets call it your 'slop') to be as small as you want by making less and less precise measurements, but the converse is also true...you can always find the slop by increasing the precision of your measurement. You cannot eliminate the slop; you cannot ever hit the line.: its a matter of precision. Just like no two snowflakes are alike. Throwing more and more darts DOES increase the liklihood of hitting something more than once, but if the liklihood of hitting it is already infinitely small; throwing more and more darts doesn't make it any more likely. Throwing infinite darts creates what is called an 'indeterminate' form, and we cannot solve it that way. We get an infinitely large number of opportunities of achieving something that has an infinitely small chance of happening. Its incorrect to conclude that it will happen infinite times....Infinity x (1/infinity) does not equal infinity. It doesn't equal, or mean, anything. However, throwing a FINITE number of darts....say two, at something that we have an infintely small chance of hitting IS solvable. Its three times 'infinitely small', which is still 'infinitely small'. If you have taken any calculus, you know the value of 'infintely small' is zero. --riverman |
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"Kevin Vang" wrote:
Two darts could conceivable land equidistant from the center; however, the probability of that happening is 0. Explaining why will require a bit of less than elementary probability theory, with integral calculus as a prerequisite. We can go there, if you are up to it... Kevin I was wondering when you would pop up. We could let it go with an "It's intuitively obvious and not important in solving the problem at hand". In practice, we're talking about physical objects which are not made up of infinitesimal particles, but a finite set of rather large (relatively) molecules of stuff, so the set of positions where a dart could penetrate is a very small subset (albeit a very large set) of the set of points on a circle through the location of any dart. In practice, I would prefer to find an approximation to the solution through Monte Carlo simulation (with real darts, not a computer model), accompanied by large quantities of fermented malt. In which case I'd hazard (hah!) a guess that the probability approaches 1 as the amount of malt consumed approaches unconsciousness. -- Stan from the applied side. |
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"riverman" wrote in news:1162440298.459352.48600
@e3g2000cwe.googlegroups.com: The combinatoric arrangement merely gives all the arrangements. Yes, and has next to nothing to do with the probability of any outcomes. In five years, there are two possible outcomes to the experiment "Who has more money, me, or Donald Trump". I could have more money, or he can have more money. By your logic, I therefore have a 50% chance of having more money than The Donald. Good to know. -- Scott Reverse name to reply |
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"Stan Gula" wrote in message news:Fik2h.10075$gf5.7278@trndny01... "Kevin Vang" wrote: Two darts could conceivable land equidistant from the center; however, the probability of that happening is 0. Explaining why will require a bit of less than elementary probability theory, with integral calculus as a prerequisite. We can go there, if you are up to it... Kevin I was wondering when you would pop up. We could let it go with an "It's intuitively obvious and not important in solving the problem at hand". In practice, we're talking about physical objects which are not made up of infinitesimal particles, but a finite set of rather large (relatively) molecules of stuff, so the set of positions where a dart could penetrate is a very small subset (albeit a very large set) of the set of points on a circle through the location of any dart. In practice, I would prefer to find an approximation to the solution through Monte Carlo simulation (with real darts, not a computer model), accompanied by large quantities of fermented malt. In which case I'd hazard (hah!) a guess that the probability approaches 1 as the amount of malt consumed approaches unconsciousness. -- Stan from the applied side. I concur. But I was also about to propose to Jon (who IS a computer geek) to write a computer simulation and run it a few hundred million times. --riverman |
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On 2 Nov 2006 12:37:19 GMT, Scott Seidman wrote:
"riverman" wrote in news:1162440298.459352.48600 : The combinatoric arrangement merely gives all the arrangements. Yes, and has next to nothing to do with the probability of any outcomes. In five years, there are two possible outcomes to the experiment "Who has more money, me, or Donald Trump". I could have more money, or he can have more money. By your logic, I therefore have a 50% chance of having more money than The Donald. Good to know. You might as well run for president, too. You have a 50% chance of being elected. -- Charlie... http://www.chocphoto.com |
