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I have a bag containing $n$ coins. Each coin has a unique integer on it in the range $[1, m]$. As a corollary, $n \le m$.

We may sample $k$ coins without replacement where $k < n$. Given that we have knowledge of the numbers on $k$ coins but we don't know the value of $n$ or $m$:

  • Is it possible to estimate $m$ and if so, what is the formula?

  • Is it possible to estimate $n$ and what is that formula?

This is a variation on the German Tank Problem, and as far as I can tell, it models it exactly if $n = m$ but I am wondering specifically about the cases where $n < m$

My attempts so far

Regarding the estimation of m

My first instinct is to say that it depends heavily on what guarantees we have on how the n coins are distributed.

For instance, if $m = 5$ and the bag is filled sequentially (e.g. ${1,2,3}$), there is no way for you to know that it was filled that way. If you did know, you could maybe say that every subsequent number after the highest number you draw has a slightly lower chance of being $m$.

If I know that the bag is filled randomly, it seems like it would be easier to estimate $m$ though I am not sure how to put it into words why.

Regarding the estimation of n

I imagine that this could be a function of $m$. Once you have an estimator for $m$, you could make some guesses between the largest sample and $m$.

No doubt, since it is a function of $m$, it would also vary based on how the bag was filled. I have no idea where to start though.

NeRoboto
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  • It seems that all you know about $n$ is that it is at least $k$. It is as if I gave you a bag of coins and told you to take $10$ out and then estimate how many coins remain in the bag – Henry Sep 04 '23 at 07:57
  • What if the coins are something like $1, 2, 3, .., 99, 100$ and $10^{27}$? I don't think you can get any non-trivial estimate for $m$ just from a sample. – quarague Sep 04 '23 at 09:53
  • @quarague What do you mean by non-trivial in this case? – NeRoboto Sep 04 '23 at 11:06
  • The trivial estimate is just looking at the k coins I got and estimating m to be the biggest denomination you see there. I don't think you can do better than that in general even if you also know n. – quarague Sep 04 '23 at 11:08

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