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What's a boy to do?
wrote in message ... 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 Hee, hee, hee. Wolfgang is it just me or has anyone else noticed that the probability of surrenders in general (and dicklet's in particular) being gracious tends to decrease over time? |
What's a boy to do?
wrote in message ... 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? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. 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... Maybe. I'm sure we are all eager to hear the results of the experiment While from a humanity standpoint I hope that helps, from a keeping-down-lunch standpoint, I don't care to know if it did, Humanity has nothing to do with it.......toys aren't people. Wolfgang |
What's a boy to do?
"Wolfgang" wrote in message ... "Kevin Vang" wrote in message It's an illusion. Obviously, you can never reach the wall because you can subdivide the remaining distance to it in half infinitely. Naturally, it follows that you are infintely adding increments of distance to that already travelled. This will, eventually, result in travelling an infinite distance. This is all very tiring. There is also the mental strain of dealing with the proposition of an infinite distance yet to go. You are actually dropping of sheer exhaustion about half way to the wall. Lively discussion then ensues... You should direct some of your less gifted students to leave the classroom and come here. This will have the salutary effect of raising the IQ in both places. :) Unfortunately, by your logic, they can't get here either. :-( --riverman (and exactly where they ARE has historically been the source of lively teacher's room discussions for an eternity, or more.) |
What's a boy to do?
On Thu, 2 Nov 2006 12:27:39 -0600, Kevin Vang wrote:
In article t, says... 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. Then "mathman" clearly didn't know what he was talking about. I do this in my Calc II class every year when we get to the chapter on infinite series. I don't have any nekkid cheerleaders, Yeah, the odds of most mathematicians having any of THEM is infinitely small... but I stand at one wall Now, that, they often do... TC, R ....oh, please...fine, fine, fine: G... |
What's a boy to do?
On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote:
Again, I accept your proposal...wanna bet on the outcome? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. Hang on a minute....first things first. How much would you like to lose? 2500USD? 1000EURO? 1000GBP? |
What's a boy to do?
On Thu, 02 Nov 2006 18:14:56 GMT, "Calif Bill"
wrote: 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. And while these two geniuses were figuring, calculating, and such, any actual man present would say "rules my ass," walk on down and make her see God a few times, help her up on her shaking, bowed legs, get her a real nice dress, and take her to drinks and dinner...and if things continue to go well, probably spend the night at her place... HTH, R ....I mean, really...what guy with any sense at all worries about all this theoretical gibberish when there's real, good-looking pussy just 300 feet away... |
What's a boy to do?
wrote in message ... On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote: Again, I accept your proposal...wanna bet on the outcome? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. Hang on a minute....first things first. How much would you like to lose? 2500USD? 1000EURO? 1000GBP? Sure. What time shall I expect you? Wolfgang oh, we're going to have so much fun! |
What's a boy to do?
On Thu, 2 Nov 2006 14:26:25 -0600, "Wolfgang" wrote:
wrote in message .. . On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote: Again, I accept your proposal...wanna bet on the outcome? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. Hang on a minute....first things first. How much would you like to lose? 2500USD? 1000EURO? 1000GBP? Sure. What time shall I expect you? Yeah, that's what I thought...you puss out before you're forced to welsh on the bet...run back under the porch, lil' pup... |
What's a boy to do?
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What's a boy to do?
