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Post by Eva Yojimbo on Feb 21, 2020 2:43:48 GMT
Still waiting for Arlon10 , the guy who perfectly understands probability/Bayes and how useless they are, to have a go at this. I'm still waiting for you to decide what exactly you want to know. Are the three "other" bags considered separately or collectively? The thirteen people here, usually less than four at a time, are not my teacher and never will be. If you are too stupid to understand that, and it appears you are, you are in for more disappointment than this. You don't make the rules. Your grades are meaningless. You are not a teacher here or in real life. It is amazing how having internet access makes people think they are teachers. A very good question is what in real life does picking things out of bags prepare a person to do? It's not even geology. I see you already lost the taxicab problem. You know that one makes no sense, right? Yes. the math follows a formula, but those "given" conditions that you never justify do need to be justified. You gave up on that, I disregard this. Here is the "new" information for your scenario, no one does or should care how such a failure as you entertains himself. The internet is full of people like you who are failing real life. That's why the political news is so absurd. The worst part though is how you think you're the teachers. You are not. You are the grunts who follow orders you can't understand and expect others to follow your orders. Wait, national debt skyrocketing? Not to worry, you have confidence in your absurd guesses. I asked precisely what I wanted to know with my questions and I gave you all the information necessary to answer it. Phludowin and general313 have already solved it. Rest of your post is tangential rambling that nobody cares about. Apparently you think flinging insults protects from your obvious failures on this and other such subjects you pretend to be an expert on. The very fact that you feel compelled to call me "stupid," a "failure," and accuse me of having "socks" all because you can't do a simple math problem says a great deal about you. It would be much better for your soul, Arlon, to simply admit that sometimes you don't know/understand something, and that's OK, rather than projecting all your failings onto others.
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Post by goz on Feb 21, 2020 2:53:40 GMT
He's my hero. I bet he was even in MASH as an extra!
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Post by general313 on Feb 21, 2020 14:54:40 GMT
I'm still waiting for you to decide what exactly you want to know. Are the three "other" bags considered separately or collectively? The thirteen people here, usually less than four at a time, are not my teacher and never will be. If you are too stupid to understand that, and it appears you are, you are in for more disappointment than this. You don't make the rules. Your grades are meaningless. You are not a teacher here or in real life. It is amazing how having internet access makes people think they are teachers. A very good question is what in real life does picking things out of bags prepare a person to do? It's not even geology. I see you already lost the taxicab problem. You know that one makes no sense, right? Yes. the math follows a formula, but those "given" conditions that you never justify do need to be justified. You gave up on that, I disregard this. Here is the "new" information for your scenario, no one does or should care how such a failure as you entertains himself. The internet is full of people like you who are failing real life. That's why the political news is so absurd. The worst part though is how you think you're the teachers. You are not. You are the grunts who follow orders you can't understand and expect others to follow your orders. Wait, national debt skyrocketing? Not to worry, you have confidence in your absurd guesses. I asked precisely what I wanted to know with my questions and I gave you all the information necessary to answer it. Phludowin and general313 have already solved it. Rest of your post is tangential rambling that nobody cares about. Apparently you think flinging insults protects from your obvious failures on this and other such subjects you pretend to be an expert on. The very fact that you feel compelled to call me "stupid," a "failure," and accuse me of having "socks" all because you can't do a simple math problem says a great deal about you. It would be much better for your soul, Arlon, to simply admit that sometimes you don't know/understand something, and that's OK, rather than projecting all your failings onto others. This could be the basis for another probability problem: What is the probability that Arlon will submit an answer to the Ball Problem in the OP? And if he does, what is the probability that it will be correct?
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Post by general313 on Feb 21, 2020 14:56:09 GMT
He's my hero. I bet he was even in MASH as an extra! At least you didn't say I was an extra on The Andy Griffith Show!
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Post by Eva Yojimbo on Feb 21, 2020 15:23:23 GMT
I asked precisely what I wanted to know with my questions and I gave you all the information necessary to answer it. Phludowin and general313 have already solved it. Rest of your post is tangential rambling that nobody cares about. Apparently you think flinging insults protects from your obvious failures on this and other such subjects you pretend to be an expert on. The very fact that you feel compelled to call me "stupid," a "failure," and accuse me of having "socks" all because you can't do a simple math problem says a great deal about you. It would be much better for your soul, Arlon, to simply admit that sometimes you don't know/understand something, and that's OK, rather than projecting all your failings onto others. This could be the basis for another probability problem: What is the probability that Arlon will submit an answer to the Ball Problem in the OP? And if he does, what is the probability that it will be correct? I might give him this last day before I post my own solution. When I posted this I was genuinely unsure about whether Arlon would be able to solve it. I put my own "prior" at about 60/40 that he wouldn't, but he has been able to solve some math problems in the past. Thing is that I knew he didn't understand Bayes, and I knew his "solution" to Monty Hall was a kind of intuitive cheat, so I wanted to see if he could utilize that same intuitive method on a more complex problem. I genuinely didn't know if he could, but I was interested to find out. I was, at least, hoping he'd try before I posted the answer. Very happy that at least you and phludowin solved it. I expected as much, though, since y'all are smart cats. 
