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So now he says he's willing to put 100 dollars on the 1 in 5... I'm swayed back to the 1 in 3... So here's the original question-- If you have 4 marbles in a bag, 2 are red, 1 is blue, 1 is white You grab 2 marbles from the bag at the same time One (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Didn't mean to offend. Sorry 'bout that. I read the site thru again and your synopsis of the explanation above and I concur that it should be 2/3, but I like to work things thru... :) If I were to walk in off the street right after one of the (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Ok, here's where I'll take it from. The car is behind one of the doors, so the probabilities must add up to 1. 3 doors, 1/3 of a chance for each door. You pick door 1. There is a 1 in 3 chance the car is behind that door. There is a 2 in 3 (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) I think it's a great idea! Dave K (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) This analysis assumes there is a percievable difference between the red balls, which as originally stated, there isn't. It's similar to the difference between combinations and permutations. The two permutations you've listed with a red ball in (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) If you keep going on and on and on... well I probably still will, unless someone explains why I should not. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Bleargh! typo. r2 w (r1 is the only red ball, it is removed) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Why not? There are two different red balls, so each of those red balls is equally likely as the white ball to be in my hand. (...) No, because you only tried to remove r2. We are only removing *a* red ball from the bag; because there are 2 red (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Clear as mud, for then if the red is irrelevant, then there cannot be (...) there can only be (...) for if we remove either red ball from the bag, that very same red ball in this specific instance cannot be in the person's hand. We cannot take (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) Basically, each column decreases in liklihood as more choices are added. Start with one colum: Prob. Prize 1/3 1 1/3 2 1/3 3 Now, you get to choose a door. Since the prize door doesn't affect the probability: Prob. Prize Choice 1/9 1 1 1/9 1 2 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Nope, you're assuming that the marble pulled out of the bag is r2. It might be r1. Better to leave off the numbers, as the red marbles are not distinct. hand bag r r w r r w w r r remove a red from the bag and it reduces to: hand bag r w r w w (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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| | Re: math question (or pattern... whatever...)
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(...) I think I may have found the fallacy... (indulge me if you please) Here's a bag with 3 marbles in it--2 are red (goats, whatever), one is white (brand new car) Pick a random one out of the bag and hold it in your hand--don't look at it! What's (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) This one here? (URL) many trials? It is a *probability* after all, no guarantees about a short run of trials. I ran 30 "stay" and 30 "switch" and got roughly 40 and 60 as the two percentage results. Moving to the serious number cruncher, which (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) think that way, though I'll note I passed Calculus). I come up with 2/3 if you switch. A 1/3 chance that you have the right choice initially. The other two doors are an aggragate 2/3 chance. Even though one is eliminated, it is a wrong one, so (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) though I admit it has been a long time since I have taken my probablity and statistics class. (...) Gosh, everyone is seizing on this: I didn't say it was and it has nothing to do with my answers!!! :-) (...) Sigh. Yes, I know that. (...) Yes, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) If I were to think about it like a sequence of events, logically leading up to the end choice, it would go something like this-- The event starts-- Contestant picks a door -- 1, 2, or 3 If this is all there is to the contest, we have 33 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) See this one for a clearer picture: (URL)No, because he won't use the same algorithm since he then becomes (...) That's not the problem presented, though. According to the problem, it's KNOWN that he ALWAYS reveals a zonk prize door that you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Thank you for repeating the message I already read and applied to the situation. I did illuminate that it is not a valid Let's Make a Deal scenario, but it's still the one I'm talking about. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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