Jonathan Cook wrote: Jonathan Cook wrote: "you threw" versus "you throw"... I suppose I should have wrote "you threw" versus "you are going to throw". Actually, it doesn't matter. Are we in agreement that the statement and my statement are supposed to be identical except for tense? If so, then its definitely not relevant. The probability of some outcome does not change before or after the event occurs. Stating 'you are going to throw three darts' presupposes a range of possible outcomes. Stating 'you threw three darts' without specifying what any of the specific results were does not eliminate any of the presupposed outcomes. Stating 'you have thrown two darts, and now you are about to throw the third' likewise, in the absence of giving any specifics about the results of the previous two tosses, does not eliminate any of the presupposed outcomes. Scott is creating his own unsolvable problem by asserting that there is an honest-to-goodness value for the position of dart A, and then asserting that he has to know what it is to solve the problem. You have to consider ALL the continuous values for dart A, which happen to be all the values of dart B (minus the singularity where they occupy the same point). As such, this is a completely symmetrical problem, and all results are equally likely over the array of possible outcomes. There is a similar problem using discrete variables rather than continuous. A man says he has two kids, and mentions that one is a girl. What is the probability that the other is also a girl? The answer is NOT 50%, since some information has been disclosed. But if he says "I have two kids, and here is one of them", the the probability that the other is a girl IS 50%. All events are symmetric. These problems; the Monty Hall one, the three darts one, the two kids ones, and several more like them, are well-understood and often used to illustrate what I mentioned early on: that its easy (especially when dealing with infinities) to create indeterminate forms, or when dealing with discrete outcomes, to try to solve problems using methods that lead to dead ends. However, there are often methods that reduce the problems to much more solvable forms, without prejudice. In fact, most of probability is just that; finding a simpler, accurate way to model a problem. The key to solving the three darts problem is to see that each event is independant and has the same distribution (regardless of what the distribution is). As such, the darts are interchangable without prejudice, and all those continuous random variables and infinite possible arrangements simplify to a counting problem. Tense has nothing to do with it, because absolutely no infomation is disclosed if a dart or two or three have already been thrown. If information IS disclosed, then you can no longer assume that three events are symmetrical, as the P(dartB = dartA) becomes zero. Try modeling it with a Monte Carlo program. You'll get 2/3. --riverman |
What's a boy to do?
"Kevin Vang" wrote:
snip I don't have any nekkid cheerleaders... Life isn't perfectg , but I stand at one wall and tell the class I plan to walk half of the distance to the other wall, then walk half of the remaining difference, then walk half the remaining difference, and so on. I ask "Will I make it to the other wall?" I ask for a show of hands, and most students raise their hands for "No". Then I say, "Watch me," and walk directly across the room until I bump into the far wall. Lively discussion then ensues... See, the problem of modeling a theoretical problem with real world finite sized objects. When I was in school the last time, they had just built a 30 story library tower with a roughly square cross section and notched corners. You could walk into one of these notches and have walls of brick about 10 feet wide and 30 stories tall in front of you. I had a little demonstration for friends inspired by Zeno's Paradox (and as you might expect, this was inspired by, and usually demonstrated after considerable beer had been disappeared). I would start them off about 100 yards away from the corner and have them walk toward the notch, fixing their vision at the top of the building. The 'theory' was that the apparent height of the building would increase as they got closer and the rate of the increase would appear to accelerate as they got closer. If you could get right into the corner you would see the effect of the building 'growing' very fast. It was hilarious that almost everybody would walk smack into the brick wall, transfixed by the illusion. Hilarious to me, that is. Unfortunately, being a State building, the bricks started falling off the facade, and they closed access to the building. |
What's a boy to do?
"riverman" wrote in message
There is a similar problem using discrete variables rather than continuous. A man says he has two kids, and mentions that one is a girl. What is the probability that the other is also a girl? The answer is NOT 50%, since some information has been disclosed. But if he says "I have two kids, and here is one of them", the the probability that the other is a girl IS 50%. All events are symmetric. I'm having a problem with the wording of the questions, I suppose. In your example above, I believe that once the first girl is revealed, the question more accurately becomes, "What is the probability that both children are girls?" I agree that's not .50. I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin, the probability of heads is .50. If I toss it again, what is the probability of heads? It's .50. How is that different from the man with two children? Joe F. |
What's a boy to do?
On 2 Nov 2006 15:13:08 -0800, "Wolfgang" wrote:
wrote: On Thu, 2 Nov 2006 14:26:25 -0600, "Wolfgang" wrote: wrote in message .. . On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote: Again, I accept your proposal...wanna bet on the outcome? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. Hang on a minute....first things first. How much would you like to lose? 2500USD? 1000EURO? 1000GBP? Sure. What time shall I expect you? Yeah, that's what I thought...you puss out before you're forced to welsh on the bet...run back under the porch, lil' pup... O.k., let me see if I've got this straight. You want people to believe that you were seriously offering to make a bet about something or other that is a complete mystery to you and that I had taken you seriously and accepted the bet and that somehow or other somebody was going to do something or other to settle the bet in front of reliable witnesses to everyone's satisfaction and that the results would then be duly published here and that I would be appropriately contrite and confess that I really DID know all along that you are God......and.....something or other more. That about it? Nope. While your claimed hesitance to take advantage of someone who didn't understand what they were doing would be oh-so-noble, it's BS. If one wagers, one risks losing. All you needed to do is put your money where your mouth is. What other people believe or don't believe is up to them. You made a proposition (which I note you snipped from above): "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." and I twice made it clear that I accepted it I didn't qualify or otherwise modify it. I then offered to wager and gave an idea of the stakes in which I was willing to risk. If those stakes were too much for you, you could have offered lesser with no shame. If you swung as big a dick as you'd like to think, you would have simply accepted or declined like a man, put the ball in my court, and have done as I did - not worry about what "people" might or might not believe. Instead, you pussed. In fact, I'd offer that I was wagering on that, too... Like I said - run back under the porch, lil' pup, R |
What's a boy to do?
