I pick two distinct real numbers at random from some arbitrary distribution. Then I flip a coin and if it comes up heads, I tell you the larger one, if it comes up tails I tell you the smaller one. You must guess whether it was the larger one or the smaller one.
Give a strategy that wins with a probability greater than \(\frac{1}{2}\).
If you see an \(x\), guess “larger” with probability \(p(x)\) and “smaller” with probability \(1-p(x)\), where \(p(x)\) is a monotonically increasing function from \(\mathbb{R}\) to \((0,1)\) . A possible function is:
\[p(x) = \frac{1}{2} + \frac{\arctan(x)}{\pi}\]Proof
Suppose the distinct numbers are \(a\) and \(b\) with \(a < b\) .
The probability, \(P\), of guessing correctly is:
\[\begin{align} P & = \frac{1}{2}(1 - p(a)) + \frac{1}{2}p(b) \\ & = \frac{1}{2} + \frac{1}{2} (p(b) - p(a)) \end{align}\]From this we can see that:
\[P > \frac{1}{2} \text{ if } p(b) > p(a)\]But since \(p(x)\) is monotonically increasing and \(b > a\), we have \(p(b) > p(a)\). Thus the overall probability of winning is greater than \(\frac{1}{2}\).