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 20:08:33 GMT
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Viewed:
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363 times
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In lugnet.off-topic.geek, David Eaton writes:
> In lugnet.off-topic.geek, Bruce Schlickbernd writes:
> > I think your analysis is incorrect, precisely because you did *not* list all
> > outcomes.
> >
> > Choose Prize Door Revealed Possible choices
> > 1 1 2 Stay (win). Switch (lose).
> > 1 1 3 Stay (win). Switch (lose).
> > 1 2 3 Stay (lose). Switch (win).
> > 1 3 2 Stay (lose). Switch (win).
> >
> > Choose Prize Door Revealed Possible choices
> > 2 2 1 Stay (win). Switch (lose).
> > 2 2 3 Stay (win). Switch (lose).
> > 2 1 3 Stay (lose). Switch (win).
> > 2 3 1 Stay (lose). Switch (win).
> >
> > Choose Prize Door Revealed Possible choices
> > 3 3 1 Stay (win). Switch (lose).
> > 3 3 2 Stay (win). Switch (lose).
> > 3 1 2 Stay (lose). Switch (win).
> > 3 2 1 Stay (lose). Switch (win).
> >
> > 12 different door combinations with 24 outcomes: 12 win if you stay, 12 win
> > if you switch. The nine you list is very incomplete.
>
> 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 "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:
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.
>
> > Choose Prize Door Revealed Possible choices
> > 1 1 2 Stay (win). Switch (lose). • and 1 1 2
> > 1 1 3 Stay (win). Switch (lose). • and 1 1 3
> > 1 2 3 Stay (lose). Switch (win).
> > 1 2 3 Stay (lose). Switch (win).
> > 1 3 2 Stay (lose). Switch (win).
> > 1 3 2 Stay (lose). Switch (win).
(etc. snip)
> In essense, the thing that's actually contributing to the permutation is the
> choice you made and where the prize is. Which door specifically is revealed
> is irrelevant to the permutation, since EITHER way the door is chosen, the
> outcome is the same (effectively, it's a sub-permutation).
>
> Regardless, because we *know* that the host *always* obeys certain rules (IE
> he'll always reveal a goat-door that you didn't pick), we can equivalently
> say that he always picks the lowest-numbered doors to reveal in case of a
> tie, since he's ALREADY using an algorithm.
No, because he won't use the same algorithm since he then becomes
predictable, so he will vary the door revealed (and also the whole question
is nonsense since it would be a choice of a "zonk" (goat, cow, etc.) and a
lesser prize, and a greater prize (or even two equivalent prizes) since
Let's Make a Deal didn't use double-zonks like that)..
>
> If you're still not getting it, seriously-- try it yourself. Try with
> keeping your original choice a bunch of times, then try with switching a
> bunch of times. Actually, I believe Larry posted a site with links to
> simulators (not that I tried them) where you can play with it as much as you
> want.
I tried the simulator: it supported my conclusions. Really!
-->Bruce<--
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Message has 2 Replies:
 | | Re: math question (or pattern... whatever...)
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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)
| | | Re: math question (or pattern... whatever...)
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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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Message is in Reply To:
| | Re: math question (or pattern... whatever...)
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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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