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"Scott Seidman" wrote in message . 1.4... "riverman" wrote in news:1162440298.459352.48600 @e3g2000cwe.googlegroups.com: The combinatoric arrangement merely gives all the arrangements. Yes, and has next to nothing to do with the probability of any outcomes. In five years, there are two possible outcomes to the experiment "Who has more money, me, or Donald Trump". I could have more money, or he can have more money. By your logic, I therefore have a 50% chance of having more money than The Donald. Good to know. That's a seriously lame strawman argument. Speaking of straws....put three straws in your hand...one short, one medium, one long. What's the probability of drawing the shortest one last? And what's the probability of drawing the longest one first, and the shortest one after that? --riverman |
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"riverman" wrote in :
And what's the probability of drawing the longest one first, and the shortest one after that? --riverman Yes, that problem is a matter of combinatorials, but that isn't the problem you posed. You toss three darts at a target. Dart A misses the target, then Dart B misses by even more. What is the probability that Dart C will miss by more than Dart A? I just need to know the distribution of dart C, and the location of Dart A. I don't even care if darts A,B, and C are independent. It is not a combinatorial problem. If dart A is 1mm from the target, the probability is very good that dart C will miss by more. If dart A is a mile from the target, the probability is very poor. You could ask your question in a different way, to get the answer you want, which is "you are going to throw three darts at a target. What is the probability that the third dart will miss by more than the first dart?" This is a VERY different question, but the answer is the one you are describing. -- Scott Reverse name to reply |
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Scott Seidman wrote in
. 1.4: If dart A is 1mm from the target, the probability is very good that dart C will miss by more. If dart A is a mile from the target, the probability is very poor. To express this better, the location of the dart is a continuous random variable, but you aren't treating it that way. -- Scott Reverse name to reply |
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"Scott Seidman" wrote in message . 1.4... "riverman" wrote in : And what's the probability of drawing the longest one first, and the shortest one after that? --riverman Yes, that problem is a matter of combinatorials, but that isn't the problem you posed. You toss three darts at a target. Dart A misses the target, then Dart B misses by even more. What is the probability that Dart C will miss by more than Dart A? I just need to know the distribution of dart C, and the location of Dart A. I don't even care if darts A,B, and C are independent. It is not a combinatorial problem. If dart A is 1mm from the target, the probability is very good that dart C will miss by more. If dart A is a mile from the target, the probability is very poor. You're getting close. Considering that dart A and dart C have the same distribution, then either dart can occupy whatever spot the other dart occupies (minus the situation where they occupy the same point). As you said, the probability changes according to the location of dart A. Whatever high probability exists if A is close is countered by the low probability if A is far. Thus, all the possible positions of A and C equal all the possible positions of C and A....its a combinatoric problem. Specifically because I DON'T give the position of dart A. If its any comfort, I'm not making this problem up. --riverman |
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"Charlie Choc" wrote in message -- joe responds glad to see you returned --don"t we get a trip report about your western summer? even if no fishing is included--most of the messages here do not. spent a couple of day in Graham co last week with Jeff----then Dene and I spent a week bumming around snowbird area--one big rain so tried to improve my nymphing ability with slight success, but did dig out one big rainbow in big pool just upstream from snowbird cabin. They call them steelhead up that way as they come into streams from lakes to lay eggs. Went to eye doctor yesterday and after he said I was still 20/20 I explained that tennis balls I once returned now go flying by, and Jeff had to tie on small flies for me. Dr said perhaps I need to clean my glasses and gave me a bill for $121 Kayaking in backwater salt most every day--not catching much but just cannot say inside and do homework when Carolina in Oct is smiling Indian Joe |
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"riverman" wrote in :
