Subject:
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Re: Probability: (Was: Re: Chaotic Systems...)
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Newsgroups:
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lugnet.off-topic.debate
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Date:
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Fri, 26 Jan 2001 13:51:02 GMT
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Viewed:
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1282 times
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David Eaton wrote:
>
> In lugnet.off-topic.debate, Scott Arthur writes:
> >
> > > Here's an interesting one probability: (totally non sequeter):
> > >
> > > There used to be a game show on TV where they'd have 3 doors. Behind ONE of
> > > the doors was a prize. Behind the OTHER two doors, there was either nothing
> > > or just something stupid... I think when I heard it, it was a 'goat' as the
> > > 'gag' prize.
> > >
> > > Anyway, the contestant would pick a door. And of course the host, trying to
> > > build up suspense would open one of the remaining doors with a gag prize,
> > > and give the contestant one last chance to change their mind.
> > >
> > > So the question: is it statistically to your advantage to switch your choice
> > > to the remaining unopened door or to stick with your original choice?
> >
> >
> > I'd get rid of the 1/3 chance and take the 1/2. Regardless of the laws of
> > probability, sods law still says I will not win!
>
> Actually, the odds that it's in the OTHER door (the one you didn't pick) are
> now up to 2/3, not just 1/2!
>
> I remember that this question actually generated a couple debates from a
> magazine and several colleges who were disputing the probability on the
> issue... But you really actually are twice as likely to get the prize by
> switching doors! :)
Yea, one place it was a big deal in was Ask Marylyn (sp?) in Parade.
There's several ways to analyze it and get to the 2/3 chance. The one I
realized yesterday is the simplest (but perhaps not most intuitive) is
to realize that by switching, you're in effect getting to chose BOTH of
the doors you didn't choose in the first place. In other words, by
switching, you're getting to buy 2 tickets for the price of one. Another
way to look at it is to exhaustively enumerate the possibilities:
DONT SWITCH POSSIBILITIES:
1. you pick A, correct is A - score 1
2. you pick A, correct is B - score 0
3. you pick A, correct is C - score 0
4. you pick B, correct is A - score 0
5. you pick B, correct is B - score 1
6. you pick B, correct is C - score 0
7. you pick C, correct is A - score 0
8. you pick C, correct is B - score 0
9. you pick C, correct is C - score 1
9 possibilities, 3 score, thus chance is 3/9 or 1/3
YOU SWITCH POSSIBILITIES:
1. you pick A, correct is A - score 0 *
2. you pick A, correct is B - score 1
3. you pick A, correct is C - score 1
4. you pick B, correct is A - score 1
5. you pick B, correct is B - score 0 *
6. you pick B, correct is C - score 1
7. you pick C, correct is A - score 1
8. you pick C, correct is B - score 1
9. you pick C, correct is C - score 0 *
9 possibilities, 6 score, thus chance is 6/9 or 2/3
* Note that in these cases, there are two sub-possibilities for which
door is opened for you, which both loose. It is important to realize
that each of those possibilities has half the chance of the other
possibilities occuring. Not doing so results in the answer being 50-50
for switching, but that is an incorrect analysis. The way to completely
analyze it to not create problems is to take each decision in order,
assigning a probability to each and multiplying.
In other words (expanding out possibility 1 to it's two sub
possibilities):
You pick A (1/3 probability)
GIVEN that you pick A, the correct answer is A (1/3 probability -
combined probability is now 1/9)
GIVEN that you picked A and the correct answer is A, the host shows you
B (1/2 probability - he is chosing between B or C - combined probability
1/18)
You switch and lose.
Total probability 1/18
You pick A (1/3 probability)
GIVEN that you pick A, the correct answer is A (1/3 probability -
combined probability is now 1/9)
GIVEN that you picked A and the correct answer is A, the host shows you
C (1/2 probability - he is chosing between B or C - combined probability
1/18)
You switch and lose.
Total probability 1/18.
Note that since it doesn't matter which of B or C the host picks, we can
just ignore that choice, and reduce it to:
You pick A (1/3 probability)
GIVEN that you pick A, the correct answer is A (1/3 probability -
combined probability is now 1/9)
When you switch you will lose.
Total probability 1/9
--
Frank Filz
-----------------------------
Work: mailto:ffilz@us.ibm.com (business only please)
Home: mailto:ffilz@mindspring.com
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Message is in Reply To:
 | | Re: Probability: (Was: Re: Chaotic Systems...)
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| (...) Actually, the odds that it's in the OTHER door (the one you didn't pick) are now up to 2/3, not just 1/2! I remember that this question actually generated a couple debates from a magazine and several colleges who were disputing the probability (...) (23 years ago, 26-Jan-01, to lugnet.off-topic.debate)
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