I was recently reminded of this probability "paradox" (in quotes because in fact there's nothing really paradoxical about it, but it's surprising to the intuition), which I first encountered back in the rec.puzzles Usenet group. I want to know if it has a name. Here it is:

Suppose I write any two distinct (real) numbers on two blank cards that I then put face down. You get to look at the number on one of the cards, of your choice. Now, I ask you whether that number is the lower or the higher of the two. Can you answer in such a way that your probability of being correct is strictly greater than $\frac12$?

At first, it seems incredible that this is even being asked. I can use any method I choose to select the two numbers, and you have no way of knowing what the other number is. Nonetheless, you can answer in such a way that your response is more likely than not to be correct.

Method: Select any cumulative distribution function $F(x)$ that is strictly increasing; that is, pick a function $F(x)$ such that

- $F(x) > F(y)$ for all $x > y$
- $\lim_{x \to -\infty} F(x) = 0$
- $\lim_{x \to +\infty} F(x) = 1$

To make things simple, a suitable function is $F(x) = \frac{2^x}{1 + 2^x}$. (This is similar to the standard logistic function, but with base $2$ instead of base $e$.)

Now, if the number you look at is $x$, you say "higher" with probability $F(x)$, and "lower" with probability $1-F(x)$. For instance, if you turn over a card and see the number $3$, you say "higher" with probability $F(3) = \frac{2^3}{1+2^3} = \frac89$, and "lower" with probability $1 – F(3) = \frac19$.

Your probability of getting the right answer can now be determined as follows: Suppose the two numbers I selected were $a$ and $b$, with $b > a$. Your answer is correct if either you looked at $a$ and said "lower" (with probability $1-F(a)$), or you looked at $b$ and said "higher" (with probability $F(b)$). Since each of these two scenarios is equally likely—remember, you got to choose the card you looked at, with no prior information—your overall probability of guessing correctly is

$$ P(\text{correct}) = \frac{1-F(a)+F(b)}{2} = \frac12 + \frac{F(b)-F(a)}{2} $$

But since $F(x)$ is strictly increasing, and $b > a$, we must have $F(b) > F(a)$, and so your probability of being correct must be strictly greater than $\frac12$.

Has anyone else heard of this seeming paradox, along with a name?

Update (2019-10-16-0413Z): Math.SE didn't provide this tip before I posted the question, but now this shows up as a "related question": Who discovered this number-guessing paradox?

I'll have to take a closer look at that!