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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Tue, 4 Mar 2003 20:17:39 GMT
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Viewed:
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372 times
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In lugnet.off-topic.geek, David Koudys writes:
> In lugnet.off-topic.geek, David Eaton writes:
> > 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 "Choose 2, Prize 3"-- which in your list, only gets
> > one entry. Basically, if you want to count "Choose 1, Prize 1, Door Revealed
> > 2" and "Choose 1, Prize 1, Door Revealed 3" as *different*, then you have to
> > count each *other* selection twice as well, resulting in:
>
> 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 - one possibility
>
> If I choose door one and the prize is behind door 3, the host can *only*
> show me door 2 - one possibility
>
> So if I choose door 1, there are only 4 possibilities, not 6, as you listed
> below.
>
> Now whether I switch or stay, as listed above, is 50/50
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 switch!
2 2 stay!
2 3 switch!
3 1 switch!
3 2 switch!
3 3 stay!
Effectively, there's a 1/3 chance that the prize is in door #1. And there's
a 1/9 chance that you *PICK* door number 1 AND that the prize is back there.
Further, there's a 1/18 chance that it's Prize 1, Choice 1, door revealed 2,
and another 1/18 chance that it's Prize 1, Choice 2, door revealed 3.
HOWEVER. It's a 1/3 chance that the prize is in door #1. 1/9 chance that you
pick door #2. And since there's only ONE choice of doors to be revealed,
it's a 1/9 chance that it's Prize 1, Choice 2, door revealed 3.
Hence, probability wise (using Bruce's model):
Prob. Choose Prize Door Revealed Possible choices
1/18 1 1 2 Stay (win). Switch (lose).
1/18 1 1 3 Stay (win). Switch (lose).
1/9 1 2 3 Stay (lose). Switch (win).
1/9 1 3 2 Stay (lose). Switch (win).
Prob. Choose Prize Door Revealed Possible choices
1/18 2 2 1 Stay (win). Switch (lose).
1/18 2 2 3 Stay (win). Switch (lose).
1/9 2 1 3 Stay (lose). Switch (win).
1/9 2 3 1 Stay (lose). Switch (win).
Prob. Choose Prize Door Revealed Possible choices
1/18 3 3 1 Stay (win). Switch (lose).
1/18 3 3 2 Stay (win). Switch (lose).
1/9 3 1 2 Stay (lose). Switch (win).
1/9 3 2 1 Stay (lose). Switch (win).
DaveE
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Message has 1 Reply:
 | | Re: math question (or pattern... whatever...)
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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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Message is in Reply To:
| | Re: math question (or pattern... whatever...)
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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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