Picking balls from a bag until one colour is missing A bag contains balls of three different colours namely $A，B，C$. The number of each types of balls is $n$.

Assume you pick 3 balls randomly from the bag every round, without replacement. And you stop if you have picked up all balls of one colour.  What's the probability distribution of number of the rounds you spend?

I think we can identify this with the same question except you pick one ball a time，but I still cannot handle it.
My answer is not as detailed as Marko's and it does not give the probability of rounds but probability of draws before a ball of one colour to become extinct first.  But a simple enumeration of the combination of draws could lead to the below and all X's are the number of draws that yield the combination. Now let us denote each of them with sub-scripts.

$AAA -X_1$

$AAB -X_2$

$AAC -X_3$

$ABB -X_4$

$ACC -X_5$

$ABC -X_6$

$BBB -X_7$

$BBC -X_8$

$BCC -X_9$

$CCC -X_10$

Let us analyse Ball A

$3X_1+2X_2+2X_3+X_4+X_5+X_6 = n$

If you notice it is same for B and C too.

Let us analyze for Ball B

$3X_7+2X_4+2X_8+X_2+X_6+X_9 = n$

So is for Ball C

Summation of the solution of all three will make sure that one of those balls will get to run out the first.

Now use generating function to find the solution for any one of the equations. And the coefficient of $X^n$ is the number of ways you can draw all n balls of any type first.
Let us assume $n=20$

Thus
$(1+x^3+x^6+x^9+x^{12}+x^{15}+x^{18}+x^{21})(1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}+x^{16}+x^{18}+x^{20})^2(\sum_{i=0}^{20}x^i)^3$

The coefficient of $x^{20}$ in the above generating function is $6602$.

Total number of ways all such draws could result in a sum of 60 
$3(X_1+X_2+X_3+X_4+X_5+X_6 +X_7+X_8+X_9+X_{10})= 60$

$(X_1+X_2+X_3+X_4+X_5+X_6 +X_7+X_8+X_9+X_{10})= 20$
and the coefficient of $x^{20}$ in the below generation function

$(\sum_{i=0}^{20}x^i)^{10}$ is $10015005$.

Thus the required probability = $\frac{6602\times 3}{10015005}$

different n, you will have to go through this procedure to obtain the probability.

I hope I have given you a simple solution.

Goodluck.
Here  is  a  basic  contribution,   working  with  a  closely  related
question. We solve the problem where  we have $j$ instances of each of
$n$ types of  coupons and draw without replacement until  we have seen
all $j$ coupons  of some type.  Using the notation  from the following
MSE    link    we
introduce the marked generating function

$$\left(\sum_{k=0}^{j-2} 
\frac{j!}{(j-k)!} \frac{z^k}{k!} + j w z^{j-1}\right)^n.$$

The coefficient on $[z^m]$  here represents distributions of sequences
of $m$  draws from the $n$  types according to probability,  where the
ones that occur  $j-1$ times have been marked. Each  of the latter may
be augmented to a  complete set of some color where  the weight is one
because $j-1$  coupons have already  been drawn.  As we only  need the
count we differentiate with respect to $w$ and set $w=1$, getting

$$n\times \left(\sum_{k=0}^{j-1} 
\frac{j!}{(j-k)!} \frac{z^k}{k!}\right)^{n-1}
\times j z^{j-1}.$$

With  the  method  from  the  linked  post  we  thus  obtain  for  the
probability

$$P[T = m] = \frac{1}{m!} {nj\choose m}^{-1}
(m-1)! [z^{m-1}] n j z^{j-1} \left(\sum_{k=0}^{j-1} 
\frac{j!}{(j-k)!} \frac{z^k}{k!}\right)^{n-1}
\\ = \frac{1}{m!} {nj\choose m}^{-1} \times n \times j \times
(m-1)! [z^{m-1}] z^{j-1} (-z^j + (1+z)^j)^{n-1}.$$

