>> Do a search for Coupon Collector's Problem. For equal random drawings 
>> it's roughly nHn that is n * Sum k=1-n of 1/k multiplied by n
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> Brain death! - n * Sum k=1-n of 1/k

Thanks Phil, when you know what it's called it's easy to find.

So, my answer is that the expected number of drawings to get all balls is 
given by:

E[T] = n*(1/1 + 1/2 + 1/3 + ... 1/n)

Cool neat little formula :-)

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