Imagine a group of people. Each person has a number (>0) of different objects with different properties. The complete set of objects is of fix value.
Example: There are K objects and J people. Person X can have a blue, green and yellow object; Person Y can have an orange object.
How do I determine the probability of two people having a common object?
Thank you!
Let us do away with the requirement that each person must own an object. This seems to make the puzzle harder.
In this simpler case: $1-(\frac{J}{2^J})^K$... is what I think. I could be wrong.
Let us say we have K objects ${T_1, \dots T_K}$ and we align them in a row and we have J people and we align in a column. $\matrix{ & T_1 &T_2 \dots T_K \\ \hline P_1 | \\ P_2| \\ \vdots \\ P_J| }$
Now let us write $a_{m,n}$ in $m$th row and the $n$th column. We will write $a_{m,n}=1$ when $P_m$ owns object $T_n$. Otherwise we write $0$.
Now our array is filled with binary entries. And each matrix is possible. That is, we have $2^{JK}$ possible arrays. The array with all zeros would mean that nobody own anything and the array with all $1$s would be where every person owned every object. Lovely.
If I have understood your question: Each array is equally likely.
The question is what's the probability that each column has only $1$ non zero entry. The are $2^J$ possible ways to write column of length $J$ with binary entries and $J$ of these ways have exactly one nonzero entry. Then $\frac{J}{2^J}$ is the probability that it has exactly one nonzero entry. And the probability that there are $K$ such rows would be $(\frac{J}{2^J})^K$.
Now to ask the same question but to demand that we only consider the cases where our array has the property that each row has at least one non zero entry was your original question. Maybe this framework is helpful for attacking that question.