Extracting the coefficient we find

$${nj-1\choose m-1}^{-1} [z^{m-j}] 
\sum_{q=0}^{n-1} {n-1\choose q} (-1)^{n-1-q} z^{j(n-1-q)}
(1+z)^{qj}
\\ = {nj-1\choose m-1}^{-1} 
\sum_{q=0}^{n-1} [z^{m-j(n-q)}] {n-1\choose q} (-1)^{n-1-q}
(1+z)^{qj}
\\ = {nj-1\choose m-1}^{-1} 
\sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{qj\choose m-j(n-q)}
\\ = {nj-1\choose m-1}^{-1} 
\sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{qj\choose nj-m}.$$

Observe that

$${qj\choose nj-m} {nj-1\choose m-1}^{-1} 
= \frac{(qj)! (m-1)! } {(nj-1)! (m-(n-q)j)! }
\\ = {nj-1\choose qj}^{-1} {m-1 \choose m-(n-q)j}.$$

We record for the probabilities the formula

$$\bbox[5px,border:2px solid #00A000]{
P[T=m] = \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj}^{-1} {m-1 \choose (n-q)j-1}.}$$

We now verify that this is a probability distribution. 
This requires the value of

$$\sum_{m=j}^{n(j-1)+1} {m-1 \choose (n-q)j-1}
= \sum_{m=j-1}^{n(j-1)} {m\choose (n-q)j-1}
\\ = [z^{(n-q)j-1}] \sum_{m=j-1}^{n(j-1)} (1+z)^m
= [z^{(n-q)j}] ( (1+z)^{n(j-1)+1} - (1+z)^{j-1} ).$$

With $0\le  q\le n-1$ the second  term does not contribute  and we may
continue with

$$ \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj}^{-1} {n(j-1)+1\choose (n-q)j}
\\ = \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj}^{-1} {nj+1-n\choose qj+1-n}
\\ = \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q} \frac{qj}{nj-qj}
{nj-1\choose qj-1}^{-1} {nj+1-n\choose qj+1-n}
\\ = \sum_{q=1}^{n-1}  {n-1\choose q-1} (-1)^{n-1-q}
{nj-1\choose qj-1}^{-1} {nj+1-n\choose qj+1-n}.$$

Observe once more that

$${nj-1\choose qj-1}^{-1} {nj+1-n\choose qj+1-n}
= \frac{(nj+1-n)! \times (qj-1)!}{(nj-1)! \times (qj+1-n)!}
\\ = {nj-1\choose n-2}^{-1} {qj-1\choose n-2}.$$

We thus find for the sum of the probabilities

$${nj-1\choose n-2}^{-1} 
\sum_{q=1}^{n-1}  {n-1\choose q-1} (-1)^{n-1-q}
{qj-1\choose n-2}
\\ = {nj-1\choose n-2}^{-1} 
\sum_{q=0}^{n-2}  {n-1\choose q} (-1)^{n-q}
{qj+j-1\choose n-2}
\\ =  1 + \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-q}
{qj+j-1\choose n-2}.$$

The sum vanishes, as in

$$\sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-q}
[z^{n-2}] (1+z)^{qj+j-1}
\\ = [z^{n-2}] (1+z)^{j-1}
\sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-q} (1+z)^{qj}
\\ = [z^{n-2}] (1+z)^{j-1} (1-(1+z)^j)^{n-1},$$

but $(1-(1+z)^j)^{n-1} = (-1)^{n-1} j^{n-1} z^{n-1} + \cdots$ and
there is no contribution.  This confirms it being a probability
distribution. 

