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Posted: 11/28/2003 5:16:23 PM EDT
[Last Edit: 11/28/2003 5:20:15 PM EDT by The_Macallan]

You have three dice. On a single roll of all three, what is the probability that you will get a 5 or a 6 on any of the die?

Show how you arrived at your answer.
Link Posted: 11/28/2003 5:38:29 PM EDT
It has been a long time since I have used my probs and stats, but here is my guess. You have a 1 in 3 chance of rolling a 5 or a 6 on any single die or a 2 in 3 chance of not rolling 5 or 6. To not roll a 5 or 6 on any of the three dice is a 2/3*2/3*2/3 probability (8/27 or .296). To figure the inverse probability (rolling a 5 or 6 at least once) subtract .296 from 1 which equals .704 or 70.4% chance of rolling AT LEAST ONE 5 or 6.
Link Posted: 11/28/2003 5:59:51 PM EDT
Probability should be 100%. You have a 2/6 chance on any 1 die & 3 of them, so 6/6. At least I think, the most math I had was algebra2.
Link Posted: 11/28/2003 6:03:37 PM EDT
2/6 + 2/6 + 2/6
Link Posted: 11/28/2003 6:04:07 PM EDT
lol
Link Posted: 11/28/2003 6:09:00 PM EDT
here's another one once you guys are done with mac's problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to yo ur advantage to switch your choice? I'll give you the answer: yes you should switch you explain why
Link Posted: 11/28/2003 6:11:24 PM EDT
[Last Edit: 11/28/2003 6:13:12 PM EDT by Cypher214]
Because the guy is an asshole and assumes you'll think he's trying to trick you and pick door #1.
Link Posted: 11/28/2003 6:12:12 PM EDT
2/3
Link Posted: 11/28/2003 6:14:21 PM EDT
Originally Posted By Red_Beard: here's another one once you guys are done with mac's problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to yo ur advantage to switch your choice? I'll give you the answer: yes you should switch you explain why
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2/3 chance you picked a goat on your first choice. Opening door #3 reinforces that since #3 was NOT the car. So it's more likely that door #2 (the one you didn't pick) is the car and YOUR door is the 2nd goat. Now solve my question please. [:D]
Link Posted: 11/28/2003 6:18:40 PM EDT
Originally Posted By Red_Beard: here's another one once you guys are done with mac's problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to yo ur advantage to switch your choice? I'll give you the answer: yes you should switch you explain why
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Yes. RE: Bayes' Theorem
Link Posted: 11/28/2003 6:28:27 PM EDT
Originally Posted By Red_Beard: here's another one once you guys are done with mac's problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to yo ur advantage to switch your choice? I'll give you the answer: yes you should switch you explain why
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You should switch. It's Bayes Theorem right? I don't remember how to prove it, but I remember studying it in college. If you hold you still have a 1/3 chance, but if you switch it's a 2/3 chance IIRC.
Link Posted: 11/28/2003 6:38:19 PM EDT
Im sick of that crap. From my first exam this semester:
An experiment consists of tossing a fair die (six sides and the numbers 1 through 6 on its faces) until the number, 3, occurs or the die is tossed 4 times, whichever comes first. The number of tosses is a random variable, N. What is the numerical expected value of N, E(N)?
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And from the latest test:
A circuit consists of a resistor, R, at an absolute temperature, T, and a capacitor, C, in parallel. (The double-sided power spectral density of the short-circuit Johnson noise current of a resistor is 2kT/R and the double sided PSD of the open circuit Johnson noise voltage is 2kTR.) a) Find a general formula for the mean-squared voltage across the capacitor (and resistor) in terms of k, R, T and C. This formula should contain no integral signs, Fourier transforms or convolution operators. b) You should find that the mean-squared voltage is independent of resistance, R. By sketching the PSD of the noise voltage across the capacitor for the two cases, R=1, C=1, T=300 and R=2, C=1, T=300, explain how that can happen even though the PSD of the resistor Johnson noise current (or voltage) noise is a function of R.