"rb608" wrote in message news:WWF2h.7479$OK3.540@trndny09... "riverman" wrote in message There is a similar problem using discrete variables rather than continuous. A man says he has two kids, and mentions that one is a girl. What is the probability that the other is also a girl? The answer is NOT 50%, since some information has been disclosed. But if he says "I have two kids, and here is one of them", the the probability that the other is a girl IS 50%. All events are symmetric. I'm having a problem with the wording of the questions, I suppose. In your example above, I believe that once the first girl is revealed, the question more accurately becomes, "What is the probability that both children are girls?" I agree that's not .50. I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin, the probability of heads is .50. If I toss it again, what is the probability of heads? It's .50. How is that different from the man with two children? The main difference is that one is a conditional probability, the other is not. The possible sample space for someone having two children is BB BG GB GG. Since man gave us the condition that 'at least one is a girl', you know that BB is not a possibility. Thus, the remaining possible sample space is BG, GB or GG (these all meet his condition), and the only sucessful outcome is GG, so the probability that "the other child is also a girl, given that at least one is a girl" is 1/3. The Rosencrantz/Guilderstern example differs. If you ask "what's the probability that the second coin toss is H" the answer is .50, regardless of what took place before---its not conditional. If you ask "whats the probability of getting two heads, given that the FIRST toss is heads", then we have to limit our possible outcomes to HT or HH (these are the only outcomes that meet the condition), and the only successful outcome is HH, with a probability of 1/2. That would be similar to the man identifying the girl as his eldest, or his youngest. That serves to remove one of the BG/GB options. But if you ask "whats the probability of tossing a coin twice and getting two Heads, given that you get at least one Heads", then the two problems are identical. The possible outcomes are HH, HT and TH, and the successful outcome is HH. --riverman |
What's a boy to do?
wrote in message ... On 2 Nov 2006 15:13:08 -0800, "Wolfgang" wrote: wrote: On Thu, 2 Nov 2006 14:26:25 -0600, "Wolfgang" wrote: wrote in message .. . On Thu, 2 Nov 2006 13:00:17 -0600, "Wolfgang" wrote: Again, I accept your proposal...wanna bet on the outcome? Sure. You bring the dartboard......and the absinthe. I got Oprah and Emeril on DVD. Hang on a minute....first things first. How much would you like to lose? 2500USD? 1000EURO? 1000GBP? Sure. What time shall I expect you? Yeah, that's what I thought...you puss out before you're forced to welsh on the bet...run back under the porch, lil' pup... O.k., let me see if I've got this straight. You want people to believe that you were seriously offering to make a bet about something or other that is a complete mystery to you and that I had taken you seriously and accepted the bet and that somehow or other somebody was going to do something or other to settle the bet in front of reliable witnesses to everyone's satisfaction and that the results would then be duly published here and that I would be appropriately contrite and confess that I really DID know all along that you are God......and.....something or other more. That about it? Nope. While your claimed hesitance to take advantage of someone who didn't understand what they were doing would be oh-so-noble, it's BS. If one wagers, one risks losing. All you needed to do is put your money where your mouth is. What other people believe or don't believe is up to them. You made a proposition (which I note you snipped from above): "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." and I twice made it clear that I accepted it I didn't qualify or otherwise modify it. I then offered to wager and gave an idea of the stakes in which I was willing to risk. If those stakes were too much for you, you could have offered lesser with no shame. If you swung as big a dick as you'd like to think, you would have simply accepted or declined like a man, put the ball in my court, and have done as I did - not worry about what "people" might or might not believe. Instead, you pussed. In fact, I'd offer that I was wagering on that, too... Like I said - run back under the porch, lil' pup, So, you're saying that you seriously DO expect people to believe that you were seriously offering to make a bet about something or other that is a complete mystery to you and that I had taken you seriously and accepted the bet and that somehow or other somebody was going to do something or other to settle the bet in front of reliable witnesses to everyone's satisfaction and that the results would then be duly published here and that I would be appropriately contrite and confess that I really DID know all along that you are God......and.....something or other more?!! You don't get any brighter as the week gets older, do you? :) O.k., let's pretend (just for the moment) that you could find your way to the real world for a brief visit. How would you propose that your little bet be settled? Wolfgang |
What's a boy to do?
