Apparently, the solution to the Card Doubling Paradox is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite limited, given that such a simple distribution is impossible)?

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    $\begingroup$ This may be of help. $\endgroup$ – user3533 Dec 18 '10 at 14:17

A uniform probability over the positive real numbers would not satisfy all three axioms of probability. Specifically, there is a conflict between the second axiom ($P(\Omega) = 1$) and the third axiom (countable additivity).

Check out the wikipedia entry http://en.wikipedia.org/wiki/Axioms_of_probability. I think it should be clear how a uniform probability over the positive reals would cause problems.


If a uniform probability distribution did exist, then for any integer $n$ the probability a real number $x$ satisfies $n \leq x < n+1$ would have to be the same for all $n$. Call this probability $p$.

One of the rules that a probability distribution has to satisfy is that if $\{E_n\}$ are a countably infinite collection of disjoint events then $P(\cup_n E_n) = \sum_n P(E_n)$. Let $E_n$ be the event "the real number $x$ satisfies $n \leq x < n+1$". Since every real number is between some $n$ and $n + 1$, $P(\cup_n E_n) = 1$. On the other hand, $\sum_n P(E_n) = p + p + p + ....$. If $p > 0$, this gives infinity. If $p = 0$ it gives zero. In either case, you'll never add up to $1$. Hence you cannot have a uniform probability distribution over the reals.


For every probability density $f$, $\lim_{x\to \pm \infty} f(x) = 0$ must hold.

It is clear that for a uniform distribution, the density has to be constant on the considered interval.

Combining both requirements, only $g(x) = 0$ remains as a choice for a uniform density on $(-\infty, \infty)$. But $\int_{-\infty}^\infty g(x)dx = 0 \neq 1$, that is $g$ is no proper probability density.


"Uniform probability distribution" always refers to a pre-existent natural measure on a space with a transitive group of "translations", so that any two points resp. neighbourhoods of two points are comparable which each other. The simplest space of this sort would be the set $\mathbb Z$ of integers, and already there the problem you address becomes manifest. Of course it is easy to prove that there doesn't exist a uniform probability distribution on $\mathbb Z$; but you want an intuitive reasoning. While it is easy to come up with a random mechanism that selects any number between $-10^6$ and $10^6$ with equal probability, it is just unimaginable to fathom a mechanism that selects 5 with the same probability as any power of the prime with number $67212345$.


The problem stems from measure theory. Let $V^1, V^2, V^3, ...$ be a sequence of Vitali sets such that the collection $\{V^n : n \in \mathbb{N}\}$ partitions $[ \,0, 1) \,$.

It is then possible to create a mapping $k$ from each set $V^n$ onto $n$ such that $( \,\forall x \in V^n)\,(\,k(x) = n) \,$). An example of this is described in detail here:

Is this fraction undefined? Infinite Probability Question.

At this point, it's easy to see that if $a, b \in \mathbb{N}$ and $x$ is selected uniformly at random from $[ \,0, 1) \,$, then the probability that $x$ will fall in $V^a$ is the same as that of $x$ falling in $V^b$. There is one problem, however, in that these probabilities are undefined.

This is because Vitali sets are non-measurable. If they had a positive real measure, then the sum of their probabilities in this case would be infinite. If they had a $0$ measure, then their sum would be $0$. We are left with no definable cumulative distribution function over the naturals.

Given the above clarification, we could use the above model to select a real in $\mathbb{R}$ such that all real numbers have an equal, but undefined, probability of being selected:

1) Let $f : \mathbb{N} \rightarrow \mathbb{Z}$ be bijective.

2) Let $x, y$ be selected uniformly at random from $[ \,0, 1) \,$.

3) We then have $r = f(k(x)) + y$.


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