Thus, all the possible positions of A and C equal all the possible positions of C and A... But the question as you pose it has nothing to do with all the possible positions of dart A-- it has to do with one specific position of dart A! Let's say that A is 5cm away. Then you are looking for p(C5), which has a value that depends only on the distribution of dart C. Whatever high probability exists if A is close is countered by the low probability if A is far. True before dart A is thrown, but not after dart A is thrown. Now, you have a real honest to goodness value for dart A. Thus, all the possible positions of A and C equal all the possible positions of C and A....its a combinatoric problem. Specifically because I DON'T give the position of dart A. This doesn't mean that you can just ignore the fact that Dart A is stuck at a precise location in the dartboard, and it's why there isn't enough info to offer a p-value. If its any comfort, I'm not making this problem up. Then the person who did got it wrong. -- Scott Reverse name to reply |
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On Thu, 02 Nov 2006 13:37:57 GMT, "Joe McIntosh"
wrote: "Charlie Choc" wrote in message -- joe responds glad to see you returned --don"t we get a trip report about your western summer? even if no fishing is included--most of the messages here do not. I took a lot of pictures and helped build some houses, went to a pow-wow and rodeo, and generally had a ball, although 8 weeks is plenty long enough to sleep in what amounts to a tent with wheels. As for the fishing part, I floated the Bighorn and Yellowstone with 'Bouncer' and we had a fine time and caught some nice fish. I also fished in Yellowstone, but water was low and the streams were crowded - especially the Firehole and in Lamar Valley. The park seemed more crowded the 1st 2 weeks of September than it has been in July, and that combined with the rutting elk and bison make for some dangerous encounters. Several people were hospitalized while I was there, mostly by virtue of getting too close to bull elk. My best 'catching' was probably in Grand Teton and just outside it on the Gros Ventre and the Snake. Not as many fishermen down there but the place was thick with photographers, and you had to be careful and stay away from the rutting moose by the rivers. -- Charlie... http://www.chocphoto.com |
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"Scott Seidman" wrote in message . 1.4... "riverman" wrote in : Thus, all the possible positions of A and C equal all the possible positions of C and A... But the question as you pose it has nothing to do with all the possible positions of dart A-- it has to do with one specific position of dart A! Let's say that A is 5cm away. Then you are looking for p(C5), which has a value that depends only on the distribution of dart C. Whatever high probability exists if A is close is countered by the low probability if A is far. True before dart A is thrown, but not after dart A is thrown. Now, you have a real honest to goodness value for dart A. Yes, but the possbilities for the value for A are limitless. You cannot calculate all the different arrangements. Remember, the original question did not specify where A landed, only that it did. Thus, all the possible positions of A and C equal all the possible positions of C and A....its a combinatoric problem. Specifically because I DON'T give the position of dart A. This doesn't mean that you can just ignore the fact that Dart A is stuck at a precise location in the dartboard, and it's why there isn't enough info to offer a p-value. No its not. Its stuck in any of an infinite number of locations. Each one has a different effect on the probability of B being closer. If its any comfort, I'm not making this problem up. Then the person who did got it wrong. Well, here's the results of one person who ran 1 million trials each, using uniform, normal, gaussian and random distributions. http://www.gatago.com/rec/puzzles/25594126.html He came up with 2/3 every time. I just have an aversion to Monte Carlo methods, but go ahead and try it yourself and let us know what the result is. --riverman |
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"Stan Gula" wrote in message news:Fik2h.10075$gf5.7278@trndny01... ...In practice, I would prefer to find an approximation to the solution through Monte Carlo simulation (with real darts, not a computer model), accompanied by large quantities of fermented malt. In which case I'd hazard (hah!) a guess that the probability approaches 1 as the amount of malt consumed approaches unconsciousness. All well and good as a purely intellectual exercise but in real world situations one cannot afford a cavalier disregard (such as you have demonstrated here) for the effects of important variables like (in this instance) time. If the probability's approach to 1 is proportional to the approach of looming unconsciousness (a proposition I am not prepared to contest) it may nevertheless prove to be chimerical, depending, obviously, on whether or not (and, if so, when) the blessed union with the cosmic consciousness occurs. This, in turn, is a function of rate of consumption over time.* Wolfgang *precise analyses do, of course, also depend on mitigating factors like body mass, whether the malt is consumed in a raw or distilled form, etc., but these details need not concern us here as they can (and, indeed, must) be determined empirically in each experimental run. |
What's a boy to do?
