Subject:
|
Re: math question (or pattern... whatever...)
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Tue, 4 Mar 2003 20:03:23 GMT
|
Viewed:
|
375 times
|
| |
| |
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:
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
Dave K
-who is just getting more and more confus-ed by the second, but loving the
problem....
<snip>
> DaveE
|
|
Message has 1 Reply: | | Re: math question (or pattern... whatever...)
|
| (...) 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 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
Message is in Reply To:
| | Re: math question (or pattern... whatever...)
|
| (...) 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)
|
61 Messages in This Thread:
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
This Message and its Replies on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|