Probability of combinatoric explosion. Someone posted a puzzle on a forum.

Now this is fairly easy. Probability of reverse: The crate doesn't contain a unusual = the crate contains a weapon or none of two found crates contains a unusual. This produces a quadratic equation, I discard the solution outside the [0,1] range and substract the other one from 1 to get the answer.
The interesting part is the 5% of the 55% chance for 2 crates. If you open 1000 crates, you.  will end 
up with about 1100 new crates on the average and about no chance ever to reduce this number. It will only grow. You WILL get infinite number of crates if you start off with enough. Still, starting with just one you're quite likely not to get any specials.
Now what is the chance you will get the infinite number of crates (and as result, unusuals) if you start off with only one crate?

This is the extinction problem in branching process
The probabilty of extinction by generation (try) $m$ is given recursively by
$d_m=p_0 + p_2 (d_{m-2})^2$ with $p_0=0.45$ and $p_2=0.55$
The probabily of eventual extinction is obtained by finding a smaller than one root for $d=p_0 + p_2 (d)^2$. In our case: $d=9/11=0.8181...$
Then, the probability of opening infinite crates is $2/11=0.1818...$
