Balls and bins - picking drawing one ball from each bin and calculating probability that all balls are black 
Assume we have $3$ bins: bins $L_1, L_2, L_3$, in each bin, there are $3$ balls. Alice is picking one ball from each bin at random. The balls can be either white or black. Alice does not know how many black or white balls are in each bin. She only knows the total number of black balls (denoted as $B$) and white (denoted as $W$) in the bins. Let's assume that the total number of black balls is $6$. We want to compute the probability that all selected balls are black (let's denote it $p$).

Intuitively, the distribution of white and black balls can end up as follows:
Case 1:

*

*$L_1$ has $B = 1 $,

*$L_2$ has $B = 2$,

*$L_3$ has $B = 3$
or, Case 2:

*

*$L_1$ has $B = 2$

*$L_2$ has $B = 2$

*$L_3$ has $B = 2$
Case 3:

*

*$L_1$ has $B = 3$

*$L_2$ has $B = 3$

*$L_3$ has $B = 0$
Thus, the total probability of picking only black balls is
$$ p = (\frac{1}{3} \cdot \frac{2}{3} \cdot 1) + (\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}) + 0 = 0.5185$$
Is the above correct?
Q2: In the second case (when each box has an equal number of black balls) Alice has the highest chances to pick only black balls (which is $0.2962$), so this is her "best-case scenario", while the $p = 0.5185$ is the total probability given that Alice cannot predict a-priori how the black balls will be distributed among the bins. Is the total probability considered an "average case" or should the average-case be computed differently?
EDIT:
Following the guidance in the comments, here is an updated version:
Assume we have $n$ balls and $k$ bins. $B$ balls are black, $W = n - B$ balls are white. We assigned the balls uniformly at random into the bins in such a way that each bin has $x = \frac{n}{k}$ elements ($k|n$). Once the balls are in the bins we pick from each bin one ball uniformly at random. What is the probability that all $k$ selected balls will be black?
Let $F$ be an event that all balls are black, and let $B_i$ denote the number of black balls in the $i'th$ bin.
From the total probability we can define it as follows:
$$
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} Pr(F | B_1 = b_1, \ldots B_k = b_k)\cdot Pr(B_1 = b_1, \ldots B_k = b_k)
$$
The first probability component is
$$ Pr(F | B_1 = b_1, \ldots B_k = b_k) = \prod_{i=1}^k \frac{b_i}{x} $$
The second is
$$
Pr(B_1 = b_1, \ldots B_k = b_k) = \frac{B!W!}{b_1!\cdot b_2! \cdot \ldots \cdot b_k! w_1! \cdot w_2! \cdot \ldots \cdot w_k!} \cdot \frac{(x!)^k}{n!} \\
= \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}
$$
Combining the two together we get
$$
(*) \quad 
Pr(F) = \sum_{b_1 + b_2 + \ldots b_k = B} \left( \prod_{i=1}^k \frac{b_i}{x} \cdot \frac{B!(n-B)!}{\prod_{i=1}^k b_i!\cdot(x-b_i)!}\cdot \frac{(x!)^k}{n!}\right)
$$
Is that correct?
Also, there was a lot of discussion in the comments that the problem which I described is equal to asking: given $n$ balls, where $B$ are black and $W$ are white if I pick $k$ balls what is the probability that all of them are black. This probability would be defined as
$$ (**)\quad 
Pr(F) = \frac{{B \choose k}}{{n \choose k}}
$$
I totally agree with the given examples, that it is the same value. However, this means the formulas $(*)$ and $(**)$ should be equal. I tried to simplify the formula (*) but I couldn't get what is in $(**)$. Is there something wrong with $(*)$?
 A: Assume the bins are filled randomly.
Then the probability of the cases are the following:

*

*$(3,2,1):\quad 6\times \frac{\binom63\binom30}{\binom93}\frac{\binom32\binom31}{\binom63}=\frac9{14}$

*$(2,2,2):\quad 1\times\frac{\binom62\binom31}{\binom93}\frac{\binom42\binom21}{\binom63}=\frac9{28}$

*$(3,3,0):\quad 3\times\frac{\binom63\binom30}{\binom93}\frac{\binom33\binom30}{\binom63}=\frac1{28}$
Correspondingly the probability to pick only the black balls is:
$$
\frac9{14}\frac33\frac23\frac13+\frac9{28}\frac23\frac23\frac23=\frac5{21}.
$$
The same probability one would obtain if the balls were drawn directly from a bin with 9 balls:
$$
\frac69\frac58\frac47=\frac5{21},
$$
which was clear also without any calculation.

EDIT:
In general the number of ways to fill the boxes with $b_1,b_2,\dots b_k$ black balls and $w_1,w_2,\dots w_k$ white balls is:
$$
\frac{b!}{b_1!b_2!\cdots b_k!}\frac{w!}{w_1!w_2!\cdots w_k!}
$$
A: The probability of getting exactly three black balls is given by:
$$
P(\text{3 B}) = \frac{6\choose 3}{9\choose 3} \simeq 0.2380952
$$
and it's the same of course of what Alice had if she extracts the 3 balls directly from a unique bin of 9 balls.

The bin subdivision doesn't affect the probability and it is obvious if you look at the following way. Imagine that Alice extracts one ball at the time, without looking, and "rank" the balls as the order of extraction. The "free" extraction corresponds in selecting balls 1,2,3; the "bins" extraction corresponds in selecting balls 1,4,7 after putting 1,2,3 on the first bin, 4,5,6 on the second and 7,8,9 on the third. It's obvious that the probability of having 3 blacks does not change.
However, if you really insist in starting with bins, things get unnecessarily complicated and tedious, because you have to calculate the probability of each bin composition. So, let's start assuming balls are numbered 1 to 9, and, as above, 1-2-3 belong to the first bin and so on. Next, we think at the "positions" of the white balls (easier to visualize, being less); there are ${9\choose 3} = 84$ different combinations they might assume. Now, we see how many of this combinations belong to the 1/2/3, 2/2/2 or 3/3/0 bin distribution.
Let's start with 2/2/2, which is easier. A white ball must stay in position 1,2 or 3; another in 4,5 or 6 and the last in 7,8 or 9. In total, you have $3\cdot 3\cdot 3 = 27$ combinations and so:
$$
P(2/2/2) = \frac{27}{84} \; .
$$
Now, we consider the 3/3/3 distribution. Now there are just 3 combinations of white balls that allow it, and namely 1-2-3, 4-5-6 and 7-8-9. So we have:
$$
P(3/3/0) = \frac{3}{84} \; .
$$
For the 3/2/1, we see that there are 6 of them, by permuting 1,2 and 3. For the specific 1/2/3 distribution we have that the first white ball must stay either in position 1,2 or 3; the other two must stay between ball 4 and 6, leaving a black ball that can be stay in any three of the remaining spot. So it means that you have $6\cdot 3 \cdot 3 = 54$ combinations for 3/2/1 and so:
$$
P(3/2/1) = \frac{54}{84} \; .
$$
Now we can go ahead and calculate your desired probability:
$$
P(\text{3 B}) = P(2/2/2) \cdot \left(\frac{2}{3}\right)^3 + P(3/2/1)\cdot\frac{1}{3}\cdot\frac{2}{3}
$$
