Probability over an infinite set. Of two numbers.

Question

Suppose there is a man, who chooses two completely random numbers. So they can be equal too. And they can be only positive real numbers. One of them can even be $285294.38285967281$ or anything bizarre. So my question is, what is the probability that the first number is greater?

So I did it like this: Probability of two numbers being equal would be $\frac{1}{\infty}$ and $\lim_{x\to\infty} \frac{1}{x}=0.$ And thus, our required probability, over an infinite space will be $\frac{1}{2}$.

But some of my fellow men are saying that probability theory cannot be extended over infinite space. But I totally disagree with them. Even, the infinite monkey theorem uses infinity. So am I right? Also one of my friends also told me that there is a way by the x-y plane. The areas of the two triangles formed on either sides of the line $y=x$ will be equal and a line doesn't have an area. So probability will be $\frac1{2}$.

Answer 1

"Two completely random numbers" is not a well-defined probability distribution, even when restricted to positive real numbers. The actual probability distribution must be something more specific, for example, if all positive real numbers are possible choices then the exponential distribution is a possibility. But notice that with an exponential distribution, the probability of choosing a number between $1000$ and $1001$ is less than the probability of choosing a number between $1$ and $2$. There is literally no possible random process with an infinite number of possible outcomes that can have a uniform probability distribution over all of its range. If you want all positive real numbers to be possible, you must let there be some numbers $m$ and $n$ such that the probability that your random number $x$ satisfies $m \leq x < m + 1$ is less than the probability that $n \leq x < n + 1$.

Since you want to be able to choose any positive real number, and you don't seem to want any one number to be much more likely to be chosen than any other number near it, I'll consider only continuous distributions over the interval $[0,\infty)$. A number selected from a continuous distribution can be any real number in the range of the distribution, but for any particular number $k$ in that range, the probability that the random number will be exactly equal to $k$ is zero. (See this earlier question and its answers for further discussion of that fact.) If you choose two numbers independently using identical continuous probability distributions, it is not forbidden for the second number to be equal to the first, but the probability of that event is zero.

So an argument by symmetry can show that the first number is greater with probability exactly $\frac 12$, provided you have first defined an actual continuous probability distribution over the positive real numbers (and have accepted the necessary fact that this distribution will be non-uniform).

There is also a way to do this by use of the $x,y$ plane. We let $x$ be chosen randomly according to a continuous distribution over $[0,\infty)$, and also let $y$ be chosen independently according to an identical distribution. It is then possible to define a joint distribution function for $(x,y)$ over the $x,y$ plane such that the probability of the point $(x,y)$ falling in any particular region is the integral of the joint distribution function over that region. So define a suitable probability distribution, find the joint distribution, integrated it over the (infinite) region of the $x,y$ plane where $x < y$ and $y > 0$, and show that this integral equals $\frac 12.$ (The easiest way to show this may again be by a symmetry argument.)

Answer 2

You are right and wrong.

First off you can't give a uniform probability to a countably infinite range (the discrete case), but you can for an uncountable domain (the continuous case).

But in the countable case you can always give it some probability distribution and the probability that the first one is greater than the second is never going to be exactly $1/2$, regardless of the distribution, because you have to account for the probability that they are equal (in the continuous case this probability can be zero but not in the discrete case). There are three possibilities, $x<y$, $x>y$, but also $x=y$. That last probability does depend on the distribution. It can only be zero in the continuous case.

Suppose for example the set of numbers is $\{1,2,3,\dots\}$. And the probability of picking number $n$ is $\frac{1}{2^n}$. Then since $\sum_{n=1}^\infty\frac{1}{2^n}=1$ this is a valid probability distribution. Then if we choose two numbers $x$ and $y$ according to these probabilities, three things can happen: event $A$: $x<y$, or event $B$: $x>y$ or event $C$: $x=y$. These are three distinct and exhaustive probabilities, so $pr(A)+pr(B)+pr(C)=1$. Thus we cannot have $pr(A)=pr(B)=1/2$ otherwise we'd have to have $pr(C)=0$, which would mean it is impossible that the two numbers are equal. But it's not impossible. It is in fact equal to $\frac12\cdot\frac12 + \frac14+\frac14+\cdots = 1/3$. Thus since $pr(A)=pr(B)$, we have that the probability of any one of the three events is $1/3$.

On the other hand if there were an uncountable number of possibilities, like your example in the plane, so that any given outcome has probability zero, then it will be exactly $1/2$. So your partially right, it can be $1/2$ but doesn't have to be.