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(...) You're also being incompletely rigorous. :) If you want to do an "all possible outcomes" probability check, you have to add in all the 'door revealed=prize' options which exist for the sake of the probabilities involved, but are never invoked (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, it supports the 50/50. Host will always offers regardless of whether you chose correctly or not (basic premise - choose the "host does not know"): simulator runs 50/50. If it's a screw you scenario, then yes, you switch, but I (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Liking it or not is irrelevant. The probabilities are what they are. And the answer is still 2/3 if you switch, given the premises. No matter whether you like it or not. Go to the site I gave you and read what is written there. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Bruce Schlickbernd writes: <snip> (...) So, like others suggest about knowing that one of the marbles is red, the chance that the other one is red... the one that we know becomes basically irrelevant to the chance--there (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I think your analysis is incorrect, precisely because you did *not* list all outcomes. Choose Prize Door Revealed Possible choices 1 1 2 Stay (win). Switch (lose). 1 1 3 Stay (win). Switch (lose). 1 2 3 Stay (lose). Switch (win). 1 3 2 Stay (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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Dave, you rock. I am finally convinced. :) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) For any still not getting this one (it's tricky!) here's another way to analyze the problem. List all outcomes: Prize Choice Door Revealed Stay or Switch? 1 1 2 stay! 1 2 3 switch! 1 3 2 switch! 2 1 3 switch! 2 2 1 stay! 2 3 1 switch! 3 1 2 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) See the link I posted and follow some of the links it has. You are in good company, lots of mathemeticians came to the same conclusion. (and roasted Marilyn Vos Savant about it at the time it first got a lot of publicity) That was before they (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Inasmuch as you get to pick a second time, the odds are 50/50. If you did not get to pick a second time, the odds would be 1/3, not 1/2. Of course, this is leaving out the poker aspect: if your initial pick was incorrect, would they even offer (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Nope. The analysis that I have seen that argues for a different outcome than 1/2 has to do with the fact that you're being (or so it is argued) conveyed more information than it seems. That is, they KNOW which door is right, and by showing you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I already proved that I should stay out of this... but. When you had 3 doors to pick from, it was 1/3 that whatever door you choose is the correct one. However, after opening one of the doors you didn't pick (assuming the host is not opening (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, there's a 2/3 chance that it's behind door Z! A very sneaky problem, in fact. It was shown to us in high school as the subject of a very minor controversy. After being published as "2/3 chance" in a math publication, some college (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I think the two answers depend on a more accurate statement of the problem. If the problem is stated that you pick two marbles, and given that the first one you pick is red, what is the chance the second one is red, then order matters, and the (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) doh, you're right :) Also, I was calculating the chance of getting 2 red marbles, in general, not checking the chance given the first one was red... :) That's what you get when you do an experiment, and have an expected solution in your (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Dave Schuler writes: <snip> (...) I'd say that me drawing a red one is 1 in 3, and you drawing a red one is one in 3 so the chances are for us to draw a red at the same time is 1 in 9. The chances to draw a blue one at the (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) I don't quite get why this works-- if you're doing rand($#bag), won't that always pick r, r, or w for the 1st value? Hence, theoretically, you could pick "wb", which isn't in your sample set... on the other hand, you're increasing the chance (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, David Eaton writes: <snip> (...) I'd say 1 in 2, or 50 percent, but maybe I'm missing something. Good question though. Dave K (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Ross Crawford writes: <snip> (...) I like driving home from work--gives me time to think about stuff... And after reading all the great replies, I'm going to propose the following solution: If you have a bag with 4 marbles, (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Your friend has assumed that there is a difference between the reds, and you know which one you have seen in your hand. As there is no information in the original problem to substantiate this assumption, you have to assume it's not the case, (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Larry Pieniazek writes: <snip> (...) Thanks Larry, I'll add it to the rationale. I failed gr. 12 advanced math once--was too busy being the editor of the yearbook that particular year (and playing cards in the cafeteria) It (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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