The problem
This is a normal $n$-balls-$m$-bins problem: suppose you have $n$ numbered balls that will be placed at random in $m$ bins. But the twist is: this experiment will be repeated $d$ times.
What is the probability that every ball end up alone in a bin at least once after an experiment?
What I have done so far
I managed to prove a lower bound using simple probability,
The probability of a single ball not being alone in a single experiment, is inverse of it being alone, that is, the other $n-1$ balls missing the same bin. That probability is $1 - (1 - 1/m)^{n-1}$.
The probability of a single ball not being alone in all $d$ experiments is $(1 - (1 - 1/m)^{n-1})^d$. The union of this value for $n$ balls is at most $n(1 - (1 - 1/m)^{n-1})^d$. This is an upper bound because the events aren't independent, so the real probability may be less than that.
The final probability is the inverse of the above, and that is a lower bound (as the other was an upper bound).
While this helps me improving some other lower bound I already had, it seems to me it would be possible to achieve an exact value. Is it right?