Subject:
|
Re: math question (or pattern... whatever...)
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Wed, 5 Mar 2003 01:35:50 GMT
|
Viewed:
|
464 times
|
| |
|
|
In lugnet.off-topic.geek, James Brown writes:
> In lugnet.off-topic.geek, James Brown writes:
>
> > No, because you only tried to remove r2. We are only removing *a* red ball
> > from the bag; because there are 2 red balls, and only 1 is ever in our hand,
> > there is always a red ball in the bag to remove. It is irrelevant whether
> > that ball is r1 or r2.
> >
> > Redoing it with numbers:
> >
> > hand bag
> > r1 r2 w
> > r2 r1 w
> > w r1 r2
> >
> > Now removing a red ball from the bag we get:
> >
> > hand bag
> > r1 w (r2 is the only red ball, it is removed)
> > r2 w (r2 is the only red ball, it is removed)
>
> Bleargh! typo.
> r2 w (r1 is the only red ball, it is removed)
>
> > w r1 OR r2 (either may be removed, as both are in the bag)
> >
> > James
Saw that ;) but understood...
but if we're talking one instance--one event--and that's what we're, indeed,
talking about--you are on a game show, one event--you have one chance of
winning one car behind one door....
so you have 3 doors
I won't bother to number them...
you choose one of those doors--you have a 1 in 3 chance of picking the right
one.
They open up either one of the two remaining doors--it is now open--it
cannot be the one you have chosen in this particular instance, nor can it be
the one with the car in it as laid out by the rules of the event.
So if the door is opened, it's a done deal--you cannot have possibly chosen
that door, nor does it contain the car--
So does your choice of doors contain the car? or do you want to switch to
the only other unopened door?
In a hypothetical stance, where we can say we don't know which door was
opened, I can concede the point that there are different possibilities, such
as the 2/3rds--but that's saying that we don't know which door is opened.
At the actual event, the stake in the ground is that one door is
opened--that cannot be the one you have chosen, nor does it contain the car.
After you have grabbed a marble, one *other* marble is then taken from the
bag--that specific marble taken from the bag, in this specific event, cannot
possibly be the one in your hand, nor can it be the other marble still in
the bag.
The question, as stated, is that if you were on this game show, should you
switch--and the answer, even though I don't completely agree with it (for I
am swayed by the 2/3), is it doesn't matter--you have a 50/50 chance.
If we were to use your logic, then my friend is right with the 4 marble
scenario (that started this debate), and that there is a 1 in 5 chance that
both marbles are red--but I believe, as do many others, that it is only 33
percent because the one known marble cannot possibly still be in the bag,
nor can that specific known marble be in the hidden space in your hand.
r r b w
in your hand:
Exposed Not Exposed
r1 r2
r1 b
r1 w
r2 b
r2 w
b w
If we have a known quantity--r1--exposed to us, we cannot then say that it
may be the other red marble exposed to us--that possibility is invalid
because we can 'see' one red marble--it cannot be 'the other' red marble for
then we would be swapping the red marbles. The second you have a known
quantity, the exposed marble or the open door, it's a done deal.
Exposed Not Exposed
r1 r2
r1 b
r1 w
these following cannot be possibilities in this given scenario--
r2 b
r2 w
b w
so the chance of having 2 red marbles in your hand,if one is shown to be
red, is reduced to the following chance
Exposed Not Exposed
r1 r2
r1 b
r1 w
or 1 in 3
Let's reduce the picking to one door--I pick door number 1--I think the
arguement will hold up no matter which door I picked, so lets see the results
All possibilities listed:
I pick
Door The Prize Door opened Stay with my chosen door?
1 1 1 yes
1 1 1 no-switch
1 1 2 yes
1 1 2 no-switch
1 1 3 yes
1 1 3 no-switch
1 2 1 yes
1 2 1 no-switch
1 2 2 yes
1 2 2 no-switch
1 2 3 yes
1 2 3 no-switch
1 3 1 yes
1 3 1 no-switch
1 3 2 yes
1 3 2 no-switch
1 3 3 yes
1 3 3 no-switch
I think those are *all* the possible outcomes for me just choosing door
number 1 at the beginning.
Now eliminate those that don't adhere to the quesition--as in they won't
open the door I picked or they won't open the door with the prize behind it...
Whats left--
Door The Prize Door opened Stay?
1 1 2 yes
1 1 2 no-switch
1 1 3 yes
1 1 3 no-switch
1 2 3 yes
1 2 3 no-switch
1 3 2 yes
1 3 2 no-switch
Tally the winnings--
Door The Prize Door opened Stay? Win?
1 1 2 yes yes
1 1 2 no-switch no
1 1 3 yes yes
1 1 3 no-switch no
1 2 3 yes yes
1 2 3 no-switch no
1 3 2 yes yes
1 3 2 no-switch no
50/50
Just taking the first two lines to further clarify another point--
Door The Prize Door opened Stay? Win?
1 1 2 yes yes
1 1 2 no-switch no
Sequence of events--
If I picked door 1 and the prize is behind door 1,
Then door number 2 is revealed to have a goat,
Then they ask me if I want to switch,
I cannot possibly choose door #2 for it's already open--If I want to switch,
I can only switch to door #3--there is no other option. Either I stay or I
switch--these two possibilities, and only these two possibilities, exist for
me to choose from. If that is the case, then I have 50/50 chance staying or
switching...
Again, If we follow thru an actual event, this is what I come up with.
I'm going to re-read the site... :)
Dave K
|
|
Message is in Reply To:
61 Messages in This Thread:
  
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