Continuing with the expectation we require the value of

$$\sum_{m=j}^{n(j-1)+1} m {m-1 \choose (n-q)j-1}
= (n-q)j \sum_{m=j}^{n(j-1)+1} {m \choose (n-q)j}
\\ = (n-q)j [z^{(n-q)j}] \sum_{m=j}^{n(j-1)+1} (1+z)^m
\\ = (n-q)j [z^{(n-q)j+1}] ((1+z)^{n(j-1)+2} - (1+z)^j)
= (n-q)j {n(j-1)+2\choose (n-q)j+1}.$$

The second term did not contribute since we have $(n-q)j+1\gt j.$
We thus have for the expectation

$$j \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj}^{-1} (n-q) {n(j-1)+2\choose (n-q)j+1}
\\ = j \sum_{q=0}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj}^{-1} (n-q) {nj+2-n\choose qj+1-n}
\\ = j \sum_{q=1}^{n-1}  {n-1\choose q} (-1)^{n-1-q}
\frac{qj}{nj-qj} {nj-1\choose qj-1}^{-1} (n-q) 
\\ \times \frac{nj+2-n}{nj-qj+1}
{nj+1-n\choose qj+1-n}
\\ = j (nj+2-n) \sum_{q=1}^{n-1} q {n-1\choose q} (-1)^{n-1-q}
{nj-1\choose qj-1}^{-1}
\frac{1}{nj-qj+1}
{nj+1-n\choose qj+1-n}.$$

Re-using the earlier factorization we get

$$j (nj+2-n) {nj-1\choose n-2}^{-1} 
\sum_{q=1}^{n-1} q {n-1\choose q} (-1)^{n-1-q}
\frac{1}{nj-qj+1}
{qj-1\choose n-2}
\\ = j (nj+2-n) (n-1) {nj-1\choose n-2}^{-1} 
\sum_{q=1}^{n-1} {n-2\choose q-1} (-1)^{n-1-q}
\frac{1}{nj-qj+1}
{qj-1\choose n-2}
\\ = j^2 n (nj+1) {nj+1\choose n-1}^{-1} 
\\ \times
\sum_{q=0}^{n-2} {n-2\choose q} (-1)^{n-q}
\frac{1}{nj-qj-j+1}
{qj+j-1\choose n-2}.$$

Working with the sum term we have

$${n-2\choose q} (-1)^{n-q}
\frac{1}{nj-qj-j+1}
{qj+j-1\choose n-2}
\\ = \mathrm{Res}_{z=q}
\frac{1}{nj-j+1-zj}
\prod_{p=0}^{n-3} (zj+j-1-p)
\prod_{p=0}^{n-2} \frac{1}{z-p}.$$

Now since $\lim_{R\to\infty} 2\pi R  \times R^{n-2}/R/R^{n-1} = 0$ and
residues sum  to zero we may  evaluate this by taking  the negative of
the residue at $z=n-1+1/j$. This is the computation:

$$-\mathrm{Res}_{z=n-1+1/j}
\frac{1}{nj-j+1-zj}
\prod_{p=0}^{n-3} (zj+j-1-p)
\prod_{p=0}^{n-2} \frac{1}{z-p}
\\ = \frac{1}{j} \mathrm{Res}_{z=n-1+1/j}
\frac{1}{z-(n-1+1/j)}
\prod_{p=0}^{n-3} (zj+j-1-p)
\prod_{p=0}^{n-2} \frac{1}{z-p}
\\ = \frac{1}{j}
\prod_{p=0}^{n-3} (nj-p)
\prod_{p=0}^{n-2} \frac{1}{n-1+1/j-p}
\\ = \frac{1}{j} \times (n-2)! \times {nj\choose n-2} \times
j^{n-1} \prod_{p=0}^{n-2} \frac{1}{nj-j+1-pj}.$$

With

$${nj\choose n-2} {nj+1\choose n-1}^{-1}
= \frac{n-1}{nj+1}$$ we finally have the closed form

$$\bbox[5px,border:2px solid #00A000]{
E[T] = n! \times
j^n \times \prod_{p=0}^{n-2} \frac{1}{nj-j+1-pj}.}$$

To see what the asymptotics are we use the alternate form

$$\bbox[5px,border:2px solid #00A000]{
E[T] = n \times j \times
\frac{\Gamma(n)\Gamma(1+1/j)}{\Gamma(n+1/j)}.}$$

Keeping $j$ fixed and letting $n$ go to infinity yields the asymptotic

$$n\times j\times \Gamma(1+1/j) \times n^{-1/j}
= n^{1-1/j} \times j\times \Gamma(1+1/j).$$

There is  an enumeration routine  that may  be compared to  the closed
forms both of which were implemented in the following Maple code.