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Im [b]not[/b] looking forward to the final exam on the 4th. Jonathan
Link Posted: 11/28/2003 7:00:24 PM EDT
i like pie
Link Posted: 11/28/2003 7:26:38 PM EDT
Originally Posted By The_Macallan:
Originally Posted By Red_Beard: here's another one once you guys are done with mac's problem: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to yo ur advantage to switch your choice? I'll give you the answer: yes you should switch you explain why
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2/3 chance you picked a goat on your first choice. Opening door #3 reinforces that since #3 was NOT the car. So it's more likely that door #2 (the one you didn't pick) is the car and YOUR door is the 2nd goat. Now solve my question please. [:D]
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That's arbitrary. Yes, you may have a 2/3 chance of choosing a door with a goat behind it for your first choice; however, once the game show host shows you door number 3 has a goat behind it, and he gives you the opportunity to choose again you have a 50/50 chance at guessing on the car. I understand the logic of the theorem, but personally I always go with my gut in situations like this. Where probability really comes in handy is when you have the time/chance to really get it working for you. With so few choices though, I personally wouldn't bother unless I felt it specifically benefited the situation. You want probability to work for you, gather up a nice pile of cash and go play some cards or slots. You do it right you win big, you just need a bankroll big enough to stay until the payoff.
Link Posted: 11/28/2003 7:33:15 PM EDT
[Last Edit: 11/28/2003 7:33:46 PM EDT by The_Macallan]
Originally Posted By Corporal_Chaos: Yes, you may have a 2/3 chance of choosing a door with a goat behind it for your first choice; however, [red]once the game show host shows you door number 3 has a goat behind it, and he gives you the opportunity to choose again you have a 50/50 chance at guessing on the car[/red]. I understand the logic of the theorem, but personally I always go with my gut in situations like this.
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After the host opens door #3 - you're not being asked to choose a 50/50 choice, but rather you're being asked if you still want to keep your original choice (which has a 2/3 chance of being a loser) or switch (which gives you a 2/3 chance of being a winner).
Link Posted: 11/28/2003 7:36:48 PM EDT
Punters. Here's the REAL test. Sharpen your pencils, kids. The Final Exam Instructions: Read each question carefully. Answer all questions. Time Limit: 4 hours. Begin immediately. 1) H I S T O R Y Describe the history of the papacy from its origins to the present day, concentrating especially, but not exclusively, on its social, political, economic, religious, and philosophical impact on Europe, Asia, America, and Africa. Be brief, concise, and specific. 2) M E D I C I N E You have been provided with a razor blade, a piece of gauze, and a bottle of Scotch. Remove your appendix. Do not suture until your work has been inspected. You have 15 minutes. 3) P U B L I C S P E A K I N G Twenty-five hundred riot-crazed aborigines are storming the classroom. Calm them. You may use any ancient language except Latin or Greek. 4) B I O L O G Y Create life. Estimate the differences in subsequent human culture if this form of life had developed 500 million years earlier, with special attention to its probable effect on the English parliamentary system. Prove your thesis. 5) M U S I C Write a piano concerto. Orchestrate and perform it with flute and drum. You will find a piano under your seat. 6) P S Y C H O L O G Y Based on your degree of knowledge of their works, evaluate the emotional stability, degree of adjustment, and repressed frustrations of each of the following: Alexander of Aphrodisias, Rameses II, Gregory of Nicea, Hammurabi. Support your evaluations with quotations from each man's work, making appropriate references. It is not necessary to translate. 7) S O C I O L O G Y Estimate the sociological problems which might accompany the end of the world. Construct an experiment to test your theory. 8) M A N A G E M E N T S C I E N C E Define management. Define science. How do they relate? Why? Create a generalized algorithm to optimize all managerial decisions. Assuming an 1130 CPU supporting 50 terminals, each terminal to activate your algorithm; design the communications interface and all necessary control programs. 9) E N G I N E E R I N G The disassembled parts of a high-powered rifle have been placed in a box on your desk. You will also find an instruction manual, printed in Swahili. In ten minutes a hungry Bengal tiger will be admitted to the room. Take whatever action you feel is appropriate. Be prepared to justify your decision. 10) E C O N O M I C S Develop a realistic plan for refinancing the national debt. Trace the possible effects of your plan in the following areas: Cubism, the Donatist controversy, the wave theory of light. Outline a method for preventing these effects. Criticize this method from all possible points of view. Point out the deficiencies in your point of view, as demonstrated in your answer to the last question. 11) P O L I T I C A L S C I E N C E There is a red telephone on the desk beside you. Start World War III. Report at length on its socio-political effects, if any. 12) E P I S T E M O L O G Y Take a position for or against truth. Prove the validity of your position. 13) P H Y S I C S Explain the nature of matter. Include in your answer an evaluation of the impact of the development of mathematics on science. 14) P H I L O S O P H Y Sketch the development of human thought; estimate its significance. Compare with the development of any other kind of thought. 15) G E N E R A L K N O W L E D G E Describe in detail. Be objective and specific. * * E X T R A C R E D I T * * Define the universe; give three examples.