On 1 Nov 2006 16:27:51 -0800, "riverman" wrote:
Calif Bill wrote: You could have an almost infinite amount of darts the exact same distance from the center. The only limiting number is how big the circle is from the center and how big of diameter is the dart. There are an infinite number of points equidistant from the center point. And it depends on neither the precision or accuracy of the measurement. And in your measurement of the distance it would be more accuracy and not precision. Precision only gives more numbers after the decimal point. Bill: I'm not sure where to start, but there are a lot of little details in your assertations that are erroneous. There's some truth also, so don't lose hope :-) First of all, yes the definition of a circle states there are an infinite number of points in a plane that are equidistant from a given point, but the liklihood of getting even two darts to land on that circle is slim. (Just how slim is discussed in the second paragraph below.) We don't even have to agree on how slim for now, but the the more darts you want to have land on that circle, the less likely it is to happen, and it approaches zero as the number of darts gets larger and larger. Although the phrase 'almost infinite' is actually meaningless, I assume you mean we are looking at numbers that are growing huge beyond comprehension, so the liklihood of it happening is shrinking tiny beyond comprehension. Secondly, it IS a matter of precision, not accuracy. We don't care what the actual distance from the center is, what we do care about is whether or not two darts have the same measurement from the center, even if that measurement is wrong. If we use an inaccurate tool, then we might get a wrong amount (a broken ruler might show each dart to be 10.55 cm from the center, while they are both actually much less that that). That's 'inaccurate', but if the numbers match, then we can still assert that they are the same distance. If we use a ruler with really fat indicator lines, we might get both measuring 10.55 cm, however if we used a vernier caliper, calibrated or not, we might get one of them measuring 10.550000000000001 cm and the other measuring 10.550000000000002 cm. Those are measures of high PRECISION, and my assertation is that, no matter how the darts land, we can always use more precise measuring devices until we find where the numbers vary. And they always will, even if we have to go to electron microscope levels. Just as no two snowflakes are alike, no two darts can land the same distance from the center. Now, I appreciate that some people might have an ingrained prejudice against math because it doesn't always conform to their intuition (and this might be you, or it might not). But when faced with something that doesn't seem to 'fit' what we want to believe, there are two choices: find out the rules of math and learn to analyze things according to those rules, including learning the constraints and limitations and the meaning of those, or else continue to assert that what we believe is right because it 'feels right' to us, and use poorly structured arguments or misnomers to claim that nothing has any validity, so we can't possibly be wrong. That way lies madness. --riverman Whether or not _you_ can measure to _your_ satisfaction that two points on a 2-dimensional plane are _absolutely_ the same distance from an initially-chosen point (in this case, a "target"_), those two points certainly exist. The random selection of a second point (the landing of Dart A) "x" distance from the first point (the "target") creates a radius from which a circumference may be scribed. The second dart (Dart B) and its landing point have no relevance and can be ignored. A third dart is thrown (Dart C). According to your theory, that dart can easily and readily strike any point on the disk or any point outside of the circumference created by the selection of the first and second points, up to and including "un-measurably" close to the inside or the outside of the circumference, but can never actually strike a point on the circumference. IOW, the third point (Dart C) can only create a second radius that must be less than or greater than the first radius. With not being able to select a second point on the circumference, arcs, in such a world, don't exist. If arcs don't exist, geometry, trig, etc. begins to break down. In the failure cascade of interrelated bits , it takes all math down with it. Congratulations, you've talked your way out of a fairly decent, secure job...yep, you're a Democrat...ah, well, perhaps there's a job on Kerry's staff for ya... On the practical side, it seems rather curious that you can measure to your own satisfaction that it isn't the same distance, yet you cannot measure to your satisfaction that it is the same distance. HTH, R |
What's a boy to do?
wrote in message ... Whether or not _you_ can measure to _your_ satisfaction that two points on a 2-dimensional plane are _absolutely_ the same distance from an initially-chosen point (in this case, a "target"_), those two points certainly exist. The random selection of a second point (the landing of Dart A) "x" distance from the first point (the "target") creates a radius from which a circumference may be scribed. The second dart (Dart B) and its landing point have no relevance and can be ignored. A third dart is thrown (Dart C). According to your theory, that dart can easily and readily strike any point on the disk or any point outside of the circumference created by the selection of the first and second points, up to and including "un-measurably" close to the inside or the outside of the circumference, but can never actually strike a point on the circumference. IOW, the third point (Dart C) can only create a second radius that must be less than or greater than the first radius. With not being able to select a second point on the circumference, arcs, in such a world, don't exist. If arcs don't exist, geometry, trig, etc. begins to break down. In the failure cascade of interrelated bits , it takes all math down with it. Congratulations, you've talked your way out of a fairly decent, secure job...yep, you're a Democrat...ah, well, perhaps there's a job on Kerry's staff for ya... On the practical side, it seems rather curious that you can measure to your own satisfaction that it isn't the same distance, yet you cannot measure to your satisfaction that it is the same distance. Thus forcing one to conclude that forty bucks worth does not constitute an "expensive education" so much as it does outright robbery. Wolfgang on the other hand, given what we get here for free, forty bucks worth of entertainment would almost certainly kill any mere mortal. :) |
What's a boy to do?
