>>> Now, if we perform K trails, the probability of seeing all the balls is 
>>> (N! + (K-N)!) / N^N. (Assuming K >= N.)
>>
>> Can you explain how you got that formula?
>
> Intuitively, assuming N trials the formula is just N! / N^N, so if we add 
> (K-N) extra trails, you get the formula above.
>
> On reflection, I'm not sure that's actually correct...

What if K=5 and N=2? I think the answer should be a 30/32 chance of seeing 
both balls, but your formula gives 8/4 !

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