Minimal number of ball draws in order to get 0.999 probability I recently got a very confusing question regarding probability. I read it a lot of times but I just could not understand it, so I will try to write it in the best way possible. 
We have a group of natural numbers $A$ $=$ $\{1,2,...,N\}$. We have a box full of balls and the box contains $n$ balls so that $n \in A$.
In the box there is 1 black ball and the rest are white balls. We draw a random ball and we repeat. What is the minimal number of ball draws (with returning the ball back to the box) do we need in order to conclude that in a probability of at least 0.999 we have $n = 1$ or $n > 1$.
Notes from the question:- 
$1.$ The meaning of the word "conclude" in this context is that if all the balls that we drew were black that means $n = 1$ and if we drew at least $1$ white ball then that means $n > 1$.
$2.$ The lecturer told me that it is enough to find the minimal number of ball draws if $n = 2$.
My Approach:-
I thought of the ball draws as a Geometric variable $X$ with a success probability of $1/2$ because in case of $n = 2$ we have 1 black ball and 1 white ball. The variable $X$ counts the number times we drew a black ball until we draw a white ball for the first time. I tried to find the minimal number of draws from this equation that I thought of $P(X = k) \ge 0.999$ and k is my answer but from this I get that k is $0$.
Any help with this?
 A: So there are totally $n$ balls, $1$ black and $n-1$ white, that you draw randomly with replacement.
You draw a predetermined number $m$ of balls and for the result you distinguish just two outcomes:
 - a) all $m$ balls are black;
 - b) at least one of the balls is white, i.e. $b = \neg \,a$;   
Clearly for outcome b) it doesn't matter whether you stop at the first white ball, or continue with all the $m$ draws.
Assume that $n$ and $m$ are given, then
$$
P(a\;\left| {\;n \wedge m} \right.) = {1 \over {n^{\,m} }}\quad P(b\;\left| {\;n \wedge m} \right.) = 1 - {1 \over {n^{\,m} }}
$$
that is
$$
P(a\;\left| {\;n \wedge m} \right.) = {{P(a \wedge n\;\left| {\;m} \right.)} \over {P(n\;\left| {\;m} \right.)}} = {1 \over {n^{\,m} }}
$$
$n$ is not known, and we are going to estimate it from the draws (at least this is my understanding).
What is known is that $n \in \{1,2,\cdots,N\}$, $N$ is supposedly known and
we may assume that $n$ is uniformly distributed in that range.
Since $m$ and $n$ are independent, then
$$
P(n\;\left| {\;m} \right.) = P(n) = {1 \over N}
$$
Denoting by $A$ the event $n=1$, and by $B= \neg \,A$ the complementary event $1 < n \le N$, we have
$$
\left\{ \matrix{
  P(a \wedge A\;\left| {\;m} \right.) = {1 \over {1^{\,m} }}P(A\;\left| {\;m} \right.) = {1 \over N} \hfill \cr 
  P(b \wedge A\;\left| {\;m} \right.) = 0 \hfill \cr 
  P(a \wedge B\;\left| {\;m} \right.) = \sum\limits_{2\, \le \,n\, \le \,N} {{1 \over {n^{\,m} }}P(n\;\left| {\;m} \right.)}
  = {1 \over N}\sum\limits_{2\, \le \,n\, \le \,N} {{1 \over {n^{\,m} }}}  = {1 \over N}\left( {H_{\,N}^{\,\left( m \right)}  - 1} \right) \hfill \cr 
  P(b \wedge B\;\left| {\;m} \right.) = \sum\limits_{2\, \le \,n\, \le \,N} {\left( {1 - {1 \over {n^{\,m} }}} \right)P(n\;\left| {\;m} \right.)}
  = 1 - {1 \over N}H_{\,N}^{\,\left( m \right)}  \hfill \cr 
  P(a\;\left| {\;m} \right.) = {1 \over N}H_{\,N}^{\,\left( m \right)} \quad \quad 
  P(b\;\left| {\;m} \right.) = 1 - {1 \over N}H_{\,N}^{\,\left( m \right)}  \hfill \cr}  \right.
$$
where the sum is expressed through the Generalized Harmonic number.
We can therefore conclude that
$$
\eqalign{
  & P(A\;\left| {\;a \wedge m} \right.) = {{P(a \wedge A\;\left| {\;m} \right.)} \over {P(a\;\left| {\;m} \right.)}} = {1 \over {H_{\,N}^{\,\left( m \right)} }}  \cr 
  & P(B\;\left| {\;a \wedge m} \right.) = {{P(a \wedge B\;\left| {\;m} \right.)} \over {P(a\;\left| {\;m} \right.)}}
 = \left( {{{H_{\,N}^{\,\left( m \right)}  - 1} \over {H_{\,N}^{\,\left( m \right)} }}} \right) = 1 - P(A\;\left| {\;a \wedge m} \right.) \cr} 
$$
i.e. that, upon having drawn $m$ black balls, we can say that the probability
that the urn contains just one black ball is $1 / {H_{\,N}^{\,\left( m \right)}}$.
For example we get for $m=1,\cdots,12$ :
$N=2$
$$\frac{2}{3},\frac{ 4}{5},\frac{ 8}{9},\frac{ 16}{17},\frac{ 32}{33},\frac{ 64}{65},\frac{ 128}{129},\frac{ 256}{257},\frac{ 512}{513},
\frac{ 1024}{1025},\frac{ 2048}{2049},\frac{ 4096}{4097}$$
$N=3$
$$\frac{6}{11},\frac{ 36}{49},\frac{ 216}{251},\frac{ 1296}{1393},0.96584, 0.98329, 0.9918, 0.99596, 0.998, 0.99901, 0.99951, 0.99975$$
(rounded to 5 digits)
It looks that $m=10$ is the  No. of "all-black" draws sufficient to state at $99.9%$ that the urn contains just one ball,
for whichever $N \ge 2$.
In fact
$$
\left\{ \matrix{
  {1 \over {H_{\,N - 1} ^{\left( m \right)} }} > {1 \over {H_{\,N} ^{\left( m \right)} }} \hfill \cr 
  \mathop {\lim }\limits_{N\, \to \,\infty } H_{\,N} ^{\left( m \right)}  = \zeta (m)\quad \quad  \hfill \cr 
  {1 \over {\zeta (10)}} = {{93555} \over {\pi ^{\,10} }} > 0.999 \hfill \cr}  \right.
$$