On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote:
SNI-I-I-I-IP I will simply confine myself Well, no, you didn't do either, but perhaps you should... to making a proposition open to anyone. Give me three darts and a prediction of where they will land relative to one another in terms of distance from the center of the target, and I will prove you wrong EVERY time. :) Gee, it seems like this might be an attempt at a sucker bet...OK. I accept. And I'd offer that you couldn't even do it ONE time... and that you couldn't do it even if given a 3-dimensional "dartboard"...but don't pee all over yourself, here's another hint: the taxpayers of Olathe, Kansas are probably very glad you can't do it even that one time...why, heck, one might say that's the essence of an industry... HTH, R ....I feel generous, here's another hint: ya better go back to sucker-bet development school - with the "bet" above, it doesn't matter how, when, or if you throw them... Wolfgang |
What's a boy to do?
wrote in message ... On 1 Nov 2006 16:27:51 -0800, "riverman" wrote: Whether or not _you_ can measure to _your_ satisfaction that two points on a 2-dimensional plane are _absolutely_ the same distance from an initially-chosen point (in this case, a "target"_), those two points certainly exist. Yes, they do. The random selection of a second point (the landing of Dart A) "x" distance from the first point (the "target") creates a radius from which a circumference may be scribed. The second dart (Dart B) and its landing point have no relevance and can be ignored. Not necessarily, it depends on what is being asked. "Conditional probabilities" do exist. But in the case of what you are discussing (the existance of arcs), I concur; we can ignore the second dart for now. A third dart is thrown (Dart C). According to your theory, that dart can easily and readily strike any point on the disk or any point outside of the circumference created by the selection of the first and second points, up to and including "un-measurably" close to the inside or the outside of the circumference, but can never actually strike a point on the circumference. IOW, the third point (Dart C) can only create a second radius that must be less than or greater than the first radius. Yes, that's correct also. There is a statement in calculus that asserts that no matter what two numbers you choose on the number line, there is always another number between them. No matter how close to the circumference you get, you can always get closer. But you cannot get there unless you, well, get there. With not being able to select a second point on the circumference, arcs, in such a world, don't exist. No one said you cannot select a second point. What is being said is that the probability of another dart hitting that point, or any other point on that circle, is zero. Thats because the point is infinitely small. The probability of hitting something infinitely small is infinitely small....zero, in fact. If arcs don't exist, ....and everything after this antecedant is YOUR proposition, since I know they do exist. geometry, trig, etc. begins to break down. In the failure cascade of interrelated bits , it takes all math down with it. Congratulations, you've talked your way out of a fairly decent, secure job...yep, you're a Democrat...ah, well, perhaps there's a job on Kerry's staff for ya... Not me. Say hello to John for us, willya? :-) The way you are posing your interpretation of this is risky. I ask you; How big is a point? If you answer 'infinitely small', then I ask; 'How is it possible to construct anything out of points, then?" Certainly the one-dimensionality of points and the existance of 2- and 3-dimensional objects constructed of points are not exclusive. Math exists, life goes on, and you get to remain a Republican. --riverman |
What's a boy to do?
wrote in message ... On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote: SNI-I-I-I-IP I will simply confine myself Well, no, you didn't do either, but perhaps you should... to making a proposition open to anyone. Give me three darts and a prediction of where they will land relative to one another in terms of distance from the center of the target, and I will prove you wrong EVERY time. :) Gee, it seems like this might be an attempt at a sucker bet...OK. I accept. And I'd offer that you couldn't even do it ONE time... and that you couldn't do it even if given a 3-dimensional "dartboard"...but don't pee all over yourself, here's another hint: the taxpayers of Olathe, Kansas are probably very glad you can't do it even that one time...why, heck, one might say that's the essence of an industry... HTH, R ...I feel generous, here's another hint: ya better go back to sucker-bet development school - with the "bet" above, it doesn't matter how, when, or if you throw them... The beauty of saying nothing is that you can never be proved wrong and that you never have to retract a statement, ainna? One can only suppose that someone suggested this strategy to you and that you stick to it without a hint as to its efficacy out of sheer dogged inability to think of anything else to do. Well, that and the fact that so many play so gently with you. :) Wolfgang who, it must be admitted, has always been a bit rougher with his toys than the other kids. |
What's a boy to do?
