> If we select K balls, the probability of ball #1 appearing becomes 1/N + 
> 1/N + 1/N + 1/N + ... K times over, minus the probability of ball #1 
> appearing more than once.
>
> In other words, we have K/N - E, and I can't figure out how to compute E.

You can think of it the other way round, ie the probability of ball #1 
appearing after K tries is 1 minus the probability of ball #1 not occurring 
after K tries.  That is 1-((N-1)/N)^K.

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