Subject:
|
Re: math question (or pattern... whatever...)
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Wed, 5 Mar 2003 14:41:27 GMT
|
Viewed:
|
505 times
|
| |
|
|
In lugnet.off-topic.geek, James Brown writes:
> In lugnet.off-topic.geek, David Koudys writes:
> > 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.
>
> Ok, here's where I'll take it from. The car is behind one of the doors, so
> the probabilities must add up to 1. 3 doors, 1/3 of a chance for each door.
>
> You pick door 1. There is a 1 in 3 chance the car is behind that door.
> There is a 2 in 3 chance that the car is behind one of the other doors.
> Follow me?
>
> Now, here's the thing - ** those probabilities DO NOT CHANGE. ** The odds
> of your door being correct are still (and always) 1 in 3.
>
> If at this point you were offered "you can stick with what you've got, or
> you can have the other two doors?" what you would take? The other two doors
> - that's a 2/3 probability of having your car.
>
> Revealing one of those two doors as a goat changes nothing - We already know
> that one of those two doors has a goat behind it! Revealing what's actually
> behind doors at this point is an academic excercise; the probabilities have
> already been determined.
>
> James
> (gah. I can't believe I'm still arguing this.)
Didn't mean to offend. Sorry 'bout that.
I read the site thru again and your synopsis of the explanation above and I
concur that it should be 2/3, but I like to work things thru... :)
If I were to walk in off the street right after one of the doors was opened,
and I was asked to choose between two of the remaining doors, it would be
50/50 that I'd get the car.
If all 3 doors are closed, and I have to pick a door, I have a 1 in 3 chance
of getting the car.
If I don't pick a door and they open one, then ask me to pick, I'd have the
choice between two doors--50/50
If I pick a door, I have a 1 in 3 chance that I'm right, therefore the other
two doors combined have a 2 in 3 chance that one of them has the car.
If they open one of those unchosen doors to reveal a goat, the entire 2/3
chance that a car is behind one of those two doors is 'shifted' to the
remaining door.
It works.
And I like it--is sitting well with me.
But (and there's always a but...)
The discussion that started this...
If I have 4 marbles in a bag, 2 red, 1 white, 1 blue
And I reach in and grab 2 of those marbles, and I find out that one of those
marbles in my hand is red, what is the chance that the other marble is also red.
By the above logic, the probability is set when I grab the marbles out of
the bag--
so the probable combinations in my hand...
r1 r2
r1 w
r1 b
r2 w
r2 b
w b
Those are the only probable combinations in my hand--I don't think it
matters what the order is in my hand)
Without opening my hand, the chance that both marbles are red is 1 in 6.
If I reveal 1 of the marbles and see that it's red, what's the chance the
other marble is red as well?
I used to say 1 in 3, but that'd be wrong using the above logic--the
probability is set at the grab, and the only thing we did when we revealed
that one of the marbles is red, is lose the last option (the w b)...
So fundamentally, the question hasn't changed--what is the chance that both
marbles are red? After revealing that one of the marbles is red, the
probability is 1 in 5, or 20 percent.
So I was wrong earlier with my 33 percent.
Eh, whatreya gonna do?
:)
Dave K
|
|
Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
|
| (...) Ok, here's where I'll take it from. The car is behind one of the doors, so the probabilities must add up to 1. 3 doors, 1/3 of a chance for each door. You pick door 1. There is a 1 in 3 chance the car is behind that door. There is a 2 in 3 (...) (22 years ago, 5-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
|
|
|
|
