What is the likelihood of getting the same result from /dev/random?

  • $\begingroup$ Is it 1 in 10000^n where n is the length of bytes (or bits) pulled? $\endgroup$
    – AAllgood
    Apr 5, 2022 at 16:30
  • $\begingroup$ Bits have 2 options per bit, so you would expect 1 in $2^n$ if the output is well distributed (to match any specific n-bit value per extraction, previously generated or not). Generally you'd expect it to be well distributed, but in the end that's an implementation question. $\endgroup$
    – Maarten Bodewes
    Apr 5, 2022 at 16:42
  • 2
    $\begingroup$ "getting the same result" --> same as what? $\endgroup$
    – Mikero
    Apr 5, 2022 at 16:49
  • 1
    $\begingroup$ @Mikero the same twice in a row. $\endgroup$
    – AAllgood
    Apr 5, 2022 at 16:57

1 Answer 1


I'll read the question as: we draw two bitstrings $S$ and $S'$ each of $b$ bit(s) from /dev/{u}random, assumed an ideal random generator (which is it's aim). What's the probability that $S$ and $S'$ are identical, noted $\Pr(S=S')$ ?

Note: if $b$ is a multiple of $8$, $S$ and $S'$ can be thought as bytestrings each of $b/8$ bytes.

A simple way to solve this is to consider that $S'$ was chosen after $S$, and uniformly at random, independently of $S$, among the values $S'$ can get. Since $S'$ is $b$-bit, there are $2^b$ such values, and each has probability $p=1/2^b=2^{-b}$ to be chosen (since the sum of all probabilities must be $1$, and each of the $2^b$ values has the same probability). Since $S$ is $b$-bit, $S$ is one of these $2^b$ values. Therefore $$\Pr(S=S')\,=\,1/2^b\,=\,2^{-b}$$

Note: at first I considered the question so simple that it's best answered by who asked the question, and closed it. Perhaps that was a mistake. I realized there is pedagogical interest in a precise argumentation with standard notation for that elementary question (and, as an aside, that I had made a sign mistake in an initial comment, now deleted).


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