Link Posted: 11/28/2003 7:42:26 PM EDT
[Last Edit: 11/28/2003 7:44:14 PM EDT by The_Macallan]
[b]DzlBenz[/b], I actually added that to the end of a final exam as a joke one semester, some students actually tried to answer them. [:D] [b]Okay now, back on topic... [size=4]Can someone please answer my original question DAMMIT!!!!???[/size=4][/b] [img]http://us.news2.yimg.com/us.yimg.com/p/afp/20031124/capt.sge.hvh28.241103070028.photo00.default-384x235.jpg[/img]
Link Posted: 11/28/2003 7:46:13 PM EDT
Originally Posted By The_Macallan:
Originally Posted By Corporal_Chaos: Yes, you may have a 2/3 chance of choosing a door with a goat behind it for your first choice; however, [red]once the game show host shows you door number 3 has a goat behind it, and he gives you the opportunity to choose again you have a 50/50 chance at guessing on the car[/red]. I understand the logic of the theorem, but personally I always go with my gut in situations like this.
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After the host opens door #3 - you're not being asked to choose a 50/50 choice, but rather you're being asked if you still want to keep your original choice (which has a 2/3 chance of being a loser) or switch (which gives you a 2/3 chance of being a winner).
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I disagree. Once the host opens door #3, you only have 2 options. At this point, since 2 doors have goats and one door has a car, and you already know that door 3 has a goat, you get to chose between one door with a car and one door with a goat. giving you a 50/50 chance. Switching will niether help or hurt your odds.
Link Posted: 11/28/2003 7:50:25 PM EDT
[Last Edit: 11/28/2003 7:57:47 PM EDT by Corporal_Chaos]
Originally Posted By gribble:
Originally Posted By The_Macallan:
Originally Posted By Corporal_Chaos: Yes, you may have a 2/3 chance of choosing a door with a goat behind it for your first choice; however, [red]once the game show host shows you door number 3 has a goat behind it, and he gives you the opportunity to choose again you have a 50/50 chance at guessing on the car[/red]. I understand the logic of the theorem, but personally I always go with my gut in situations like this.
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After the host opens door #3 - you're not being asked to choose a 50/50 choice, but rather you're being asked if you still want to keep your original choice (which has a 2/3 chance of being a loser) or switch (which gives you a 2/3 chance of being a winner).
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I disagree. Once the host opens door #3, you only have 2 options. At this point, since 2 doors have goats and one door has a car, and you already know that door 3 has a goat, you get to chose between one door with a car and one door with a goat. giving you a 50/50 chance. Switching will niether help or hurt your odds.
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Exactly. Two choices, you can choose either. You know one door of the two has a goat behind it, and the other a car. The second decision is 50/50[:)] because one of the goats has been removed from the equation. **EDIT** Like I said earlier. I understand the logic of the theorem, but personally, with so few choices, I don't consider it a viable theorem. Now, if there were 5 doors, 4 goats, 1 car, and 2 of the doors and goats were removed from the equation, then I'd place my trust in probability[:)]. Having absolutly nothing to do with probability, in my experience, I prefer to stick with my first choice. Almost everytime I second guess myself, I end up having to kick myself in the ass. Trust your gut!
Link Posted: 11/28/2003 7:52:49 PM EDT
[Last Edit: 11/28/2003 7:55:15 PM EDT by The_Macallan]
Originally Posted By gribble: I disagree. Once the host opens door #3, you only have 2 options. At this point, since 2 doors have goats and one door has a car, and you already know that door 3 has a goat, you get to chose between one door with a car and one door with a goat. giving you a 50/50 chance. Switching will niether help or hurt your odds.
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If the host sneaks a peak and doesn't tell you, have the 2/3 odds against your original choice changed any? No. Of course not. You're still holding a 2/3 loser. Him telling you or not doesn't improve the 2/3 odds that you picked a loser. You're not being asked to choose between door #1 and door #2, you're being asked if you want to keep your original choice which was a 2/3 loser.