Scott Seidman wrote:
... You could ask your question in a different way, to get the answer you want, which is "you are going to throw three darts at a target. What is the probability that the third dart will miss by more than the first dart?" This is a VERY different question, but the answer is the one you are describing. Exactly correct. Good luck trying to convince the roffian gaggle. ;-) -- Ken Fortenberry |
What's a boy to do?
"Jonathan Cook" wrote in message ... Scott Seidman wrote: You could ask your question in a different way, to get the answer you want, which is "you are going to throw three darts at a target. What is the probability that the third dart will miss by more than the first dart?" This is a VERY different question, but the answer is the one you are describing. Yep. I would agree that the probability above is 1/3 if you added the conditional 'given that the second dart missed by more than the first'. Otherwise, I'm inclined (but not convinced) that the probability above is 1/2. Since the outcome of the second dart is not taken into account, your question is identical to 'you throw a dart, watch some TV and eat lunch, then throw another. Whats the probability that the second is closer than the first?' Where I am unsure is because of the existance of the middle dart...I have to think that through more. But I am certain that if you add that conditional, then you have my question, with a probability of 1/3. --riverman |
choc tr
"Charlie Choc" wrote hey, duc, let us know when you post some images, either at your place or ab.--you are coming along well, for a digiboy. :) yfitons wayno |
What's a boy to do?
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What's a boy to do?
On Thu, 2 Nov 2006 23:37:44 +0800, "riverman" wrote:
wrote in message .. . On 1 Nov 2006 16:27:51 -0800, "riverman" wrote: Whether or not _you_ can measure to _your_ satisfaction that two points on a 2-dimensional plane are _absolutely_ the same distance from an initially-chosen point (in this case, a "target"_), those two points certainly exist. Yes, they do. The random selection of a second point (the landing of Dart A) "x" distance from the first point (the "target") creates a radius from which a circumference may be scribed. The second dart (Dart B) and its landing point have no relevance and can be ignored. Not necessarily, it depends on what is being asked. "Conditional probabilities" do exist. But in the case of what you are discussing (the existance of arcs), I concur; we can ignore the second dart for now. Well, I guess it's good that at least some of the time, you don't argue with yourself... A third dart is thrown (Dart C). According to your theory, that dart can easily and readily strike any point on the disk or any point outside of the circumference created by the selection of the first and second points, up to and including "un-measurably" close to the inside or the outside of the circumference, but can never actually strike a point on the circumference. IOW, the third point (Dart C) can only create a second radius that must be less than or greater than the first radius. Yes, that's correct also. There is a statement in calculus that asserts that no matter what two numbers you choose on the number line, there is always another number between them. No matter how close to the circumference you get, you can always get closer. But you cannot get there unless you, well, get there. Oh, geez...if there's a statement and all...well, anyone thinking about math better cut it out...just think of all the books that'll need to be changed if someone ****s up and comes up with something new... With not being able to select a second point on the circumference, arcs, in such a world, don't exist. No one said you cannot select a second point. What is being said is that the probability of another dart hitting that point, or any other point on that circle, is zero. Thats because the point is infinitely small. The probability of hitting something infinitely small is infinitely small....zero, in fact. Infinitely small is not "zero." One can choose to "round it off" and just call it "zero," but it isn't, in fact, non-existent. Here's another hint: consider the points in a tangent to point/Dart A and the points in lines perpendicular to that tangent and...why, shoot, sooner or later, one might account for all the points in the plane, and then, uh-oh... HTH, R |
choc tr
On Thu, 2 Nov 2006 11:36:05 -0500, "Wayne Harrison" wrote:
"Charlie Choc" wrote hey, duc, let us know when you post some images, either at your place or ab.--you are coming along well, for a digiboy. :) There are some new ones on my web site now in the Arches, Grand Teton and Yellowstone folders, but I've got more shots that I haven't 'processed' yet. -- Charlie... http://www.chocphoto.com |
What's a boy to do?
