 | | Re: math question (or pattern... whatever...) —David Koudys
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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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| | | | | Re: math question (or pattern... whatever...) —James Brown
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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...) —David Koudys
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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...) —James Brown
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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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| | | | | | Re: math question (or pattern... whatever...) —James Brown
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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...) —David Koudys
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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...) —David Koudys
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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...) —James Brown
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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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| | | | | | | | Re: math question (or pattern... whatever...) —David Koudys
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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...) —Ross Crawford
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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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