Subject:
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Re: math question (or pattern... whatever...)
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 3 Mar 2003 23:32:38 GMT
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Viewed:
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279 times
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In lugnet.off-topic.geek, David Eaton writes:
<snip>
> 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. You're a game show contestant. Bob
> the host shows you 3 doors. You win whatever's behind the door you pick.
> Behind 2 of the doors are goats. Behind the other door, there's a brand new
> sports car.
>
> You pick door X. Now, before Bob shows you what's behind door X, he opens up
> door Y, and shows you... a goat! Bob now says you have a chance to change
> your mind, and instead of picking door X, you can switch to door Z.
>
> What's the probability that the sports car is behind door Z versus door X?
>
> DaveE
I'd say 1 in 2, or 50 percent, but maybe I'm missing something.
Good question though.
Dave K
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Message has 1 Reply:
 | | Re: math question (or pattern... whatever...)
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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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Message is in Reply To:
| | Re: math question (or pattern... whatever...)
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| (...) Yep, you're right, your friend is wrong. The problem is that you *wouldn't* count "r(2) r(1)" if order didn't matter. But then again, the thing he's forgetting is that if you don't count "r(2) r(1)", then you also CAN'T count *both* "r(1) b" (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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