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 21:35:55 GMT
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Viewed:
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390 times
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In lugnet.off-topic.geek, Bruce Schlickbernd writes:
> In lugnet.off-topic.geek, David Eaton writes:
> > In lugnet.off-topic.geek, Bruce Schlickbernd writes:
> > > 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.
> >
> > See this one for a clearer picture:
> > http://news.lugnet.com/off-topic/geek/?n=4210
>
> I don't believe that to be a correct analysis so it clears up nothing,
> though I admit it has been a long time since I have taken my probablity and
> statistics class.
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
1/9 1 3
1/9 2 1
1/9 2 2
1/9 2 3
1/9 3 1
1/9 3 2
1/9 3 3
Now, the doors revealed column. In SOME cases (when you pick the prize
door), there's actually more situations to consider. Hence, for each
situation, divide the probability of the row by however many choices you've
got for doors to open. If the host can only choose 1 door, the probabilty
stays the same. If the host can choose 2 doors, the probability is halved:
Prob. Prize Choice Door Revealed
1/18 1 1 2
1/18 1 1 3
1/9 1 2 3
1/9 1 3 2
1/9 2 1 3
1/18 2 2 1
1/18 2 2 3
1/9 2 3 1
1/9 3 1 2
1/9 3 2 1
1/18 3 3 1
1/18 3 3 2
Now let's go further and see probability for each door you can pick in the
final pick:
Prob. Prize Choice Door Revealed Final Choice
1/36 1 1 2 1 (win, staying)
1/36 1 1 2 3 (lose, switching)
1/36 1 1 3 1 (win, staying)
1/36 1 1 3 2 (lose, switching)
1/18 1 2 3 1 (win, switching)
1/18 1 2 3 2 (lose, staying)
1/18 1 3 2 1 (win, switching)
1/18 1 3 2 3 (lose, staying)
1/18 2 1 3 1 (lose, staying)
1/18 2 1 3 2 (win, switching)
1/36 2 2 1 2 (win, staying)
1/36 2 2 1 3 (lose, switching)
1/36 2 2 3 1 (lose, switching)
1/36 2 2 3 2 (win, staying)
1/18 2 3 1 2 (win, switching)
1/18 2 3 1 3 (lose, staying)
1/18 3 1 2 1 (lose, staying)
1/18 3 1 2 3 (win, switching)
1/18 3 2 1 2 (lose, staying)
1/18 3 2 1 3 (win, switching)
1/36 3 3 1 2 (lose, switching)
1/36 3 3 1 3 (win, staying)
1/36 3 3 2 1 (lose, switching)
1/36 3 3 2 3 (win, staying)
6/36 => 1/6 chance that you'll win by stay
6/18 => 1/3 chance that you'll win by switching
6/36 => 1/6 chance that you'll lose by switching
6/18 => 1/3 chance that you'll lose by staying
Hence, if you switch every time, you have a 2/3 chance of winning (twice as
likely as losing); and if you stay every time, you have a 1/3 chance of
winning (half as likley as losing).
And as for overall? Statistically (not counting human issues like the
mounting pressure, etc) 18/36 (1/2) people will switch (as expected), and
18/36 will win (as expected, actually, since it's a stupid game for the game
show to retain money, which is why they rarely (if ever?) did this
particular version). But of course, the interesting part is that of the half
that wins, 2/3 of them switched, and 1/3 of them stayed.
(I see as I'm about to post this that you figured it out by drawing it out--
cool! I'll just post for reference)
DaveE
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Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
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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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