Subject:
|
Re: math question (or pattern... whatever...)
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Tue, 4 Mar 2003 23:58:02 GMT
|
Viewed:
|
450 times
|
| |
|
|
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 (r1 is the only red ball, it is removed)
> w r1 OR r2 (either may be removed, as both are in the bag)
This analysis assumes there is a percievable difference between the red
balls, which as originally stated, there isn't. It's similar to the
difference between combinations and permutations. The two permutations
you've listed with a red ball in the hand are identical, and cannot be
counted as two different combinations.
ROSCO
|
|
Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
|
| (...) Why not? There are two different red balls, so each of those red balls is equally likely as the white ball to be in my hand. (...) No, because you only tried to remove r2. We are only removing *a* red ball from the bag; because there are 2 red (...) (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
|
|
|
|
