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Consider a quantum random number generator (QRNG) X, which generates integers at random.

(Apparently, due to quantum statistical properties, this type of generation is truly at random, see e.g. "The quantum number generator".)

My question is: what is the probability that X generates a given integer $N \in \mathbb{Z}$ ?

From the mathematical viewpoint there appears to be some obstruction, in that uniform distributions on a countable set do not exist: if $p > 0$ would be the positive probability assigned to each integer, then $\sum_{i \in \mathbb{Z}}p = \infty$, and not $1$.

On the other hand, it might be that $p$ simply should be taken $0$, and that the classical idea of uniform probability is not relevant in this quantum-mechanical context ?

But if it is, then random generation does not imply that integers are chosen with the same probability, which seems a very interesting feature to me ...

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The quantum random number generator described in the article does not generate a random integer $n \in \mathbb{Z}$. It generates a random bit $b \in \{0,1\}$. Although you could (in theory) use any finite number $k$ of these bit generators to generate a random integer in the range $[0,2^k-1]$, this is not the same as generating a random integer $n \in \mathbb{Z}$. Any physical random number generator that claims to generate a random integer $n \in \mathbb{Z}$ with a uniform distribution is clearly bogus because it will almost certainly generate a number that requires more bits than the number of fundamental particles in the observable universe to represent it.

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