V :=
proc(n, j)
    option remember;
    local L, recurse, results;

    results := 0;

    recurse :=
    proc(LL, sofar, prob)
    local choice, cprob;

        if numboccur(LL, 0) &gt; 0 then
            results := results + prob*u^nops(sofar);
            return;
        fi;

        for choice to nops(LL) do
            cprob := LL[choice]/(n*j-nops(sofar));

            recurse([seq(LL[q], q=1..choice-1),
                     LL[choice]-1,
                     seq(LL[q], q=choice+1..nops(LL))],
                    [op(sofar), choice],
                    prob*cprob);
        od;
    end;

    L := [seq(j, q=1..n)];
    recurse(L, [], 1);

    results;
end;

P := (n, j, m) -&gt; n*j/binomial(n*j, m)/m
*coeftayl(z^(j-1)*(-z^j+(1+z)^j)^(n-1), z=0, m-1);
PGF := (n, j) -&gt; add(P(n,j,m)*u^m, m=j..(j-1)*n+1);

EX := (n, j) -&gt; add(P(n,j,m)*m, m=j..(j-1)*n+1);

P2 := (n, j, m) -&gt;
add(binomial(n-1,q)*(-1)^(n-1-q)/binomial(n*j-1, q*j)
    *binomial(m-1, m-(n-q)*j), q=0..n-1);

PGF2 := (n, j) -&gt; add(P2(n,j,m)*u^m, m=j..(j-1)*n+1);

EX2 := (n, j) -&gt; n!*j^n*mul(1/(n*j-j+1-p*j), p=0..n-2);

EXGAMMA := (n, j) -&gt; n*j*GAMMA(n)*GAMMA(1+1/j)/GAMMA(n+1/j);
EXASYMPT := (n, j) -&gt; n^(1-1/j)*j*GAMMA(1+1/j);


With the calculation  that was presented here we want  to make sure we
have the  correct interpretation of  the problem from the  start.  The
following  basic program  will do  this by  computing the  expectation
through  simulation. Consult  for the  details of  the scenario  under
investigation.  The output is in fine agreement with the data i.e. the
closed form from above.


&#35;include &lt;stdlib.h&gt;
&#35;include &lt;stdio.h&gt;
&#35;include &lt;assert.h&gt;
&#35;include &lt;time.h&gt;
&#35;include &lt;string.h&gt;

int main(int argc, char **argv)
{
  int n = 6 , j = 3, trials = 1000; 

  if(argc &gt;= 2){
    n = atoi(argv[1]);
  }

  if(argc &gt;= 3){
    j = atoi(argv[2]);
  }

  if(argc &gt;= 4){
    trials = atoi(argv[3]);
  }

  assert(1 &lt;= n);
  assert(1 &lt;= j);
  assert(1 &lt;= trials);

  srand48(time(NULL));
  long long data = 0;

  for(int tind = 0; tind &lt; trials; tind++){
    int src[n*j];

    for(int cind = 0; cind &lt; n*j; cind++)
      src[cind] = cind/j;

    int done = 0; int steps = 0; 
    int dist[n];

    for(int cind = 0; cind &lt; n; cind++)
      dist[cind] = 0;


    while(!done){
      int cpidx = drand48() * (double)(n*j-steps);
      int coupon = src[cpidx];

      for(int cind=cpidx; cind &lt; n*j-steps-1; cind++)
        src[cind] = src[cind+1];

      steps++;
      dist[coupon]++;

      if(dist[coupon] == j)
        done = 1;
    }

    data += steps;
  }

  long double expt = (long double)data/(long double)trials;
  printf("[n = %d, j = %d, trials = %d]: %Le\n", 
         n, j, trials, expt);

  exit(0);
}