>> 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.

On an only slightly related topic, I did manage to find this one

http://en.wikipedia.org/wiki/Benford%27s_law

without knowing what it was called beforehand.

Post a reply to this message