hash collision chance using a 5 chars hash made of a-z A-Z and 0-9 (62 chars total) What is the chance for collision or resistance of a hash made out of 5 chars randomly picked from this sets a-z A-Z and 0-9 (62 chars total).

Also, what is total amount of combinations without collision that I can get, AFAIK :

$62^5 = 916.132.832$

Thank you.
There are indeed $62^5 = 916\,132\,832$ possible hashes in that syntax.  If they are uniformly distributed, then the standard birthday paradox applies to the probability of a collision in $n$ independent choices of $k$ possible hashes.  For $n &lt; \sqrt{k}/4$, the probability of collision is between $(n - 1)^2/(4k)$ and $n^2/k$.  For example, among 10 000 hashes, the probability of a collision is at least $$\frac{9\,999^2}{4\cdot916\,132\,832} &gt; 2.7\%,$$ and at most $$\frac{10\,000^2}{916\,132\,832} &lt; 11\%.$$

Of course, if these are hashes from a fixed public hash function like SHAKE128-30 (add two more characters to your alphabet like the base64 + and - to make the numbers work out), it is easy for an adversary to spend the energy to compute a few tens of thousands of hashes offline without interacting with you in order to find a collision with high probability.  The lower bound on the collision probability gives a guarantee on the adversary's chance of success after a certain amount of work.

So with numbers as small as this, not much can be said about fixed public hash functions.  On the other hand, this analysis remains relevant for a secret hash function, a pseudorandom function family (PRF), like SipHash or KMAC128 or keyed BLAKE2 under a secret key not known to the adversary.