Re: jars and marbles again...

On Wed, 31 Aug 2005 06:30:43 -0400, Torsten Hennig wrote:

>>Hi,
>>
>>Consider the problem of putting M identical marbles >randomly in J jars. I
>>was asked the expected number of non-empty jars. Using >indicator variables
>>I was able to calculate this expected number and the >corresponding
>>variance.
>>
>>Although this was good enough, I'd like to know if it's >possible to
>>calculate the exact distribution of the number of non->empty jars.
>>
>>Thanks,
>>
>>Boi
>
> Hi,
>
> let X be the number of _empty_ jars.
>
> Then,
>
> P(X=i) = C(J,i)*C(M-1,M-(J-i))/C(J+M-1,M).
>
> This can be explained as follows:
>
> First, choose i jars out of J which are meant to remain
> empty (C(J,i) possibilities).
> Then, distribute one marble on each of the remaining J-i
> jars. You are left with M-(J-i) marbles which can
> be distributed on the J-i jars
> (C(J-i+M-(J-i)-1,M-(J-i)) = C(M-1,M-(J-i)) possibilities).
> Now, put the possibilities together to get
> C(J,i)*C(M-1,M-(J-i))possibilities to distribute M
> marbles on J jars such that exactly i jars are empty.
> Finally, the total number of possibilities to distribute
> M marbles on J jars is C(J+M-1,M).
>
> The probability that exactly i jars are non-empty
> can be easily deduced from this result.
>
> Best wishes
> Torsten.

Hello,

Are you sure this reasoning is correct? Yes, you divide a certain number
of "correct" or "wanted" configurations by the total number of
configurations, but I don't think they those configurations are equally
probable, a necessary condition to apply Laplace's Law.

Writing out the case J=3, M=4, I find the distribution of X, the number of
empty jars, to be

P(X=0) = 36/81
P(X=1) = 42/81
P(X=2) = 3/81

clearly not corresponding to your values

P(X=0) = 1/5
P(X=1) = 3/5
P(X=2) = 1/5

I think you are making a mistake of the "I win the lottery or I don't, so
I have a 50% chance of winning the lottery" kind.

But I could be wrong myself. Any thoughts?

Boi





.

Relevant Pages

  • Re: jars and marbles again...
    ... >If you are looking for wierd quantum effects, ... >we have N electron states with N-1 electrons in them,>there ... >putting N-1 marbles in N jars, ...
    (sci.math)
  • Re: jars and marbles again...
    ... >It sums to 1 so it's a valid distribution, but I don't think it's right. ... >marbles are identical, that makes some sets of distributions appear the same ... Once you choose the non-empty 2 jars, ... The number of ways to divide the marbles into j nonempty sets is the ...
    (sci.math)
  • Re: jars and marbles again...
    ... >Consider the problem of putting M identical marbles>randomly in J jars. ... >was asked the expected number of non-empty jars. ... let X be the number of _empty_ jars. ...
    (sci.math)
  • Re: jars and marbles again...
    ... >>variance. ... >let X be the number of _empty_ jars. ... It sums to 1 so it's a valid distribution, but I don't think it's right. ... marbles are identical, that makes some sets of distributions appear the same ...
    (sci.math)
  • Re: Probability: Is this method correct?
    ... 10 balls are randomly distributed to 15 jars. ... expected number of jars that are empty? ... The reason is that in cases such as this the sum of the expectations ...
    (sci.stat.math)
Quantcast