It is not possible to pick a random natural number (out of all natural numbers), such that each number has the same probability.

But is it possible to define a probability distribution over the natural numbers, which is such that larger numbers get a higher probability of being picked?

  • 3
    $\begingroup$ This is vague. If you really mean that $n>m\implies p_n>p_m$, then clearly not, since that would make $\sum p_n$ diverge. Knowing that $\sum p_n=1$ implies that $\lim_{n\to \infty}p_n=0$. $\endgroup$
    – lulu
    Commented May 12 at 21:11
  • $\begingroup$ Can you write an answer @lulu ? $\endgroup$ Commented May 12 at 21:31

1 Answer 1



Suppose $p(2) =1/1000$. Then $p(x) \gt 1/1000$ for $x=3,4,5,...,1003$. But those probabilities would then add up to more than $1.0$, which is impossible for probabilities.

You should be able to easily write this in a more general form for $p(2)= \textrm { any }\epsilon \gt 0$, or if you allow some initial batch of smaller numbers to have a probability of $0$, then by picking some starting number other than $2$ that does have positive probability.

Essentially, the fact that probabilities add, and that you have subsets with an arbitrarily large number of elements, is going to kill any hope for this. Heck, in some sense, what you've asked for is even harder to do than the equal probabilities question, even thought both are impossible.

  • $\begingroup$ +1. Out of curiosity, when you say that "what you've asked for is even harder to do than the equal probabilities question", what did you have in mind? $\endgroup$
    – K. Jiang
    Commented May 13 at 2:34
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    $\begingroup$ @K.Jiang - Well, it's pretty loose, but when you assume equal probabilities, you can still make the same argument, and take enough terms so that the sum exceeds 1. But in our case here, the individual probabilities are larger, so we're going to get an even larger sum - or if you imagine yourself adding together consecutive $p(n)$'s, we might exceed 1 "earlier" than in the equal probability case. $\endgroup$
    – JonathanZ
    Commented May 13 at 3:07
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    $\begingroup$ It's kind of like crushing a soda can with either a 10 lb weight or a 200 lb weight - you may end up with identically crushed cans in the end, but the 200 lb one feels more "crushed". $\endgroup$
    – JonathanZ
    Commented May 13 at 3:11

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