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?

You found the solution for the question :
Let $p$, the probability of not having an unusual :


*

*Either you get a weapon

*or two crates without unusual


$$p=0.35+0.55*p^2$$
This polynomial have one root in $[0,1]$, hence 
$$p=\frac{10-\sqrt{23}}{11}\approx 0.4731$$

Your question can be answered the same way :
Let $q$, the probability of having an infinite number of crates :


*

*You open two crates and at least one can lead to an infinite number of crates. This is the complementary of both crates not having infinite number of crates inside.
$$q=0.55*(1-(1-q)^2)=0.55(-q^2+2q)$$
$$q(0.55q-0.1)=0 $$
That leads to two solutions : $q=0$ or $q=\frac{2}{11}$. But as you explained $q>0$ because the event "having infinite number of crates" is likely to happen if you have enough crates. Hence
$$q=\frac{2}{11}$$
Note that if you use $0.5$ or $0.45$ instead of $0.55$ the only valid solution is $0$.