Link Posted: 11/28/2003 7:58:51 PM EDT
33%
Link Posted: 11/28/2003 7:59:50 PM EDT
Originally Posted By The_Macallan: [b]DzlBenz[/b], I actually added that to the end of a final exam as a joke one semester, some students actually tried to answer them. [:D] [b]Okay now, back on topic... [size=4]Can someone please answer my original question DAMMIT!!!!???[/size=4][/b] [url]http://us.news2.yimg.com/us.yimg.com/p/afp/20031124/capt.sge.hvh28.241103070028.photo00.default-384x235.jpg[/url]
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H46Driver solved you original problem correctly. Took me a bit to recall my own math classes too. [:)]
Link Posted: 11/28/2003 8:06:04 PM EDT
[Last Edit: 11/28/2003 8:06:39 PM EDT by The_Macallan]
Originally Posted By 1911Shootist: H46Driver solved you original problem correctly. Took me a bit to recall my own math classes too. [:)]
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It didn't make sense. He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|] Still waiting for someone to help on this. Come on guys, don't make me go over to [b]Full-Auto.com[/b] to find someone who can answer it! [BD]
Link Posted: 11/28/2003 8:12:25 PM EDT
There are 6x6x6 possible ways we can roll the three dice. Of those 216 possible combinations, there are 152 (2x6x6 72 combinations with the first two + 40 with each of the other two that don't overlap with the previous 72) that will give you a 5 or six on any one die. Therefore the probability is 152/216 or 0.704 chance of it happening. I just noticed H46Driver had the right solution and an easier way of solving it. Oh well.z
Link Posted: 11/28/2003 8:12:30 PM EDT
Originally Posted By The_Macallan:
Originally Posted By gribble: I disagree. Once the host opens door #3, you only have 2 options. At this point, since 2 doors have goats and one door has a car, and you already know that door 3 has a goat, you get to chose between one door with a car and one door with a goat. giving you a 50/50 chance. Switching will niether help or hurt your odds.
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If the host sneaks a peak and doesn't tell you, have the 2/3 odds against your original choice changed any? No. Of course not. You're still holding a 2/3 loser. Him telling you or not doesn't improve the 2/3 odds that you picked a loser. You're not being asked to choose between door #1 and door #2, you're being asked if you want to keep your original choice which was a 2/3 loser.
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But there are only 2 other choices. Yes, there is still a 2/3 chance that the door you picked is the right one out of the three, but now that you know door three is wrong, you are down to 2 doors. Not changing your choice (sticking with door one) or changing your choice (the only other choice being door 2) will not affect your odds of winning.
Link Posted: 11/28/2003 8:17:39 PM EDT
Originally Posted By The_Macallan: He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i]
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Seems reasonable enough but then: 1.0 - .296 = .704
Link Posted: 11/28/2003 8:19:56 PM EDT
[Last Edit: 11/28/2003 8:21:39 PM EDT by Red_Beard]
Originally Posted By The_Macallan:He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|]
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think of it as two events event A: at least one appearance of 5 or 6 in three rolls of a dice event B: no appearances of 5 and no appearances of 6 in three rolls you want the probability of event A happening either A or B must happen, and they cannot both happen at the same time, so pr(A) + pr(B) = 1 so PR(A) = 1 - PR(B) you're getting confused by thinking that since PR(B) = (2/3)^3, then pr(A) = (1/3)^3 not the case pr(A) = 1 - (2/3)^3
Link Posted: 11/28/2003 8:21:04 PM EDT
math must die!
Link Posted: 11/28/2003 8:22:23 PM EDT
Not changing your choice (sticking with door one) or changing your choice (the only other choice being door 2) will not affect your odds of winning.
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It does. Since Monte Hall would not have opened the door you selected, you know the door that is still closed has a 50% chance of being a winner. Think about it. He had only two unpicked doors to chose from. The one you originally selected still has the original 33% chance of winning. It is better to always select a different door no matter how many doors there are.z
Link Posted: 11/28/2003 8:25:10 PM EDT
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|]
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think of it as two events event A: at least one appearance of 5 or 6 in three rolls of a dice event B: no appearances of 5 and no appearances of 6 in three rolls you want the probability of event A happening either A or B must happen, and they cannot both happen at the same time, so pr(A) + pr(B) = 1 so PR(A) = 1 - PR(B) you're getting confused by thinking that since PR(B) = (2/3)^3, then pr(A) = (1/3)^3 not the case pr(A) = 1 - (2/3)^3
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Explain how 2/3*2/3*2/3 gives a logical answer to the odds of NOT hitting 5 or 6 but 1/3*1/3*1/3 gives an illogical answer to the odds of hitting a 5 or 6. He [b]MAY[/b] have stumbled upon the correct answer (not sure yet) but his [u]method[/u] is clearly flawed.