On Thu, 2 Nov 2006 10:38:41 -0600, Kevin Vang wrote:
In article , says... According to your theory, that dart can easily and readily strike any point on the disk or any point outside of the circumference created by the selection of the first and second points, up to and including "un-measurably" close to the inside or the outside of the circumference, but can never actually strike a point on the circumference. IOW, the third point (Dart C) can only create a second radius that must be less than or greater than the first radius. With not being able to select a second point on the circumference, arcs, in such a world, don't exist. If arcs don't exist, geometry, trig, etc. begins to break down. In the failure cascade of interrelated bits , it takes all math down with it. It's not that the arc doesn't exist, and we cannot choose points on that arc. The point is that the probability of hitting that arc with a dart is 0. Intuitive explanation: Suppose your dartboard has radius 1. Throw a dart at the dartboard, and let r1 = radius from the center of the dartboard to the dart. Now throw a second dart, and let r be the radius. Then the probability that r = r1 is number of values of r for which r = r1 1 ------------------------------------------- = ------------ = 0. number of possible values for r infinity More technical (and more correct) explanation: If we assume that every point on the dartboard is equally likely to be hit, then the probability that r = r1 is: measure of the set for which r = r1 0 -------------------------------------- = ------------ = 0 measure of the dartboard pi * 1^2 because the dartboard is a 2-dimensional surface, the appropriate measure is area. The measure of the entire dartboard is the area of a circle with radius 1, so the area is pi*1^2 = 1. The set of points for which r = r1 is the circle with radius r1. Since the circle is just a curve with width 0 on the plane, it has area 0. Slightly more technical (and more correct): Not every point on the dartboard is equally likely to be hit. Apparently. The word on the street is that at least some are completely unhittable, what with the probability of doing so being zero or infinitely small or pi-r-square or, well, something all dangerously full of symbols and greek letters and ****... If p(r,theta) is the probablility density function giving the probability that the dart hits point (r,theta) in polar coordinates, then the probability that r = r1 is: / r1 | p(r,theta) dA / r1 0 ------------------------- = --- = 0 / 1 1 | p(r,theta) dA / 0 because we are integrating with respect to area, and the top integral is done over a region with area 0, so the value of the integral is 0. SEE! SEE! I WARNED YA, BUT NOO-O-O-O-O... IAC, three answers, each "and more correct" than the previous one. Interesting. Is this progression going to lead to something infinitely correct (or something to at least stick a fork in and call "done"), or is the probability of hitting that target zero, too? HTH, Kevin And I'm pretty certain that mathematics doesn't all disappear if somebody doesn't understand one bit of it. Hey, go easy on me, I'm learning...for example, I've already learned that when 2 math whiz-types and a rat-gutter answer a question, the odds that they will come up with the correct answer is like one in a gazillion or bazillion or some other REALLY big ol' number... And right back at ya, Pythagoras R |
What's a boy to do?
On Thu, 2 Nov 2006 10:03:10 -0600, "Wolfgang" wrote:
wrote in message .. . On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote: SNI-I-I-I-IP I will simply confine myself Well, no, you didn't do either, but perhaps you should... to making a proposition open to anyone. Give me three darts and a prediction of where they will land relative to one another in terms of distance from the center of the target, and I will prove you wrong EVERY time. :) Gee, it seems like this might be an attempt at a sucker bet...OK. I accept. And I'd offer that you couldn't even do it ONE time... and that you couldn't do it even if given a 3-dimensional "dartboard"...but don't pee all over yourself, here's another hint: the taxpayers of Olathe, Kansas are probably very glad you can't do it even that one time...why, heck, one might say that's the essence of an industry... HTH, R ...I feel generous, here's another hint: ya better go back to sucker-bet development school - with the "bet" above, it doesn't matter how, when, or if you throw them... The beauty of saying nothing is that you can never be proved wrong and that you never have to retract a statement, ainna? One can only suppose that someone suggested this strategy to you and that you stick to it without a hint as to its efficacy out of sheer dogged inability to think of anything else to do. Well, that and the fact that so many play so gently with you. :) Hee, hee, hee... Again, I accept your proposal...wanna bet on the outcome? Wolfgang who, it must be admitted, has always been a bit rougher with his toys than the other kids. Hmmm...maybe a big handful of Albolene would cut down on the irritation... While from a humanity standpoint I hope that helps, from a keeping-down-lunch standpoint, I don't care to know if it did, R |
