Krdytkyu

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?

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
$$
left{ matrix{
P(a wedge A;left| {;m} right.) = {1 over {1^{,m} }}P(A;left| {;m} right.) = {1 over N} hfill cr
P(b wedge A;left| {;m} right.) = 0 hfill cr
P(a wedge B;left| {;m} right.) = sumlimits_{2, le ,n, le ,N} {{1 over {n^{,m} }}P(n;left| {;m} right.)}
= {1 over N}sumlimits_{2, le ,n, le ,N} {{1 over {n^{,m} }}} = {1 over N}left( {H_{,N}^{,left( m right)} - 1} right) hfill cr
P(b wedge B;left| {;m} right.) = sumlimits_{2, le ,n, le ,N} {left( {1 - {1 over {n^{,m} }}} right)P(n;left| {;m} right.)}
= 1 - {1 over N}H_{,N}^{,left( m right)} hfill cr
P(a;left| {;m} right.) = {1 over N}H_{,N}^{,left( m right)} quad quad
P(b;left| {;m} right.) = 1 - {1 over N}H_{,N}^{,left( m right)} hfill cr} right.
$$
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
$$
left{ matrix{
{1 over {H_{,N - 1} ^{left( m right)} }} > {1 over {H_{,N} ^{left( m right)} }} hfill cr
mathop {lim }limits_{N, to ,infty } H_{,N} ^{left( m right)} = zeta (m)quad quad hfill cr
{1 over {zeta (10)}} = {{93555} over {pi ^{,10} }} > 0.999 hfill cr} right.
$$