Link Posted: 11/28/2003 8:27:49 PM EDT
[Last Edit: 11/28/2003 8:28:28 PM EDT by Red_Beard]
Originally Posted By The_Macallan:
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|]
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think of it as two events event A: at least one appearance of 5 or 6 in three rolls of a dice event B: no appearances of 5 and no appearances of 6 in three rolls you want the probability of event A happening either A or B must happen, and they cannot both happen at the same time, so pr(A) + pr(B) = 1 so PR(A) = 1 - PR(B) you're getting confused by thinking that since PR(B) = (2/3)^3, then pr(A) = (1/3)^3 not the case pr(A) = 1 - (2/3)^3
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Explain how 2/3*2/3*2/3 gives a logical answer to the odds of NOT hitting 5 or 6 but 1/3*1/3*1/3 gives an illogical answer to the odds of hitting a 5 or 6. He [b]MAY[/b] have stumbled upon the correct answer (not sure yet) but his [u]method[/u] is clearly flawed.
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ok, think of three independent rolls of a single die the event you're interested in is this: in three rolls 5 appears ZERO times and 6 appears ZERO times what is the chance of this on the first roll? it's the chance that you get a 1,2,3, or 4 i.e. 4/6 which is 2/3 the chance is the same on the other two rolls, and since rolls of the die are independent, you multiply the probabilities to get the chance of NOT( 5 or 6) happening three times
Link Posted: 11/28/2003 8:32:03 PM EDT
Originally Posted By zoom:
Not changing your choice (sticking with door one) or changing your choice (the only other choice being door 2) will not affect your odds of winning.
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It does. Since Monte Hall would not have opened the door you selected, you know the door that is still closed has a 50% chance of being a winner. Think about it. He had only two unpicked doors to chose from. The one you originally selected still has the original 33% chance of winning. It is better to always select a different door no matter how many doors there are.z
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So by that reasoning if there were 100 doors and you chose one only to find out that 98 of them had goats would the odds of the one you had chosen being the car be 1% and the one remaining be 99%? I don't think so.
Link Posted: 11/28/2003 8:39:47 PM EDT
[Last Edit: 11/28/2003 8:41:21 PM EDT by drjarhead]
Originally Posted By drjarhead:
Originally Posted By The_Macallan: He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i]
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Seems reasonable enough but then: 1.0 - .296 = .704
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the odds of getting 3 dice to come up without a 5 or 6 should be .296 as stated. The reason for this is that you could have 1 die with 5 or 6, 2 with, or even 3 with. The total possibilities represented by 1.0, of course, leaves you with 70.4% chance that one or more dice will come up with a 5 or 6.
Link Posted: 11/28/2003 8:46:20 PM EDT
[Last Edit: 11/28/2003 8:48:08 PM EDT by Red_Beard]
Originally Posted By drjarhead:
Originally Posted By zoom:
Not changing your choice (sticking with door one) or changing your choice (the only other choice being door 2) will not affect your odds of winning.
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It does. Since Monte Hall would not have opened the door you selected, you know the door that is still closed has a 50% chance of being a winner. Think about it. He had only two unpicked doors to chose from. The one you originally selected still has the original 33% chance of winning. It is better to always select a different door no matter how many doors there are.z
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So by that reasoning if there were 100 doors and you chose one only to find out that 98 of them had goats would the odds of the one you had chosen being the car be 1% and the one remaining be 99%? I don't think so.
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it is 99/100 one out of 100 times you will have picked the right door every game is going to come down to two doors, the one you picked and the one you are offered if the one you picked is correct 1 out of 100 times then it must be wrong 99/100 times. leaving the offered door.
Link Posted: 11/28/2003 8:52:36 PM EDT
So by that reasoning if there were 100 doors and you chose one only to find out that 98 of them had goats would the odds of the one you had chosen being the car be 1% and the one remaining be 99%?
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That is correct. Think about it. You had a 1% chance of selecting the correct door when you first selected, but there is now a 99% chance that the other door is the one while the chance that the first door you selected is the one is still 1%. Always select a different door.z
Link Posted: 11/28/2003 8:52:56 PM EDT
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|]
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think of it as two events event A: at least one appearance of 5 or 6 in three rolls of a dice event B: no appearances of 5 and no appearances of 6 in three rolls you want the probability of event A happening either A or B must happen, and they cannot both happen at the same time, so pr(A) + pr(B) = 1 so PR(A) = 1 - PR(B) you're getting confused by thinking that since PR(B) = (2/3)^3, then pr(A) = (1/3)^3 not the case pr(A) = 1 - (2/3)^3
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Explain how 2/3*2/3*2/3 gives a logical answer to the odds of NOT hitting 5 or 6 but 1/3*1/3*1/3 gives an illogical answer to the odds of hitting a 5 or 6. He [b]MAY[/b] have stumbled upon the correct answer (not sure yet) but his [u]method[/u] is clearly flawed.
