An adversary gives you a set of items whose total size is $x$ (he gets to choose how $x$ is distributed. e.g. there may be $k-1$ items of size $\frac{x}{k}$ and 2 items of size $\frac{x}{2k}$).
The item are now randomly distributed into $2x$ bins (you may assume $2x\in \mathbb{N}$).
What's the probability (i.e. what can we guarantee to achieve for any adversarial choice) that no bin contains items with total size > 1?
EDIT: The adversary can only give us items of sizes $[\frac{1}{r},1]$ for some integer r (hence the number of items is somewhere between $x$ and $r\cdot x$).
(It's easy to see that if we allow the adversary to use infinitesimal items or items bigger than 1, then he can push the probability to 0).
For example, if the adversary chose all items (except for the last one) to be of size $\frac{1}{2} + \epsilon$ , then we have $2x-1$ items that no two of them can fit in a single bin. The last item can fit everywhere. hence the probability is at least (by relaxing to $2x-1$ bins):
$\frac{(2x-1)!}{(2x-1)^{2x-1}} < e^{-(2x-1)} < e^{-2x}$ .
On the other hand, all I know for a general item set is that a "good" coloring exist (easy to see using the first fit algorithm).
Any ideas?
EDIT: the motivation for this question comes from a random coloring of vertex-weighted graphs. If we can assert that the probability is only exponentially small in $x$ (as opposed to $\frac{1}{x!}$ for example), it would give me a better algorithm (by try attempting to randomly color the graph $e^{O(x)}$ times) for a problem I'm working on.