Does the double-hash H(H(x)) have greater collision probability than H(x)? Let $H$ be a collision resistant hash function and $P_c[H](S)$ the collision probability about a sample set $S$ of input elements (eg. random numbers). It increases with double hash?  That is, $P_c[H\circ H](S) \ge  P_c[H](S)$ &nbsp; ?

Well... we need a precise $P_c$ definition.  Perhaps the problem here is also the choice of a good probability definition (see notes below).



NOTES

Answer's application: the answer is important in the context of checksums used for digital preservation integrity of public files. 

Imagine the checksum (the hash digest $H(x)$) of a PDF article of PubMed Central or arxiv.org... And that we need to ensure the   integrity, with no "PDF atack" or minimizing it, in nowadays and in a far future, ~20 years.

If the answer is "YES there are a little increase", the recommendation will be "please don't use double-hashing as standard for checksum for PMC or  arxivOrg".

General probaility

The general collision probability is defined as


  Given $k$ randomly generated $x$ values, where each $x$ is a non-negative integer less than $N$, what is the probability that at least two of them are equal?


and depends only of the $|S|$, so $P_c[H\circ H] = P_c[H]$... There are a hypothesis of "perfect hash function", but I want to metric the imperfections.

If there are no error on my interpretations, I need other probability definition.

Counting the collisions

Suppose a kind of "collision tax", $\frac{N_c}{|S|}$, based on the number of collisions $N_c$ occured with a specific sample. So, for many different sample sets $S_i$ with same number of elements, $k=|S_1|=|S_2|=... =|S_i|$. Suppose a set $K$ of all sets $S_i$ and  $P_c[H](K)$ as an average of this collision rate.  

This kind of probability seems better to express the problem.

... Bloom filter efficiency

Another metric, instead a probability. Perhaps using 
Bloom filter theory as reference. A kind of efficiency-benchmark of two Bloom filters, one with $H$ other with $H\circ H$.
I'm not sure what the question here is, but obviously applying the hash function twice can never decrease the number/probability of collision as all collisions in the first invocation are maintained. 

However if H is collision free( a permutation as opposed to a random function) doubling will not cause any more collision it will remain collision free. So we see the number of collision does not strictly increase. 

For a PRF, the number of collision does clearly increase, this is one reason why when we iteratively hash passwords in for key derivation we mix in the input again every round and not simply hash calculate $H^n(salt || pass)$ 

If you look at the structure of a function you will see several cycles and threads(trees) leading into these cycles:


The cycles do not generate more collisions when doubled. Each thread meeting a cycle contributes one collision, If All the threads are of length 1 no more collisions are added when applying twice in any other case the number of collisions increases.