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ok, think of three independent rolls of a single die the event you're interested in is this: in three rolls 5 appears ZERO times and 6 appears ZERO times what is the chance of this on the first roll? it's the chance that you get a 1,2,3, or 4 i.e. 4/6 which is 2/3 the chance is the same on the other two rolls, and since rolls of the die are independent, you multiply the probabilities to get the chance of NOT( 5 or 6) happening three times
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You didn't explain anything. You just restated same the faulty logic again. The probability of rolling a 5 or 6 with one die is 1/3. [b]The probability of rolling a 5 or 6 with THREE die is NOT determined by multiplying the probability of each of the three dice (i.e. it's NOT "1/3*1/3*1/3").[/b] Therefore the probability of NOT rolling a 5 or 6 is also not determined by multiplying the probabilities of each of the three dice (i.e. it's NOT "2/3*2/3*2/3") Think of flipping a coin. The odds of getting a heads on ANY of three coins tossed is NOT determined by "1/2*1/2*1/2".
Link Posted: 11/28/2003 8:53:06 PM EDT
Okay, I see your point. Not 100% sold but I understand what you guys are saying.
Link Posted: 11/28/2003 9:02:46 PM EDT
Originally Posted By The_Macallan: [b]The probability of rolling a 5 or 6 with THREE die is NOT determined by multiplying the probability of each of the three dice (i.e. it's NOT "1/3*1/3*1/3").[/b] Therefore the probability of NOT rolling a 5 or 6 is also not determined by multiplying the probabilities of each of the three dice (i.e. it's NOT "2/3*2/3*2/3")
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No, but the probability of rolling a 5 or 6 with ALL three dice IS 1/3 x 1/3 x 1/3. Therefore the probability of NOT rolling a 5o6 at all should be 2/3 x 2/3 x 2/3 with the ALL the remaining possibilities including a 5 or 6 on at least one die. So explain to me how the answer is not .704[):)]
Link Posted: 11/28/2003 9:04:32 PM EDT
You didn't explain anything. You just restated same the faulty logic again.
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ha ha ha
The probability of rolling a 5 or 6 with one die is 1/3.
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agreed
The probability of rolling a 5 or 6 with THREE die is NOT determined by multiplying the probability of each of the three dice (i.e. it's NOT "1/3*1/3*1/3").
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agreed
Therefore the probability of NOT rolling a 5 or 6 is also not determined by multiplying the probabilities of each of the three dice (i.e. it's NOT "2/3*2/3*2/3")
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WRONG. this case is "not five and not six" on ALL THREE rolls, the above case is "five or six" on ANY ONE OF THE THREE rolls. that is the essential difference that makes the probabilities multiply in one case and not the other. "independence"
Think of flipping a coin. The odds of getting a heads on ANY of three coins tossed is NOT determined by "1/2*1/2*1/2".
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what are the chances of getting tails on ALL three tosses? what is the opposite of "heads on ANY of three tosses" it is "tails on ALL three tosses" one or the other MUST happen, and it is impossible for BOTH to happen at the same time so ... probabilty of "heads on ANY of three tosses" + probability of "tails on ALL three tosses" = 1
Link Posted: 11/28/2003 9:26:35 PM EDT
Originally Posted By Red_Beard: so ... probabilty of "heads on ANY of three tosses" + probability of "tails on ALL three tosses" = 1
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No, there are other possibilities: 1/9 chance all heads 1/9 chance all tails 7/9 chance of various combos of heads and tails
Link Posted: 11/28/2003 9:31:49 PM EDT
[Last Edit: 11/28/2003 9:33:33 PM EDT by Red_Beard]
Originally Posted By drjarhead:
Originally Posted By Red_Beard: so ... probabilty of "heads on ANY of three tosses" + probability of "tails on ALL three tosses" = 1
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No, there are other possibilities: 1/9 chance all heads
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this is included in "heads on ANY of three tosses"
1/9 chance all tails
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this is "tails on ALL three tosses"
7/9 chance of various combos of heads and tails
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this is included in "heads on ANY of three tosses". "heads on ANY of three tosses". isn't the same as "heads on ANY SINGLE ONE of three tosses". i should have used more precise language. sorry.
Link Posted: 11/28/2003 9:36:33 PM EDT
Originally Posted By Red_Beard:
Originally Posted By drjarhead:
Originally Posted By Red_Beard: so ... probabilty of "heads on ANY of three tosses" + probability of "tails on ALL three tosses" = 1
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No, there are other possibilities: 1/9 chance all heads
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this is included in "heads on ANY of three tosses"
1/9 chance all tails
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this is "tails on ALL three tosses"
7/9 chance of various combos of heads and tails
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this is included in "heads on ANY of three tosses". "heads on ANY of three tosses". isn't the same as "heads on ANY SINGLE ONE of three tosses". i should have used more precise language. sorry.