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Post by general313 on Feb 21, 2020 15:52:42 GMT
This could be the basis for another probability problem: What is the probability that Arlon will submit an answer to the Ball Problem in the OP? And if he does, what is the probability that it will be correct? I might give him this last day before I post my own solution. When I posted this I was genuinely unsure about whether Arlon would be able to solve it. I put my own "prior" at about 60/40 that he wouldn't, but he has been able to solve some math problems in the past. Thing is that I knew he didn't understand Bayes, and I knew his "solution" to Monty Hall was a kind of intuitive cheat, so I wanted to see if he could utilize that same intuitive method on a more complex problem. I genuinely didn't know if he could, but I was interested to find out. I was, at least, hoping he'd try before I posted the answer. Very happy that at least you and phludowin solved it. I expected as much, though, since y'all are smart cats. First off, thanks, such a compliment coming from you is an honor. I do think if Arlon got off his high chair and got his hands dirty, he would be quite capable of solving this problem. What holds him back is not so much ability but attitude, in my opinion.
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Post by dividavi on Feb 21, 2020 19:18:03 GMT
I don't know if it's right but I came up with a probability of this:
#1,2,3,4 means bag#1,2,3,4
#1 - 10/18 = 55.56%
#2 - 5/18 = 27.78%
#3 - 2/18 = 11.11%
#4 - 1/18 = 5.55%
Make up a grid to show what's in each bag. x means a blue ball, y means a non-blue ball.
#1 - x x x x x x x x x x
#2 - x x x x x y y y y y
#3 - x x y y y y y y y y
#4 - x y y y y y y y y y
You don't know the numbering of the bag you're using or what you'll pull out. We can assign grid locations to each ball. The last (rightmost) ball for row 1 is an x and it's at location 1,10. The first y on row 3 is at 3,3. The first y on row 4 is at 4,2.
Now, the probability of selecting any particular grid location is the same as any other grid location. You've picked a ball. What are the chances it's blue? Well, there are 40 equal possibilities and I count 18 blue and 22 non-blue so the chance of getting blue is 18/40 = 45%, hence the chance of getting non-blue is 55%.
We're told that the ball was blue so how may grid selections are there? 18. There are 10 grid selections available for #1, 5 grid selections for #2 and so on for the other two bags. That's my answer.
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Post by Aj_June on Feb 21, 2020 19:25:38 GMT
I don't know if it's right but I came up with a probability of this: #1,2,3,4 means bag#1,2,3,4 #1 - 10/18 = 55.56% #2 - 5/18 = 27.78% #3 - 2/18 = 11.11% #4 - 1/18 = 5.55% Make up a grid to show what's in each bag. x means a blue ball, y means a non-blue ball. #1 - x x x x x x x x x x #2 - x x x x x y y y y y #3 - x x y y y y y y y y #4 - x y y y y y y y y y You don't know the numbering of the bag you're using or what you'll pull out. We can assign grid locations to each ball. The last (rightmost) ball for row 1 is an x and it's at location 1,10. The first y on row 3 is at 3,3. The first y on row 4 is at 4,2. Now, the probability of selecting any particular grid location is the same as any other grid location. You've picked a ball. What are the chances it's blue? Well, there are 40 equal possibilities and I count 18 blue and 22 non-blue so the chance of getting blue is 18/40 = 45%, hence the chance of getting non-blue is 55%. We're told that the ball was blue so how may grid selections are there? 18. There are 10 grid selections available for #1, 5 grid selections for #2 and so on for the other two bags. That's my answer. You are a Ramanujan's fan, aren't you?
