home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!idacrd!desj@ccr-p.ida.org
- From: desj@ccr-p.ida.org (David desJardins)
- Newsgroups: sci.math.stat
- Subject: Re: Probability
- Message-ID: <1588@idacrd.UUCP>
- Date: 29 Aug 92 16:05:30 GMT
- References: <1992Aug29.135404.10393@wuecl.wustl.edu>
- Sender: desj@idacrd.UUCP
- Organization: IDA Center for Communications Research, Princeton
- Lines: 48
-
- Peter Pui Tak Chiu <ppc1@cec2.wustl.edu> writes:
- >Assume there are m kinds of balls in an infinitely large pool (i.e.
- >there are infinitely many balls) and the distributions of different
- >kinds of balls are equal. Find the probability of getting all kinds of
- >balls if n balls are picked.
-
- >Therefore, the probability is: (n-1)C(m-1)
- > ---------------
- > (n+m-1)C(m-1)
-
- Nope. Suppose N = M. Obviously the chance of getting one ball of each
- color is M!/(M^M). Your formula reduces to M!(M-1)!/(2M-1)!, which is
- obviously not the same thing.
-
- Your result is wrong because the arrangements are not equally likely.
- The chance of getting M red balls is 1/(M^M), but the chance of getting
- one ball of each color is M!/(M^M). You have given each of these the
- same weight.
-
- Here is a correct approach based on inclusion-exclusion. Let P(K) be
- the probability that all of the balls are selected from a particular set
- of K colors, with every color occurring at least once. Thus P(M) is the
- quantity we want. Let Q(L) be the probability that all of the balls are
- selected from a particular set of L colors, but not necessarily with
- every color occurring. Clearly Q(L) = (L/M)^N.
-
- Now, of course,
-
- Q(L) = \sum_{K=0}^L B(L,K) P(K),
-
- where B(L,K) is the binomial coefficient L-choose-K. This just says
- that if all of the balls are from a set of L colors, then the set of
- colors which actually occur must be some K-set contained within the
- L-set. The number of such K-sets is B(L,K), for each K.
-
- The inclusion-exclusion formula expresses P in terms of Q:
-
- P(K) = \sum_{L=0}^K (-1)^(K-L) B(K,L) Q(L).
-
- It is a direct consequence of the previous formula. If it's not
- familiar to you, you can look it up in any standard combinatorics text,
- or just derive it for yourself.
-
- Thus,
-
- P(M) = \sum_{L=0}^M (-1)^(M-L) B(M,L) (L/M)^N.
-
- David desJardins
-