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Nah, my mistake. Should've read more carefully.
Link Posted: 11/28/2003 9:48:45 PM EDT
The probability of rolling a 5 or 6 with one die is 1/3.
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=> 3 dice is also 1/3 6 possible correct solutions/18 possible outcomes
Link Posted: 11/28/2003 9:56:57 PM EDT
I would say that if this is what you focus you life on then the probability is that you are in need of a hobby.
Link Posted: 11/28/2003 10:16:19 PM EDT
[Last Edit: 11/28/2003 10:40:19 PM EDT by DsrtEgl50]
Originally Posted By The_Macallan:
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:
Originally Posted By Red_Beard:
Originally Posted By The_Macallan:He stated that [i]"To [red]not[/red] roll a 5 or 6 on any of the three dice is a [red]2/3*2/3*2/3 probability[/red] (8/27 or .296)"[/i] which must mean then that [i]"To [red][s]not[/s][/red] roll a 5 or 6 on any of the three dice is a [red]1/3*1/3*1/3 probability[/red] (1/27 or 0.037)"[/i] [%|]
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think of it as two events event A: at least one appearance of 5 or 6 in three rolls of a dice event B: no appearances of 5 and no appearances of 6 in three rolls you want the probability of event A happening either A or B must happen, and they cannot both happen at the same time, so pr(A) + pr(B) = 1 so PR(A) = 1 - PR(B) you're getting confused by thinking that since PR(B) = (2/3)^3, then pr(A) = (1/3)^3 not the case pr(A) = 1 - (2/3)^3
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Explain how 2/3*2/3*2/3 gives a logical answer to the odds of NOT hitting 5 or 6 but 1/3*1/3*1/3 gives an illogical answer to the odds of hitting a 5 or 6. He [b]MAY[/b] have stumbled upon the correct answer (not sure yet) but his [u]method[/u] is clearly flawed.
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ok, think of three independent rolls of a single die the event you're interested in is this: in three rolls 5 appears ZERO times and 6 appears ZERO times what is the chance of this on the first roll? it's the chance that you get a 1,2,3, or 4 i.e. 4/6 which is 2/3 the chance is the same on the other two rolls, and since rolls of the die are independent, you multiply the probabilities to get the chance of NOT( 5 or 6) happening three times
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You didn't explain anything. You just restated same the faulty logic again. The probability of rolling a 5 or 6 with one die is 1/3. [b]The probability of rolling a 5 or 6 with THREE die is NOT determined by multiplying the probability of each of the three dice (i.e. it's NOT "1/3*1/3*1/3").[/b] Therefore the probability of NOT rolling a 5 or 6 is also not determined by multiplying the probabilities of each of the three dice (i.e. it's NOT "2/3*2/3*2/3") Think of flipping a coin. The odds of getting a heads on ANY of three coins tossed is NOT determined by "1/2*1/2*1/2".
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OK, pardon my near 5th of whiskey in me and my attention deficit as I try to calculate a number based on power spectral density of 15 million data points (which is due Monday along with a 95% confidence interval) and explain this in a semi reasonable way. The probability that SOMETHING happens is 1. There is a 100% chance that SOMETHING is going to happen. What you are asking is, what is the chance that 5 or 6 is rolled in 3 tosses of a fair die. So the probabilities are 5 or 6 on the first roll. 5 or 6 on the second roll, given that you DIDN't roll a 5 or 6 on the first roll. 5 or 6 on the third roll, GIVEN that you DIDN't roll a 5 or 6 on the second roll. Now I realize you are talking three independent dice rolled at the same time, but for sake of reasonable calculation you can assume that each die is independent of one-another and thus can assume that it is one die rolled three times. What you are asking for is Pr(A|B)+Pr(A|C)+Pr(A|D). Now, the probability of rolling a 5 or 6 in one roll is 1/3. The probability of rolling a 5 or 6 given that you rolled a 1 through 4 on the first roll is a bit more difficult. You must use Bayes' theorem and calculate various probabilities. Using the Engineering approach (use your brain rather than a long calculation), we can take the fact that the event that SOMETHING happens is 100% and put it to use. The event that you DON'T roll a 5 or a 6 is an event that can be discribed easily, as the events of NOT rolling a 5 or 6 are independent of each other (whereas if the events of rolling a 5 or 6 are dependent on each other, if you roll a 5 or 6 you stop rolling!). Therefore, since they are independent the formula applies that the probability of the events is the multiplication of their individual probabilites. Therefore, the event that 5 or 6 does NOT happen in 3 rolls of a fair die can be expressed as 2/3^3 or .296. Therefore, since the probability than SOME event happens is 1 and the probability that the event of rolling 3 die with no 5s or 6s is .296, you can subtract the two to find the probability that a 5 or 6 was rolled in the tossing of 3 fair die... or 70.4% If that isn't enough explaination for you, you will have to wait til I am sober... Edited cause I been drinking heavily. And to add, the probability space of the events of three dice tossed is 216 different outcomes. There are 152 outcomes that involve 5s or 6s so by a brute force enumeration you can derive the 70.4%. Try decoding a friggen chapter in the book based on the probabilities of the different characters. What a pain in the ass that homework was. Ohhh but we need to learn basic cryptography... BAH! Jonathan
Link Posted: 11/28/2003 10:25:44 PM EDT
Originally Posted By The_Macallan: You have three dice. On a [red]single roll of all three[/red], what is the probability that you will get a 5 or a 6 on [red]any[/red] of the die? Show how you arrived at your answer.