"riverman" wrote in message ... "rb608" wrote in message news:WWF2h.7479$OK3.540@trndny09... "riverman" wrote in message There is a similar problem using discrete variables rather than continuous. A man says he has two kids, and mentions that one is a girl. What is the probability that the other is also a girl? The answer is NOT 50%, since some information has been disclosed. But if he says "I have two kids, and here is one of them", the the probability that the other is a girl IS 50%. All events are symmetric. I'm having a problem with the wording of the questions, I suppose. In your example above, I believe that once the first girl is revealed, the question more accurately becomes, "What is the probability that both children are girls?" I agree that's not .50. I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin, the probability of heads is .50. If I toss it again, what is the probability of heads? It's .50. How is that different from the man with two children? The main difference is that one is a conditional probability, the other is not. The possible sample space for someone having two children is BB BG GB GG. Since man gave us the condition that 'at least one is a girl', you know that BB is not a possibility. Thus, the remaining possible sample space is BG, GB or GG (these all meet his condition), and the only sucessful outcome is GG, so the probability that "the other child is also a girl, given that at least one is a girl" is 1/3.... O.k., I will stress, once again, that I'm no mathematician, but this has a funny smell to it. Looks to me like the math is unassailable as long as BG is something different than GB......but it manifestly isn't. BG and GB are, in fact, exactly the same thing and thus we a remaining sample space of TWO, not three possibilities......unless we posit that birth order enters the equation, in which case, the whole thing falls apart and we are dealing with an entirely different problem. Wolfgang |
What's a boy to do?
Wolfgang wrote:
O.k., I will stress, once again, that I'm no mathematician, There's no need to point that out. It's self evident. -- Cut "to the chase" for my email address. |
What's a boy to do?
"rw" wrote in message m... Wolfgang wrote: O.k., I will stress, once again, that I'm no mathematician, There's no need to point that out. It's self evident. You don't learn. What is the correct solution to Myron's problem? Wolfgang |
What's a boy to do?
Wolfgang wrote:
"rw" wrote in message m... Wolfgang wrote: O.k., I will stress, once again, that I'm no mathematician, There's no need to point that out. It's self evident. You don't learn. What is the correct solution to Myron's problem? Myron gave the correct solution. If we were told that the OLDEST child is a girl, the probability of the other child being a girl is 1/2. But we weren't told that. -- Cut "to the chase" for my email address. |
What's a boy to do?
"Wolfgang" wrote in message ... "riverman" wrote in message ... "rb608" wrote in message news:WWF2h.7479$OK3.540@trndny09... "riverman" wrote in message There is a similar problem using discrete variables rather than continuous. A man says he has two kids, and mentions that one is a girl. What is the probability that the other is also a girl? The answer is NOT 50%, since some information has been disclosed. But if he says "I have two kids, and here is one of them", the the probability that the other is a girl IS 50%. All events are symmetric. I'm having a problem with the wording of the questions, I suppose. In your example above, I believe that once the first girl is revealed, the question more accurately becomes, "What is the probability that both children are girls?" I agree that's not .50. I'll toss out the Rosencrantz and Guilderstern example. If I toss a coin, the probability of heads is .50. If I toss it again, what is the probability of heads? It's .50. How is that different from the man with two children? The main difference is that one is a conditional probability, the other is not. The possible sample space for someone having two children is BB BG GB GG. Since man gave us the condition that 'at least one is a girl', you know that BB is not a possibility. Thus, the remaining possible sample space is BG, GB or GG (these all meet his condition), and the only sucessful outcome is GG, so the probability that "the other child is also a girl, given that at least one is a girl" is 1/3.... O.k., I will stress, once again, that I'm no mathematician, but this has a funny smell to it. Looks to me like the math is unassailable as long as BG is something different than GB......but it manifestly isn't. BG and GB are, in fact, exactly the same thing and thus we a remaining sample space of TWO, not three possibilities......unless we posit that birth order enters the equation, in which case, the whole thing falls apart and we are dealing with an entirely different problem. No, it is correct. Dr Math explains it better than I do at this ungodly hour :-) http://mathforum.org/dr.math/faq/faq.boy.girl.html --riverman |
What's a boy to do?
