Object $A$ can fall on one of $k$ individuals and object $B$ can fall on one of $m$ individuals. If I am a member of both group $k$ and $m$, then my payoff is:
$$ \frac{1}{km}X+\frac{1}{k}\left(1-\frac{1}{m} \right)Y+\frac{1}{m}\left(1-\frac{1}{k} \right)Z+\left(1-\frac{1}{k} \right)\left(1-\frac{1}{m} \right)U. $$
However, my payoff in the $4^{\rm th}$ case, $U$ depends on whether one person got both objects or if both objects went to separate people. So if I am the only one who is a part of both groups, then trivially there is $0$ percent chance that both objects went to the same person.
So my question is, what is the probability that the object goes to two persons conditional on the fact that I didn't win either of the two objects? Lets just say that my payoff if two objects go to the same person is $T$ and my payoff if they go to separate persons is $P$, therefore clearly $U$ is some function of $T$ and $P$.
Assume there are $r$ people apart from me who are in both groups, obviously, $r$ must be smaller than both $m$ and $k$.
Just to be clear, what is the probability that the objects fall on the same person conditional on the fact they didn't fall on me?