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The answer to that question is 1/3. Even if you roll one die at a time it is still 1/3. When it would be different is when your question would ask something in which all three die are required in the outcome such as a sum.
Link Posted: 11/28/2003 10:37:39 PM EDT
[Last Edit: 11/28/2003 10:38:05 PM EDT by The_Macallan]
Originally Posted By drjarhead: No, but the probability of rolling a 5 or 6 [red]with ALL three dice[/red] IS 1/3 x 1/3 x 1/3. Therefore the probability of NOT rolling a 5o6 at all should be 2/3 x 2/3 x 2/3 with the ALL the remaining possibilities including a 5 or 6 on at least one die. So explain to me how the answer is not .704[):)]
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Ah. I see what I did. Thanks. 70.4% it is. Okay now, so let's up the ante. What's the odds of getting a 5 or 6 on only ONE of the three dice rolled? [:D]
Link Posted: 11/28/2003 10:39:06 PM EDT
Originally Posted By drjarhead:
Originally Posted By Red_Beard: so ... probabilty of "heads on ANY of three tosses" + probability of "tails on ALL three tosses" = 1
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No, there are other possibilities: 1/9 chance all heads 1/9 chance all tails 7/9 chance of various combos of heads and tails
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[useless trivia] In the real world the probability of heads/tails isn't 50-50. Tails is slightly more likely. The heads side of a coin is typically heavier than the tails side, so heads is more likely to be on the bottom. [/useless trivia]
Link Posted: 11/28/2003 10:57:57 PM EDT
[Last Edit: 11/28/2003 11:00:09 PM EDT by DsrtEgl50]
Originally Posted By The_Macallan: Okay now, so let's up the ante. What's the odds of getting a 5 or 6 on only ONE of the three dice rolled? [:D]
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Again, pardon the whiskey, but that one is a bit more simple using Newton's binomial theorem. The event A will happen K times in N trials is (N K) * Q^K * P^(N-K) times where N is the number of trials, Q is the probability of the event in 1 trial and P is the probability that event does NOT occur in 1 trial. Therefore you have a 3!/(1!*(3-1)!) *1/3 * (2/3)^2 chance or 4/9 chance of rolling a 5 or 6 only once in three rolls. This is supported by the brute force enumeration, being that there are 96 combinations that yield a single 5 or 6 in 216 three digit hex combinations. Edited to add the word single, cause I left it out the first time... Jonathan
Link Posted: 11/28/2003 11:03:48 PM EDT
Originally Posted By DsrtEgl50:
Originally Posted By The_Macallan: Okay now, so let's up the ante. What's the odds of getting a 5 or 6 on only ONE of the three dice rolled? [:D]
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Again, pardon the whiskey, but that one is a bit more simple using Newton's binomial theorem. The event A will happen K times in N trials is (N K) * Q^K * P^(N-K) times where N is the number of trials, Q is the probability of the event in 1 trial and P is the probability that event does NOT occur in 1 trial. Therefore you have a 3!/(1!*(3-1)!) *1/3 * (2/3)^2 chance or 4/9 chance of rolling a 5 or 6 only once in three rolls. This is supported by the brute force enumeration, being that there are 96 combinations that yield a single 5 or 6 in 216 three digit hex combinations. Edited to add the word single, cause I left it out the first time... Jonathan
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[golf clap] Well done squire! Another whiskey for the man with the hand-cannon screen name. [:D]
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