What is the probability that there more rabbits than chickens in each of these three cages? Suppose we have $4$ chickens and $12$ rabbits, and put them into $3$ cages. What is the probability that there more rabbits than chickens in each of these three cages?
I was trying to solve it by using P(more rabbits than chicken) = 1-P(more chicken than rabbits). Since the number of chickens is far less than the number of rabbits and it would be easier to calculate.
Then, the cardinality of the sample space would be $3^{12+4}$. But I don't know how to continue.
 A: Let $C_i$ and $R_i$ be the number of chickens and rabbits in cage $i$.
$$P[C_1=i_1,C_2=i_2]={4\choose{i_1}}{{4-i_1}\choose{i_2}}\left(\frac{1}3 \right)^4$$
$$i_1,i_2,i_1+i_2 \in (0,1,2,3,4)$$
and
$$P[R_1=j_1,R_2=j_2]={12\choose{j_1}}{{12-j_1}\choose{j_2}}\left(\frac{1}3 \right)^{12}$$
$$j_1,j_2,j_1+j_2 \in (0,1,2,3,...,12)$$
Then, note that $12-j_1-j_2>4-i_1-i_2 \implies 7+i_1+i_2-j_1\ge j_2$.
So,
$$P[R_1>C_1,R_2>C_2,12-R_1-R_2>4-C_1-C_2]$$
$$=\sum_{i_1=0}^4 \sum_{i_2=0}^{4-i_1} \sum_{j_1=i_1+1}^{12} \sum_{j_2=i_2+1}^{7+i_1+i_2-j_1}P[C_1=i_1,C_2=i_2]P[R_1=j_1,R_2=j_2]$$
$$=\frac{112706}{177147}\approx 0.636$$
p[i1_, i2_, r1_, r2_] = 
 Binomial[4, i1] Binomial[4 - i1, i2] Binomial[12, j1] Binomial[12 - j1, j2]/
(3^4 3^12)

Sum[p[i1, i2, j1, j2], {i1, 0, 4}, {i2, 0, 4 - i1}, 
{j1, i1 + 1, 12}, {j2, i2 + 1, 7 + i1 + i2 - j1}]
112706/177147

N[%]
0.636229

A: To Reader: As discussed in the comments, and as @Matthew Pilling explained in the comments, my solution probably have an issue (read comments provided by @Matthew Pilling). I cannot delete my answer since it has been already accepted.
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Assumptions: Chickens are indistinguishable as well as Rabbits.
Note:  For the case of distinguishable items, look at the other answer provided by @John L.
Let's say $c_{i}$ and $r_{i}$ are, respectively, the number of chickens and rabits in cage $i-th$. Therefore:
$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  r_{1} + r_{2} + r_{3} = 12  $$
However, we want the number of rabbits to be more than the number of chickens in each cage. So,
$$ r_{1} = c_{1} + a_{1} + 1 $$
$$ r_{2} = c_{2} + a_{2} + 1 $$
$$ r_{3} = c_{3} + a_{3} + 1 $$
Where, $a_{i}$ is the auxiliary variable: $ a_{i} \ge 0 $
Finally, by substitution, we'll have:
$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  a_{1} + a_{2} + a_{3} = 5  $$
So, we can first distribute the chickens and put them in cages, and then for each case, we can choose a different set of $a_{1}$, $a_{2}$, and $a_{3}$.
The final answer is:
$$ {4+3-1 \choose 3-1}{5+3-1 \choose 3-1} = {6 \choose 2}{7 \choose 2}  = 315 $$
Now, you have the number of ways putting chickens and rabits to satisfy condition. Divide it by total cases and you will get the probability.
For total number of cases:
$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  r_{1} + r_{2} + r_{3} = 12  $$
total number of cases (no condition) =
$$ {4+3-1 \choose 3-1}{12+3-1 \choose 3-1} = {6 \choose 2}{14 \choose 2} = 1365 $$
$$ probaility = \frac{315}{1365} $$
