Consider the following game:
Two real numbers are independently, randomly, uniformly chosen from the interval [0, 1].
Andrew is secretly shown the two numbers. He must choose one of the numbers and reveal it to his opponent, Bndrew. Bndrew then guesses whether the revealed number was the larger of the two. If he gets it right, he wins; otherwise, Andrew wins.
Assuming optimal play, how likely is Bndrew to win?
I claim that Bndrew wins
50%
of the time.
Bndrew can guarantee winning with at least that probability with the strategy of
flip a coin to choose
Andrew can guarantee Bndrew wins with at most that probability with the strategy of
Reveal whichever number is closer to $\frac{1}{2}$.
For example, if Andrew reveals $\frac{1}{5}$ or $\frac{4}{5}$, Bndrew knows only that the other number is uniformly distributed over $[0,\frac{1}{5}] \cup [\frac{4}{5},1]$ and wins half the time.
Assuming optimal play, I think Bndrew cannot win more than
50% of the time.
Why?
Let $a$ and $b$ the two secret numbers such that $a \leq b$, and $x$ the number given by Andrew ($x=a$ or $x=b$).
If we exclude strategies that involve saying "largest" or "smallest" at random (which cannot exceed 50% win rate), Bndrew's choice can only rely on the value of $x$. That means any strategy will consist of choosing a set $S \subseteq [0,1]$ such that Andrew will say "largest" if $x \in S$, and say "smallest" otherwise.
What's the best possible choice for $S$? It would not make sense for $S$ to not be an interval with an upper bound lower than $1$, because it would mean that there exists $a \leq b$ such that Bndrew will say "highest" for $a$ but "lowest" for $b$. Hence, we can assume that $S=[s, 1]$ for some $s \in [0,1]$.
Now, what's the best possible choice for $s$? Well, to ensure a win no matter the choice of Andrew, we would like to maximise the probability of having $a \leq s \leq b$. This happens when $s=\frac{1}{2}$.
Therefore, the best strategy for Bndrew is to say "largest" if $x \geq \frac{1}{2}$ and "smallest" if $x < \frac{1}{2}$. There is a $\frac{1}{2}$ probability of having $a \leq \frac{1}{2} \leq b$ which ensures a win for Bndrew. If $a \leq b \leq \frac{1}{2}$, Andrew can prevent Bndrew from winning by giving $x=b$ (which happens with probability $\frac{1}{4}$) and if $\frac{1}{2} \leq a \leq b$, Andrew can prevent Bndrew from winning by giving $x=a$ (which also happens with probability $\frac{1}{4}$).
