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(...) Actually, I explained (albeit briefly) why this isn't correct. You see, you now have 2 listings for "Choose 1, Prize 1", "Coose 2, Prize 2", and "Choose 3, Prize 3", when each deserves *equal* probability to each other possibility, like (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) That doesn't make sense at all-- If I choose door 1 and the prize is behind door 1, the host can choose to show me door 2 *or* door 3 - 2 possibilities If I choose door one and the prize is behind door 2, the host can *only* show me door 3 - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Whew. You are way off in your probablity and statistics analysis: if you want to double everything, then you would have to list 112 and 113 twice also. They are *not* the same selection. (...) and 1 1 2 (...) and 1 1 3 (...) (etc. snip) (...) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) <mucho snippage> (...) Bruce - I see the confusion. You aren't solving the problem as presented: "Now, here's a trickier one: Ever watched the old game show "Let's Make a Deal"? Sometimes, they'd offer people things similar to this scenario. (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Doh. This is just coming back to haunt me. I should never have included the "door revealed" column. It's entirely irrelevant, and is only serving to confuse. A better list: Prize Choice Switch or Stay? 1 1 stay! 1 2 switch! 1 3 switch! 2 1 (...) (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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(...) 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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(...) 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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(...) 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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(...) 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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(...) 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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(...) 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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(...) 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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(...) I think it's a great idea! Dave K (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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