# Why are these estimates to the German tank problem different?

Suppose that I observe $k=4$ tanks with serial numbers $2,6,7,14$.
What is the best estimate for the total number of tanks $n$?

I assume the observations are drawn from a discrete uniform distribution with the interval $[1,n]$. I know that for a $[0,1]$ interval the expected maximum draw $m$ for $k$ draws is $1 – (1/(1+k))$. So I estimate $\frac {k}{k+1}$$(n-1)≈ m, rearranged so n≈ \frac {k+1 }{k}$$m+1$.

But the frequentist estimate from Wikipedia is defined as:

$n ≈ m-1 + $$\frac {m}{k} I suspect there is some flaw in the way I have extrapolated from one interval to another, but I would welcome an explanation of why I have gone wrong! #### Solutions Collecting From Web of "Why are these estimates to the German tank problem different?" Just seen what went wrong. I accidentally put in a plus sign instead of a minus sign. Ugh: n≈ \frac {k+1}{k}$$m+1$ should be $n≈$ $\frac {k+1}{k}$$m-1$.

This is the same as the frequentist formula.