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Lets say we construct a list as follows. It has f(n) 1's, and f(2) 2's and f(3) number of three's in it etc.

Let L(n,f) be this list, so if f(n)=n^2 we get L(3,n^2)=(1,2,2,2,2,3,3,3,3,3,3,3,3,3)

If f(n) is such that f(m)>(sum f(n) n=1 to n=m-1). Then the probability to pick the number n from the list L(n,f) is above 1/2.

So if we take that f(n), but reverse is such that the smallest number always has the most appearances in the list, we make it into g(f) (we can do this since it was arbitrary), then the probability to pick the nmuber "1" from L(n,g(f)) is above 1/2 for all n, but then we let n->inf, then L(inf,g(f)) contains all the distinc integers, but the probability if you pick a random number from the list that it is 1, is still 1/2, is it possible ?

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    Rereading this after a while, I do not understand how the accepted answer addresses the OP's concern. The crux of the matter seems to be that the reasoning "So if we take that f(n), but reverse is such that the smallest number always has the most appearances in the list, we make it into g(f) (we can do this since it was arbitrary), then the probability to pick the nmuber "1" from L(n,g(f)) is above 1/2 for all n" is wrong. No list h can yield L(n,h) above 1/2 for all n.2013-08-11

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By integer in your comment, it seems you mean positive integer. A probability distribution that meets your request would be $P(n)=\frac{1}{2^n}$. It sums to 1 and has positive value for all $n \gt 0$ with $P(1) \gt P(2) \gt \ldots$ If you really want all integers it can easily be reworked, but you would have to specify which ones you want more probable.

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    Just put the integers in your favorite order. So, if you like $\{0,1,-1,2,-2,\ldots\}$ you have P(0)=1/2, P(1)=1/4, P(-1)=1/8, etc.2011-02-19