I'm taking 2 medium-length strings (50-70 chars) and hash them using md5 to get results like d2ae4f4919a10958e2c603782f0ec1cc, then recording the first 5 symbols of the hash to provide a (unique) short key. If md5 distribution is (almost) random then would the chances of a collision be 16^5 ([0-9a-f]^5) = 2/1.048.576, or different? Am I extremely lucky to get 2 hashes like d2ae4f4919a10958e2c603782f0ec1cc and d2ae41c1935738ca4a06ba28aad3e555 which start with the same 5 alphanumerals, or there's something else going on? The strings start the same, but it shouldn't matter should it.


1 Answer 1


The number of possible truncated hashes is $d=16^5$. Assuming MD5 is perfectly random, by the birthday bound, your probability of seeing at least one collision is approximately

$$ 1 - \left(\frac{d-1}{d}\right)^{n \text{ choose } 2}, $$ where $n$ is the number of strings you hashed. It doesn't take many to hit 50%. In fact, the above equals 50% when $n= 1206$.

So no, you're not extraordinarily lucky. This was pretty likely.

  • $\begingroup$ so it is 16^5. and i've hashed 10 strings... but ok $\endgroup$
    – ADC
    Jul 31, 2023 at 20:58
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    $\begingroup$ $n$ never reaches $1206$. OP said they were creating hashes in groups of two, and then comparing only the two results hashed in that group. In which case, $n$ never exceeds $2$, and the probability is very close to $1$ that they should not see two hashes collide. $\endgroup$
    – aiootp
    Jul 31, 2023 at 22:46
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    $\begingroup$ @aiootp thx you're right but i guess the birthday calculation is what i needed to know i'm using this tool now bdayprob.com if i need to take more strings in future $\endgroup$
    – ADC
    Jul 31, 2023 at 23:49
  • $\begingroup$ After hashing 10 strings, probability of at least 1 collision is about 1/23302. So what's described is unlikely, but no impossibly unlikely. $\endgroup$
    – fgrieu
    Aug 4, 2023 at 10:16

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