Game between Alice and Bob involving extremal numbers Alice generates $4$ numbers in $(0,1)$ independently and uniformly at random. She discloses one of the numbers to Bob, who is requested to guess whether the disclosed number is extremal (i.e. the smallest or the greatest among all generated numbers) or not. Is there a deterministic strategy for Alice to make sure that the chances that Bob is guessing right are at most 50%.
 A: I make all 4 numbers in the range (0,1/2] by taking either x or 1-x. The smallest number is trivially extremal. 
The second smallest number is just as likely to have come from (0,0.5] as from [0.5,1). If it came from the same side as the smallest number it isn't extremal, but if it came from the other side it must be extremal.
I submit the (original) value of this number, which has a 0.5 chance of being extremal.
A: This answer is more of a long comment:
Consider a simpler but similar game:  Alice generates $2$ numbers in (0,1) independently and uniformly at random.  She discloses one of the numbers to Bob, who must guess whether it is the higher number or not.
Here, Alice has a deterministic strategy for ensuring a 50% win probability:  Present number $x$ if $(x > y) \wedge (x+y < 1) $ or if $(x < y) \wedge (x+y > 1) $.  This works because the conditional distribution of $x$ or $y$ is still flat, so that the  number presented as the higher one is always 50%.
The game as presented does not a strategy tirivally based on that same idea (for example, taking the first and third highest numbers and applying the same critieria as before), since the distribution of the n-th higheset number is anything but flat.
A second observation is that this game is similar to what I was introduced to as Spencer's Game (although Joel Spencer denied having invented it):  Alice generates 3 uniform randoms on (0,1), selects one random and changes it to whatever she wants, and presents the numbers to Bob.  Bob then chooses one of the three presented number, and receives the original value of that number. 
Here, Alice does have a  strategy that will  hold Bob's best-play expectation to exactly $\frac{1}{2}$ but the solution is highly non-trivial.  
One can easily see in this sort of game that for any $\epsilon$, if there is a good strategy, then there is a deterministic strategy which is at least within $\epsilon$ of that good (possibly impure) strategy.  This is so because one can use arbitrarily late digits of the randoms generated to choose between possible specific selections with any required choice probabilities.
