And lo On Thu, 18 Dec 2008 08:26:56 -0000, scott <sco### [at] scottcom> did  
spake thusly:

>>> 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
>>
>>
>> Brain death! - n * Sum k=1-n of 1/k
>
> Thanks Phil, when you know what it's called it's easy to find.

Indeed, one of the major difficulties in research is trying to guess what  
someone else might have named their solution ;-)

> 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 :-)

Yup, of course as you recognise it's only an estimate the lower bound  
being n and the upper bound being infinity; there's no reason why you  
can't be pulling out the same number over and over again although you  
might suspect a fiddle if it yields a high mean absolute error.

-- 
Phil Cook

--
I once tried to be apathetic, but I just couldn't be bothered
http://flipc.blogspot.com

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