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"riverman" wrote in message oups.com... Calif Bill wrote: "Kevin Vang" wrote in message t... In article t, says... Your statement was no two darts could land equidistant from the center. As I state, totally wrong. Two darts could conceivable land equidistant from the center; however, the probability of that happening is 0. Explaining why will require a bit of less than elementary probability theory, with integral calculus as a prerequisite. We can go there, if you are up to it... Kevin Explain why could not happen. I don't know if you have taken Integral Calculus, so I won't use that to explain it. The simplest 'geometrically appealing' explanation is to say that the probability of hitting some target is directly related to the size of the target. The bigger the spot you are trying to hit, the more likely it is to hit it. Now, if you are aiming for a curved line....specifically the line that describes a circle, you have to consider the width of that line. Since circles are a collection of points, and points have no width, then you are essentially aiming for something that has a width of zero. Which means the possibility of hitting it is zero. You can hit it ONCE, because its position is unknown until the first dart actually lands. Once that dart establishes the position of the circle, it becomes a target of width zero which another dart cannot possibly hit. You can reduce the size of your miss (lets call it your 'slop') to be as small as you want by making less and less precise measurements, but the converse is also true...you can always find the slop by increasing the precision of your measurement. You cannot eliminate the slop; you cannot ever hit the line.: its a matter of precision. Just like no two snowflakes are alike. Throwing more and more darts DOES increase the liklihood of hitting something more than once, but if the liklihood of hitting it is already infinitely small; throwing more and more darts doesn't make it any more likely. Throwing infinite darts creates what is called an 'indeterminate' form, and we cannot solve it that way. We get an infinitely large number of opportunities of achieving something that has an infinitely small chance of happening. Its incorrect to conclude that it will happen infinite times....Infinity x (1/infinity) does not equal infinity. It doesn't equal, or mean, anything. However, throwing a FINITE number of darts....say two, at something that we have an infintely small chance of hitting IS solvable. Its three times 'infinitely small', which is still 'infinitely small'. If you have taken any calculus, you know the value of 'infintely small' is zero. --riverman Actually I have taken integral calculus and am a retired engineer and signal processing engineer. Using your analogy that the point is infinitely small and therefore non-existent, then the first dart could not hit a point either. And would not be in existence. Your supposition that there could never be another dart at the same exact distance fails obviously. Reminds me of the old story about the difference between a mathematician and an engineer. The two are on the football field goal line and on the other goal line is the best of the Dallas cheerleaders, buck naked. The mathman and the engineer are told the first person that gets there gets to do any thing they desire with the lassie. Only rule is you can only move 1/2 the distance to the goal in any one move. The mathman says no use to start as it is an infinite series and you will never get there. The engineer, says 8 moves and I can be close enough for my purposes. |
What's a boy to do?
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What's a boy to do?
"Calif Bill" wrote in message k.net... "riverman" wrote in message oups.com... Actually I have taken integral calculus and am a retired engineer and signal processing engineer. Using your analogy that the point is infinitely small and therefore non-existent, then the first dart could not hit a point either. And would not be in existence. Ummm, I never said it was my analogy. Its part of the definition of a point, and has been around since Euclid. Points are infinitely small 1-dimensional objects; no one said they were non-existant. I said the probability of hitting a point on a defined arc is zero. This can be proven with theoretical math, or intuitively defended as a matter of precision in measurement. Like Kevin, I go through this with my own AB and BC Calculus and Honors Precalculus students yearly, and have for 14 years. And like his story of approaching the wall, I use Zeno's paradox about Achilles and the tortoise to show the dangers of trying to solve problems using indeterminate forms. Your supposition that there could never be another dart at the same exact distance fails obviously. Reminds me of the old story about the difference between a mathematician and an engineer. LOL. Even without the myriad of stories, your use of 'obviously' underscores the difference between mathematicians and engineers. :-) --riverman |
What's a boy to do?
"Calif Bill" wrote in message k.net... ...the difference between a mathematician and an engineer.... Mathematicians don't (usually) get a license to kill and they are (usually) guilty of a passing familiarity with numbers and stuff like that. Wolfgang |
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