Sabrina Kilian <"ykgp at vtSPAM.edu"> wrote:
> Warp wrote:
> > Invisible <voi### [at] devnull> wrote:
> >> If we select one ball, the probability of it being ball #1 is 1/N.
> > 
> >   You are thinking it in the wrong way.
> > 
> >   The very first time you select one ball, the probability it's a ball you
> > have never seen before is 1.
> > 
> >   The second time you select a ball, the probability it's a ball you have
> > never seen before is (N-1)/N.
> > 
> >   With the third ball, the probability is (N-2)/N. And so on.
> > 

> It is very likely that, for the third pick of a substantial N balls that 
> the probability is (N-2)/N. However, if the first two picks happened to 
> be the same, the probability would still be (N-1)/N.

  As I said in my followup post, my suggestion gives the probability of
getting all the unique balls on the first try (iow. N!/(N^N).) It doesn't
tell how many balls you have to take out in average to see them all.

-- 
                                                          - Warp

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