Every night, different meteorologist gives the probability of rain for the next day. To judge their predictions, we use the following scoring system: if a meteorologist predicts rain with the probability $\color{blue}{p^*}$ and is right, that meteorologist receives a score of $1-(1-p)^2$; if wrong, they receive a score of $(1-p)^2$. After a while we will be able to know which meteorologist is the best. Assumming one meteorologist knows the scoring system, what is the best way for them to maximise their expected value?
I know they probably should predict with a 50% accuracy everytime because that is where both fucntion intersect but what is the formula i should use to get to the right answer?
If the meteorologist's belief is $p^*$ then the expected gain from predicting $p$ is $$p^* (1-(1-p)^2)+(1-p^*)(1-p)^2$$ which can be rewritten as $$ p^*+(1-2p^*)(1-p)^2$$ and so depending on the sign of $p^*-\frac12$ it is maximised when $p=1$ or $p=0$ in the natural way (as A.S. has said in the comments).
If $p^*\gt \frac{1}{2}$ then the expected value with $p=1$ is then $p^*$, while if $p^*\lt \frac{1}{2}$ then the expected value with $p=0$ is then $1-p^*$. If $p^*= \frac{1}{2}$ then all values of $p$ give the same expected value of $\frac12$. These might be combined into a maximum expected value of $$\tfrac12 +\left|p^* - \tfrac12 \right|.$$