"rw" wrote in message m... Wolfgang wrote: "rw" wrote in message m... Wolfgang wrote: O.k., I will stress, once again, that I'm no mathematician, There's no need to point that out. It's self evident. You don't learn. What is the correct solution to Myron's problem? Myron gave the correct solution. If we were told that the OLDEST child is a girl, the probability of the other child being a girl is 1/2. But we weren't told that. And you'd be willing to bet dicklet a million dollars on that? Wolfgang the boy just WILL NOT learn! :) |
What's a boy to do?
Wolfgang wrote:
"rw" wrote in message m... What is the correct solution to Myron's problem? Myron gave the correct solution. If we were told that the OLDEST child is a girl, the probability of the other child being a girl is 1/2. But we weren't told that. And you'd be willing to bet dicklet a million dollars on that? I'm not interested in debating elementary combinatorics with mathematical ignoramuses. -- Cut "to the chase" for my email address. |
What's a boy to do?
"rw" wrote in message m... Wolfgang wrote: "rw" wrote in message m... What is the correct solution to Myron's problem? Myron gave the correct solution. If we were told that the OLDEST child is a girl, the probability of the other child being a girl is 1/2. But we weren't told that. And you'd be willing to bet dicklet a million dollars on that? I'm not interested in debating elementary combinatorics with mathematical ignoramuses. Well, given that you didn't understand my question, I'd say that's probably a very wise decision.......it must be humiliating enough to lose debates with everyone else. Wolfgang have i mentioned recently that the boy just WILL NOT learn? |
What's a boy to do?
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What's a boy to do?
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What's a boy to do?
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What's a boy to do?
"riverman" wrote in message ... The possible sample space for someone having two children is BB BG GB GG. Since man gave us the condition that 'at least one is a girl', you know that BB is not a possibility. Thus, the remaining possible sample space is BG, GB or GG (these all meet his condition), and the only sucessful outcome is GG, so the probability that "the other child is also a girl, given that at least one is a girl" is 1/3.... O.k., I will stress, once again, that I'm no mathematician, but this has a funny smell to it. Looks to me like the math is unassailable as long as BG is something different than GB......but it manifestly isn't. BG and GB are, in fact, exactly the same thing and thus we a remaining sample space of TWO, not three possibilities......unless we posit that birth order enters the equation, in which case, the whole thing falls apart and we are dealing with an entirely different problem. No, it is correct. Dr Math explains it better than I do at this ungodly hour :-) http://mathforum.org/dr.math/faq/faq.boy.girl.html Dr. Math says: "In a two-child family, there are four and only four possible combinations of children. We will label boys B and girls G; in each case the first letter represents the oldest child: {BB, BG, GB, GG}. When we know that one child is a boy, there cannot be two girls, so the sample space shrinks to: {BB, BG, GB}. Two of the possibilities in this new sample space include girls: {BG, GB}and since there are two combinations out of three that include girls, the probability that the second child is a girl is 2/3." ( I changed the spacing and added a couple of periods in the interest of saving space and maintaining clarity in the process....I don't believe I have changed any of the meaning in the process.) I'd say your explanation is no worse than Dr. Math's. They appear identical in all essentials. Still smells funny to me. In the absence of any positional element (birth order, size, where they might happen to be standing relative to one another in a photo, etc.) how does {a boy and a girl} on the one hand differ from {a girl and a boy} on the other? To be sure, {BG} LOOKS different than {GB}but except with regard to the relative positions of the labels for {B}oy and {G}irl they are identical entities. Why does one entity get counted as two possibilities? Wolfgang |
What's a boy to do?
On Fri, 3 Nov 2006 09:19:28 -0600, "Wolfgang" wrote:
O.k., let's pretend (just for the moment) that you could find your way to the real world for a brief visit. How would you propose that your little bet be settled? Since you've asked so nicely, it's only gonna cost you a little bit to find out and you won't lose much. Here's the deal, one-time offer, not negotiable. You agree to send a check for $25USD to St. Jude's. Don't like St. Jude's? Tough ****. My offer, my choice. If you agree, then I'll outline the rest, with the wager being that if you can't prove me wrong, you send a check for $100USD (total) to St. Jude's and if you can prove me wrong, you get whatever satisfaction you might get. Don't like any of the above? Again, tough ****, etc. Are you in or out? R |
What's a boy to do?
