Orchid XP v8 <voi### [at] devnull> wrote:
> >   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.

> As with most mathematical problems, it's trivially easy once you already 
> know what the answer is... :-}

  OTOH that only tells us the probability of getting distinct balls when
taking N balls out. It doesn't tell us how many balls must be taken in
average in order to get N distinct balls.

  Btw, if you didn't bother continuing that chaing of deductions, the
probability of getting all distinct balls when taking out N balls is
N!/(N^N).

-- 
                                                          - Warp

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