Calculate the probability that red balls have a majority in at least one of the buckets There are total of $n$ balls, $r$ of which are red and the rest are black.
When they are divided equally and randomly among $k$ buckets, I'm trying to find the probability that at least one bucket has a majority of red balls, assuming that $r$ is such that it is possible to get a majority in a bucket, i.e. 
$r > \frac{n}{2k}$. 
Example:
There are $100$ balls, $7$ of which are red and the rest are black.
When they are divided into $10$ equal buckets, I'm trying to calculate the probability that at least one bucket have a MAJORITY of red balls, i.e. at least one bucket should have minimum of $6$ red balls.
I have tried different ways of breaking down the problem into combinations and probability problem.
I tried calculating the combination the red balls can be arranged and then calculating the probability, but I'm not sure if I am going in the right direction.
Any help is appreciated.
 A: 
This answer assumes that balls of the same color are indistinguishable, and that buckets are distinguishable.

Let $n=ak$, so that each bucket will receive $a$ balls, and $r>a/2$.
Each distribution of balls among the $k$ buckets corresponds to a solution of $r_1+r_2+\dots+r_k = r$ with $0\leqslant r_i \leqslant a$ for all $i$.
An involved application of stars and bars and inclusion-exclusion can provide the number $S$ of such solutions, but I find it much simpler to use generating functions.
Indeed, we'll have that
\begin{align}S
&= \left[x^r\right]{\left(1+x+x^2+\dots+x^a\right)}^k
\\&=
\left[x^r\right]{\left(\frac{1-x^{a+1}}{1-x}\right)}^k
\end{align}
Now, to find the probability $p$ we're interested in, we need only calculate the number $R$ of cases for which some bucket $i$ has $r_i>a/2$.
Then, $p=R/S$.
For this, we will unfortunately resort to inclusion-exclusion.
Let $A_i$ be the set of ball-bucket distributions for which $r_i > a_2$.
Then, by inclusion exclusion:
\begin{align}
\left |\bigcup_{i=1}^k\,A_i\right|
&=
\sum_{j=1}^k\,(-1)^{j+1}\cdot\left(\sum_{1\leqslant i_1\leqslant \dots \leqslant i_j\leqslant k}\,\left|A_{i_1}\cap\dots\cap A_{i_j}\right|\right)
\\&=
\sum_{j=1}^k\,(-1)^{j+1}\cdot\binom{k}j\,\alpha_j \tag{$*$}
\end{align}
where $\alpha_j = \left|A_{i_1}\cap\dots\cap A_{i_j}\right|$ and the equality in $(*)$ is justified by symmetry.
Thus, we need only calculate the coefficients $\alpha_j$.
That would be the number of solutions to $r_1+r_2+\dots+r_k = r$ with $a/2 <r_1,\dots, r_j \leqslant a$ and $0\leqslant r_{j+1}, \dots, r_k \leqslant a$.
Once again, we use generating functions.
We'll have
\begin{align}\alpha_j
&= \left[x^r\right]
{\left(x^{\left\lceil a/2\right\rceil}+x^{\left\lceil a/2\right\rceil + 1}+\dots+x^a\right)}^j
\cdot{\left(1+x+x^2+\dots+x^a\right)}^{k-j}
\\&= \left[x^r\right]
x^{\,j\cdot\left\lceil a/2\right\rceil}{\left(1+x+\dots+x^{a-\left\lceil a/2\right\rceil}\right)}^j
\cdot{\left(\frac{1-x^{a+1}}{1-x}\right)}^{k-j}
\\&= \left[x^{r-j\cdot\left\lceil a/2\right\rceil}\right]
{\left(\frac{1-x^{\left\lfloor a/2\right\rfloor +1}}{1-x}\right)}^j
\cdot{\left(\frac{1-x^{a+1}}{1-x}\right)}^{k-j}
\\&= \left[x^{r-j\cdot\left\lceil a/2\right\rceil}\right]
\frac{{\left(1-x^{\left\lfloor a/2\right\rfloor +1}\right)}^j\,{\left(1-x^{a+1}\right)}^{k-j}}{{(1-x)}^k}
\end{align}
Of course, if $j\cdot\left\lceil a/2\right\rceil > r$, then $\alpha_j = 0$.
We need hence only consider $j\leqslant \big\lfloor r/\left\lceil a/2\right\rceil\big\rfloor $.
This gives us a somewhat ugly but nonetheless functional formula for $p$:
$$
p=
\frac{
\displaystyle\sum_{j=1}^{\big\lfloor r/\left\lceil a/2\right\rceil\big\rfloor}\,
(-1)^{j+1}\cdot\binom{k}j\,\left[x^{r-j\cdot\left\lceil a/2\right\rceil}\right]
\frac{{\left(1-x^{\left\lfloor a/2\right\rfloor +1}\right)}^j\,{\left(1-x^{a+1}\right)}^{k-j}}{{(1-x)}^k}
}{
\displaystyle\left[x^r\right]{\left(\frac{1-x^{a+1}}{1-x}\right)}^k
}
$$

Plugging the values of $n=100$, $r=7$ and $k=10$, so that $a=10$, we get $\left\lceil a/2\right\rceil = \left\lfloor a/2\right\rfloor = 5$ and $\big\lfloor r/\left\lceil a/2\right\rceil\big\rfloor = \lfloor 7/5 \rfloor =1$.
It follows that $j=1$ is the only index in the summation and hence
\begin{align}
p&=
\frac{\displaystyle
10\cdot\left[x^2\right]
\frac{{\left(1-x^{6}\right)}\,{\left(1-x^{11}\right)}^{9}}{{(1-x)}^{10}}
}{
\displaystyle\left[x^{7}\right]{\left(\frac{1-x^{11}}{1-x}\right)}^{10}
}
\\&=
\frac{\displaystyle
10\cdot 55
}{
11440
}\simeq 4.81\%
\end{align}
A: What about considering one bucket as a binomial distribution, so that for the red balls $R - Bin(s, r, \frac{1}{k})$ for $s\in \{0,..,r\}$ and black balls $B - Bin (t, n-r, \frac{1}{k})$ for $t \in \{0,..,n-r\}$
The the probability that the number of red balls is greater than the number of black balls for given $t$ is $P(s>t)=\sum_{s=t+1}^{r} \; ^rC_s \; \left( \frac{1}{k} \right)^s \left(1-\frac{1}{k}\right)^{r-s}$
And the complete probability measured over all t is:
$P(red > black) = {\sum_{t=0}^{n-r}} \; ^{n-r}C_t \; \left( \frac{1}{k} \right)^t \left(1-\frac{1}{k}\right)^{n-r-t} P(s>t) $
Sorry I don't know how to extend this multinomially to all buckets at the mom.
