Alice and Bob are playing a game. Each time, two numbers are generated uniformly in the interval $[0,1] $, Alice can see the two numbers and choose one to tell Bob. And Bob needs to determine whether it's the larger one. Bob wins if he is correct, otherwise, Alice wins.
Alice and Bob's strategies can be represented as functions. $ F : [0,1]^2 \rightarrow [0,1]$ for Alice, and $f:[0,1] \rightarrow [0,1]$ for Bob. It means that when Alice is told two numbers $p$ and $q$, she has probability of $F(p,q)$ to tell Bob number $p$. When Bob is told number $p$, he has probability of $f(p)$ to guess $p$ is bigger.
Then for given $(p,q)$, the probability for Bob to win is $$\Phi (p,q) = \left\{ \begin{array}{ll} F(p,q)f(p)+(1-F(p,q))(1-f(q)) &\text{if } p > q \\ F(p,q)(1-f(p))+(1-F(p,q))f(q) & \text{if } p < q \\ \end{array} \right.$$
And
$$E = \int \int_{[0,1]^2} \Phi (p,q) dp dq$$ is the expection for Bob to win.
It's not hard to know that if Bob takes $$f= \left\{ \begin{array}{ll} 1 &\text{if } p > \frac 1 2 \\ 0 & \text{if } p < \frac 1 2 \\ \end{array} \right.$$
There will always be $E \ge \frac 1 2$, i.e. $f$ guarantees Bob's not losing (in long term).
I want to know whether the game is fair(Alice also has a not -losing strategy), namely, there exists an $F$, s.t. for any $f$, $E \le \frac 1 2$.
If not, I also want to know whether the following is true. For any $f$, there exists an $F$, s.t. $E \le \frac 1 2$. (Alice still has a chance not to lose).
Let Alice's startegy be: Always present the number that is closer to $\frac12$. In other words, $$ F(p,q)=\begin{cases}1 & \text{if }|p-\frac12|<|q-\frac12|,\\0&\text{otherwise.}\end{cases}$$ Then if Alice presents the number $x$ to Bob and we let $\delta=|x-\frac12|$, then the other number is uniformly distributed on $[0,\frac12-\delta]\cup[0,\frac12+\delta]$, i.e., "larger" and "smaller" both have probability $\frac12$.