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Post by dividavi on Feb 21, 2020 20:07:02 GMT
You are a Ramanujan's fan, aren't you? I certainly am. Wikipedia: British mathematician G. H. Hardy when he visited Indian mathematician Srinivasa Ramanujan in hospital. He related their conversation:[1][2][3][4]
The two different ways are:
1729 = 1^3 + 12^3 = 9^3 + 10^3That was a great discovery but I admire Ramanujan even more for his practice of expressing mathematical truths through poetry instead of standard notations. The truth Ramanujan revealed to Hardy is my second favorite equation. Numero uno in my book is this masterpiece: π^4 + π^5 ≈ e^6 , about 403 if I recall correctly. Here's something else about the number 1729 I didn't know: Masahiko Fujiwara (same wikipedia article as above) showed that 1729 is one of four positive integers (with the others being 81, 1458, and the trivial case 1) which, when its digits are added together, produces a sum which, when multiplied by its reversal, yields the original number: 1 + 7 + 2 + 9 = 19 19 × 91 = 1729
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Post by Eva Yojimbo on Feb 22, 2020 1:14:45 GMT
dividavi also solved it and got the right answer, with a unique approach, I might add. Since we now have three people who've solved it, one here on the forum, I'll go ahead and post my full solution (I rounded in the original, but since everyone's posting the full probabilities I'll do that here too): My solution: Bag A is most probable. It's 55.56% likely to be the bag you pulled the blue ball from. Reasoning: with no additional information, each bag has a 25% probability of being the correct bag. Once you pull the blue ball, that probability changes: Bag A predicts you'll pull a blue ball 100% of the time, so 1*.25 = .25 Bag B predicts you'll pull a blue ball 50% of the time, so .5*.25 = .125 Bag C predicts you'll pull a blue ball 20% of the time, so .2*.25 = .05 Bag D predicts you'll pull a blue ball 10% of the time, so .1*.25 = .025 To find the probability of each bag, divide the new number by the total (.25+.125+.05+.025 = .45). Bag A: .25 / .45 = .5556 Bag B: .125 / .45 = .2778 Bag C: .05 / .45 = .1111 Bag D: .025 / .45 = .0555 In Bayes form (for Bag A): P(A|B) = P(B|A)P(A) / (P(B|A)P(A)+P(B|~A)P(~A)) P(A|B) = Probability Bag A given you pulled blue ball P(B|A) = Probability you pulled blue ball given Bag A (1) P(A) = Probability Bag A (.25) P(B|~A) = Probability you pulled blue ball given NOT Bag A (.2667) P(~A) = Probability Not Bag A (.75) P(A|B) = 1*.25 / (1*.25+.2667*.75) P(A|B) = .25 / .45 P(A|B) = .5556
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Post by Arlon10 on Feb 22, 2020 10:12:25 GMT
I don't know if it's right but I came up with a probability of this: #1,2,3,4 means bag#1,2,3,4 #1 - 10/18 = 55.56% #2 - 5/18 = 27.78% #3 - 2/18 = 11.11% #4 - 1/18 = 5.55% Make up a grid to show what's in each bag. x means a blue ball, y means a non-blue ball. #1 - x x x x x x x x x x #2 - x x x x x y y y y y #3 - x x y y y y y y y y #4 - x y y y y y y y y y You don't know the numbering of the bag you're using or what you'll pull out. We can assign grid locations to each ball. The last (rightmost) ball for row 1 is an x and it's at location 1,10. The first y on row 3 is at 3,3. The first y on row 4 is at 4,2. Now, the probability of selecting any particular grid location is the same as any other grid location. You've picked a ball. What are the chances it's blue? Well, there are 40 equal possibilities and I count 18 blue and 22 non-blue so the chance of getting blue is 18/40 = 45%, hence the chance of getting non-blue is 55%. We're told that the ball was blue so how may grid selections are there? 18. There are 10 grid selections available for #1, 5 grid selections for #2 and so on for the other two bags. That's my answer. You are a Ramanujan's fan, aren't you? Congratulations to you and dividavi on your English. I wasn't sure what the question was till I saw dividavi 's answer. I can see now what "Is one bag more probable than the other?" means. I suppose I should have guessed that, so I will say the English is fair to middling. It would however be more clear if you said "There are four identical bags with ten marbles each of various colors. (Provide a list of the colors of the marbles in each bag.) A person then picks a blue marble from one of the bags. What is the probability it is from bag 1? Bag 2? Bag 3 Bag4?" I think what often happens here is that people who follow these things are familiar with the type of problem and understand the English shorthand better than people who are not familiar. Same thing with the Monty Hall problem, it escapes the attention of some people that the host's choice is not random and how important that is. I also notice Bayes' Theorem was not used. Nor do I believe anyone could use Bayes' Theorem without solving the problem otherwise first. I realize some of you don't care what I think. Don't then.
