What strategy should Alice choose to win this game? The following game was proposed to me:


*

*Alice has a coin which has a probability $p$ of landing on heads, $1-p$ of landing on tails. Alice doesn't know $p$. Alice is not allowed to lie.

*Bob knows $p$. Bob is allowed to lie.

*Alice tells Bob two functions $f, g: [0,1]\mapsto\mathbb R$ and asks Bob for the value of $p.$ Let Bob's answer be $\pi$.

*Alice throws the coin. If it lands on heads, Bob is awarded $f(\pi)$ points, if it lands on tails, Bob is awarded $g(\pi)$ points.


Alice wants to find a pair of functions $f,g$ such that if Bob tries to maximize the expected value for the number of points he is awarded, then $\pi=p$.
Is there a simple way to describe the entire solution space?

Bob tries to maximize his win so he calculates
\begin{equation}
\max_{\pi \in[0,1]} [p f(\pi) + (1-p) g(\pi)]
\end{equation}
For the start and sake of simplicity assume that it coincides with
$\frac{d}{d \pi}[p f(\pi) + (1-p) g(\pi)] = 0$.
We get $q(p, \pi) = p f'(\pi) + (1-p) g'(\pi)$.
Now Alice wants to construct $q$ such that $\forall p ~ q({p,p}) = 0$.
Now comes the fun part finding a solution to
$  p f'(p) + (1-p) g'(p) = 0$
We can start e.g. with $f(p) = \ln(p)-p$
then we have
$ p (\frac{1}{p} -1) + (1-p) g'(p) = 0$
equivalent to
$1+g'(p)=0$
so $g = -p$
Now we can check the second derivative to be sure that it is actually a maxima
$p f''(\pi)+(1-p)g''(\pi) = -p\frac{1}{\pi^2}$
voila.
The whole solution space should be obtainable via
\begin{equation}
-\frac{1-p}{p} g'(p) = f'(p) ~\text{with constraint}~ \forall \pi : p f''(\pi)+(1-p)g''(\pi) < 0
\end{equation}