Wolfgang wrote:
To be sure, {BG} LOOKS different than {GB}but except with regard to the relative positions of the labels for {B}oy and {G}irl they are identical entities. Why does one entity get counted as two possibilities? That's what bothers me as well. To be sure, the mathematics is correct as modeled, but I think this debate arises from attempting to word a precise, non-mathematical, real-world question to fit a desired abstract (and, I would argue, a deliberatlely oblique) answer. I don't care how many kids the guy has, the chances of any one of them being a B or G is still the same as a coin flip (genetics notwithstanding). It's clear to me that Dr. Math & Myron are both correct in their analysis of the mathematics; but I take issue that the question as asked accurately describes the model subsequently analyzed. Joe F. |
What's a boy to do?
"rb608" wrote in message ups.com... Wolfgang wrote: To be sure, {BG} LOOKS different than {GB}but except with regard to the relative positions of the labels for {B}oy and {G}irl they are identical entities. Why does one entity get counted as two possibilities? That's what bothers me as well. To be sure, the mathematics is correct as modeled, but I think this debate arises from attempting to word a precise, non-mathematical, real-world question to fit a desired abstract (and, I would argue, a deliberatlely oblique) answer. I don't care how many kids the guy has, the chances of any one of them being a B or G is still the same as a coin flip (genetics notwithstanding). It's clear to me that Dr. Math & Myron are both correct in their analysis of the mathematics; but I take issue that the question as asked accurately describes the model subsequently analyzed. Joe F. Exactly......though I'd phrase it a bit differently, I think. The answer is correct. The trouble is that it is the correct answer to the wrong question. :) One doesn't need to know much about mathematics to recognize a logical error or fallacy. There is unquestionably a positional element in the answer that is absent from the problem as stated. Wolfgang |
What's a boy to do?
wrote in message ... On Fri, 3 Nov 2006 09:19:28 -0600, "Wolfgang" wrote: O.k., let's pretend (just for the moment) that you could find your way to the real world for a brief visit. How would you propose that your little bet be settled? Since you've asked so nicely, it's only gonna cost you a little bit to find out and you won't lose much. Here's the deal, one-time offer, not negotiable. You agree to send a check for $25USD to St. Jude's. O.k. Don't like St. Jude's? Tough ****. My offer, my choice. I have no problem at with St. Jude's......hell, I don't even know who St. Jude's is. If you agree, And I do. then I'll outline the rest, O.k., I'll be waiting for the outline. with the wager being that if you can't prove me wrong, you send a check for $100USD (total) to St. Jude's and if you can prove me wrong, you get whatever satisfaction you might get. Sounds good to me. Don't like any of the above? Loving every bit of it. Again, tough ****, etc. You sound angry. Was it something I said? Are you in or out? I'm in. Wolfgang and the check is in the mail....and.....um.....oh yeah, i won't come in your mouth. |
What's a boy to do?
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What's a boy to do?
On Fri, 3 Nov 2006 14:10:14 -0600, "Wolfgang" wrote:
wrote in message .. . On Fri, 3 Nov 2006 09:19:28 -0600, "Wolfgang" wrote: O.k., let's pretend (just for the moment) that you could find your way to the real world for a brief visit. How would you propose that your little bet be settled? Since you've asked so nicely, it's only gonna cost you a little bit to find out and you won't lose much. Here's the deal, one-time offer, not negotiable. You agree to send a check for $25USD to St. Jude's. O.k. Don't like St. Jude's? Tough ****. My offer, my choice. I have no problem at with St. Jude's......hell, I don't even know who St. Jude's is. If you agree, And I do. then I'll outline the rest, O.k., I'll be waiting for the outline. Simple. Here is what you proposed: "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." I hereby give to you three darts. Anytime you wish to collect them, let me know. Here's my prediction: One dart will land at or specific distance from the center of the target (COT) with no darts closer in distance to the COT. Another dart will land a distance equal to or greater than the distance of one dart from the COT, and equal to or less than the distance of one dart from the COT, and another dart will land a distance equal to or greater than two of the darts from the COT. If the "dartboard" is of the type that can never have two darts _exactly_ the same distance from the COT, one dart will land at or a specific distance from the COT with no darts closer in distance to the COT, another dart will land a distance greater from the COT than one dart but less than another dart, and another dart will land a distance greater than two of the darts from the COT. If necessary, I'm sure someone can whip out the x's and y's and Greek letters and symbols and all sorts of really boring stuff to be all planar triangulation and um, "gee...pee...essence" and stuff, but I'm not particularly interested in doing so... FWIW, I just made three imaginary throws (I imagined that I used your darts - they were handy - hope you don't mind) and they imaginarily landed as follows: 17 fat, 8 small pie, 5 double-ring Your turn to prove my prediction "wrong EVERY time." And by the way, http://www.stjude.org/donate R ....heck, I could have predicted _anything_, and you would have a rather hard time proving it wrong "EVERY" time... |
What's a boy to do?
