Correct distribution for combinatorial balls in two bags problem I have two bags that each have two sub-bags into which I can place balls. Therefore, for an incoming set of balls, there are 4 possible destinations that each ball could land. I want to create a probability mass function for this system such that each decision is binomially distributed. I want the distribution of the numbers of balls in each of the inner bags, $f(k,l,h,m)$, where there are $k$ in inner-bag 1 (and so on).
Consider a set of balls $i+j$ such that $i$ land in the first outer-bag and $j$ land in the second outer-bag. The binomial parameters $a$ and $x$ control the distribution of outer and inner placement, respectively.
$$
f(k,l,h,m)=N \cdot \left[\binom{i+j}{i}a^i(1-a)^j\binom{i}{k}x^k(1-x)^{i-k} + \binom{i+j}{j}a^j(1-a)^i\binom{j}{l}x^l(1-x)^{j-l} \right]
$$
where $h=i-k$ and $m=j-l$ and $N$ is a normalisation constant. I am struggling to find the value of $N$ for this condition.
$$
\sum_{k,l,h,m} f(k,l,h,m)  = 1
$$
Any hints would be very appreciated.
 A: For whatever reason I'm struggling to arrive to the same distribution you provide here, so I will give you two answers: how to normalise your function and my take on the problem.
First, your normalisation. I am guessing we already know the exact values of $i$ and $j$ in this case from the way you formulate your question. In that case, we don't need to sum over $h=i-k$ and $m=j-l$, since these two numbers vary accordingly with $k$ and $l$. All we need to do then is just add:
$$\sum_{l=0}^j\sum_{k=0}^i \left[\binom{i+j}{i}a^i(1-a)^j\binom{i}{k}x^k(1-x)^{i-k} + \binom{i+j}{j}a^j(1-a)^i\binom{j}{l}x^l(1-x)^{j-l} \right]=$$ $$=(j+1)\binom{i+j}{i}\sum_{k=0}^i\binom{i}{k}x^k(1-x)^{i-k} + (i+1)\binom{i+j}{j}\sum_{l=0}^j\binom{j}{l}x^l(1-x)^{j-l}=$$$$=(j+1)\binom{i+j}{i}+(i+1)\binom{i+j}{j}=(i+j+2)\binom{i+j}{i}$$
and $N$ would be the inverse of this value.
Now, here's my take. You start with $M$ total balls, $i$ of which go to the first supergroup and $j$ go to the second supergroup, $M=i+j$. This follows an easy binomial distribution (we're supposing $M=i+j$ is fixed): $$p(i,j)=\binom{i+j}{i}a^i(1-a)^j$$
Now we want to know the distribution of the $i$ balls in the first supergroup and the $j$ balls in the second supergroup. We are supposing now that both $i$ and $j$ are known and that the distribution in the first bag is independent in respect to the distribution of the second bag. We have thus a conditioned probability $p(k,l|i,j)$ where the notation is self-explanatory. The aforementioned independence gives us $$p(k,l|i,j)=\left[\binom{i}{k}x^k(1-x)^{i-k}\right]\cdot\left[\binom{j}{l}x^l(1-x)^{j-l}\right].$$
If we now want to know the total probability mass function we just have to apply the definition of conditional probability $$p(k,l|i,j)=p(i,j)p(i,j,k,l) \Longleftrightarrow p(i,j,k,l)=p(k,l,h=i-k,m=j-l)=\frac{p(k,l|i,j)}{p(i,j)}$$
and everything is normalised.
A: I have no faith in your setup and would choose for another one.
For convenience let $b=1-a$ and $y=1-x$ and for a fixed $n$ that denotes the total numbers of balls and satisfies $k+l+h+m=n$ define:$$P(K=k,L=l,H=h,M=m)=\frac{n!}{k!l!h!m!}(ax)^k(bx)^l(ay)^h(by)^m$$where $k,l,h,m$ are nonnegative integers.
We are dealing here with a "multinomial distribution".
Then $K+H$ has binomial distribution with parameters $n$ and $a$.
For fixed nonnegative integers $i,j$ with $i+j=n$ under condition $K+H=i$ random variable $K$ has binomial distribution with parameters $i$ and $x$ and random variable $L$ has binomial distribution with parameters $j$ and $y$.