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Post by Eva Yojimbo on Feb 23, 2020 4:51:56 GMT
You are a Ramanujan's fan, aren't you? Congratulations to you and dividavi on your English. I wasn't sure what the question was till I saw dividavi 's answer. I can see now what "Is one bag more probable than the other?" means. I suppose I should have guessed that, so I will say the English is fair to middling. It would however be more clear if you said "There are four identical bags with ten marbles each of various colors. (Provide a list of the colors of the marbles in each bag.) A person then picks a blue marble from one of the bags. What is the probability it is from bag 1? Bag 2? Bag 3 Bag4?" I think what often happens here is that people who follow these things are familiar with the type of problem and understand the English shorthand better than people who are not familiar. Same thing with the Monty Hall problem, it escapes the attention of some people that the host's choice is not random and how important that is. I also notice Bayes' Theorem was not used. Nor do I believe anyone could use Bayes' Theorem without solving the problem otherwise first. I realize some of you don't care what I think. Don't then. Are you serious right now? "Is one bag more probable than the other" means... literally precisely that. I can't think of a way to phrase that any more clearly. If all bags are equally likely, the answer is "no." If one bag is more probable than the others, the answer is "yes." "How much more probable" is a way of saying "do the math and give each bag its proper probability." Bayes's Theorem was used here: IMDB2.freeforums.net/post/3693057/thread That's precisely what I used to solve the problem well before dividavi (in a PM I sent to Aj). Phludowin and general313 also did so before him. I can screenshot these solutions with timestamps if you don't believe me.
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Post by Arlon10 on Feb 23, 2020 10:21:08 GMT
Congratulations to you and dividavi on your English. I wasn't sure what the question was till I saw dividavi 's answer. I can see now what "Is one bag more probable than the other?" means. I suppose I should have guessed that, so I will say the English is fair to middling. It would however be more clear if you said "There are four identical bags with ten marbles each of various colors. (Provide a list of the colors of the marbles in each bag.) A person then picks a blue marble from one of the bags. What is the probability it is from bag 1? Bag 2? Bag 3 Bag4?" I think what often happens here is that people who follow these things are familiar with the type of problem and understand the English shorthand better than people who are not familiar. Same thing with the Monty Hall problem, it escapes the attention of some people that the host's choice is not random and how important that is. I also notice Bayes' Theorem was not used. Nor do I believe anyone could use Bayes' Theorem without solving the problem otherwise first. I realize some of you don't care what I think. Don't then. Are you serious right now? "Is one bag more probable than the other" means... literally precisely that. I can't think of a way to phrase that any more clearly. If all bags are equally likely, the answer is "no." If one bag is more probable than the others, the answer is "yes." "How much more probable" is a way of saying "do the math and give each bag its proper probability." Bayes's Theorem was used here: IMDB2.freeforums.net/post/3693057/thread That's precisely what I used to solve the problem well before dividavi (in a PM I sent to Aj). Phludowin and general313 also did so before him. I can screenshot these solutions with timestamps if you don't believe me. Here's a probability question for you, "What is the probability timestamps from this forum reflect real life sequences of events?" For 30,000 people? For 300 people? For 3 people? For one person with 3 socks? Here's another question. Link. Go ahead and answer it. I won't be upset if you get the answer right. It's not as difficult as it is fascinating.
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Post by Eva Yojimbo on Feb 23, 2020 13:08:28 GMT
Are you serious right now? "Is one bag more probable than the other" means... literally precisely that. I can't think of a way to phrase that any more clearly. If all bags are equally likely, the answer is "no." If one bag is more probable than the others, the answer is "yes." "How much more probable" is a way of saying "do the math and give each bag its proper probability." Bayes's Theorem was used here: IMDB2.freeforums.net/post/3693057/thread That's precisely what I used to solve the problem well before dividavi (in a PM I sent to Aj). Phludowin and general313 also did so before him. I can screenshot these solutions with timestamps if you don't believe me. Here's a probability question for you, "What is the probability timestamps from this forum reflect real life sequences of events?" For 30,000 people? For 300 people? For 3 people? For one person with 3 socks? Here's another question. Link. Go ahead and answer it. I won't be upset if you get the answer right. It's not as difficult as it is fascinating. I don't know; Admin, what's the probability that somebody posted a solution to this problem in this thread before I posted the solution to the problem in a PM, which I sent before I posted this thread? Maybe admin knows more about time-traveling paradoxes than I do. In any case, your increasingly convoluted reasoning as to why someone couldn't have solved this (and other) problems with Bayes is becoming extremely hilarious. Earlier it was me "copying" others, now it's that others are time traveling wizards, or that I somehow know how to manipulate this board's time-stamping system to make it seem like I posted a solution to the problem before others in this thread. Funny stuff. I posted a guess in your thread. Unlike you, I'm capable of admitting when I don't know something.
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