Efficient random guessing 
Let $(x,y)$ be a random point on the plane with some unknown continuous distribution. Your opponent randomly chooses one of the coordinates and tells you. You shall guess whether this coordinate is bigger than another one or not. How can you do this with probability strictly bigger than $\frac12$?

The case of probability being exactly $\frac12$ can be realized via the randomized choice: you say that a given coordinate is the biggest with probability $\frac12$. However, I do not know how to make the truth of the answer being of strictly higher probability.
 A: Take any continuous random variable $Z$ whose density is positive everywhere.
If the opponent tells you $t$, guess "yes" (i.e. that this one is the greater) 
if $t > Z$ and "no" if $t \le Z$.
A: First consider the strategy of saying "larger" if and only the presented number is positive.  Your prediction is guaranteed to be correct when $x$ and $y$ have opposite signs (i.e., one is positive and the other is zero or negative).  When $x$ and $y$ have the same sign, your prediction is correct with probability $1/2$.  Therefore, your overall probability of being correct is $1/2 + p(0)/2$, where $p(0)$ is the probability that $x$ and $y$ have opposite signs.  This is at least $1/2$; but you don't know that it's strictly greater than $1/2$, because you don't know for sure that $x$ and $y$ can have opposite signs.  Maybe both are guaranteed to be positive, for instance.  The way to fix this is to choose your cutoff at random (let it be $Z$, a Gaussian random variable, say), rather than fixing it to be zero.  Then your overall probability of being correct is
$$
\frac{1}{2} + \frac{1}{2}P[\min(x,y)\le Z< \max(x,y)].
$$
Since $Z$ has a nonzero probability of being in any interval $[\min (x,y),\max(x,y))$, the right-hand term is strictly positive.