wrote: "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." I hereby give to you three darts. Where? I don't see 'em. Anytime you wish to collect them, let me know. Ah! Tricksy little hobbit! Say, I'll bet sixteen thousand gazillion dollars that you're a near graduate of the southern Mississippi school of law and numbers and stuff. Here's my prediction: One dart will land at or specific distance from the center of the target (COT) with no darts closer in distance to the COT. Another dart will land a distance equal to or greater than the distance of one dart from the COT, and equal to or less than the distance of one dart from the COT, and another dart will land a distance equal to or greater than two of the darts from the COT. If the "dartboard" is of the type that can never have two darts _exactly_ the same distance from the COT, one dart will land at or a specific distance from the COT with no darts closer in distance to the COT, another dart will land a distance greater from the COT than one dart but less than another dart, and another dart will land a distance greater than two of the darts from the COT. If necessary, I'm sure someone can whip out the x's and y's and Greek letters and symbols and all sorts of really boring stuff to be all planar triangulation and um, "gee...pee...essence" and stuff, but I'm not particularly interested in doing so... FWIW, I just made three imaginary throws (I imagined that I used your darts - they were handy - hope you don't mind) and they imaginarily landed as follows: 17 fat, 8 small pie, 5 double-ring Your turn to prove my prediction "wrong EVERY time." Well all that is.....um.....hm......what's the word I'm looking for here.....mmmmm.....oh yeah, dull. Looks like sementics is a bit more than you can swallow all at one sitting. And by the way, http://www.stjude.org/donate What's a stjude? R ...heck, I could have predicted _anything_, and you would have a rather hard time proving it wrong "EVERY" time... You remind me a lot of some of the other boys here. Has anyone ever told you that if you pick up everything you can't identify and put it in your mouth you WILL eventually regret it? :) Anyway, just this once, try to tell us the truth. You're.....what?.....ten?.....maybe eleven years old? And dad lets you play on the computer in his secretary's office all day long(much to his/her disgust) because even the PUBLIC school teachers have said you're to big to be kept in a cage and besides the children tease you unmercifully (understandable, one must admit) and you just screech all the livelong day......right? :) Wolfgang jeezus.....it really IS dumber than kennie, kennie and stevie combined! |
What's a boy to do?
riverman wrote: "Wolfgang" wrote in message ... You should direct some of your less gifted students to leave the classroom and come here. This will have the salutary effect of raising the IQ in both places. :) Unfortunately, by your logic, they can't get here either. :-( No problem. We can meet them half way. :) --riverman (and exactly where they ARE has historically been the source of lively teacher's room discussions for an eternity, or more.) Tell the teachers that the answer to this mystery must remain forever beyond their grasp. Where those students ARE is where IT'S AT. Wolfgang |
What's a boy to do?
"Wolfgang" wrote in message
Well all that is.....um.....hm......what's the word I'm looking for here.....mmmmm.....oh yeah, dull. No; the word you were looking for is "lame". HTH, Joe F. |
What's a boy to do?
Ken Fortenberry wrote: 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. ;-) Don'tcha just love those "this is a great opportunity to say something seemingly cryptic that NO****in'body will understand or challenge" moments? :) Wolfgang |
What's a boy to do?
rb608 wrote: "Wolfgang" wrote in message Well all that is.....um.....hm......what's the word I'm looking for here.....mmmmm.....oh yeah, dull. No; the word you were looking for is "lame". Hm.......yeah, that WAS it. HTH, It does. Thank you. :) Wolfgang |
What's a boy to do?
"Wolfgang" wrote in message oups.com... Ken Fortenberry wrote: 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. ;-) Don'tcha just love those "this is a great opportunity to say something seemingly cryptic that NO****in'body will understand or challenge" moments? :) Oh, absolutely. It happens all the time, won't they? --riverman |
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