I am trying to determine the expected value of the following game:
Let $u$ be drawn from a uniform distribution on $[0,1]$. We write down $u$ on one side of a piece of paper and $u^2$ on the other side. We then randomly draw one side of the paper and win one dollar if we correctly guessed whether the observed number is $u$ or $u^2$.
My intial intuition tells me that there should be a cutoff value $u^*$ such that the probability of getting the guess right can be separated into two cases depending on whether $u>u^*$ or $u<u^*$. To find the cutoff value we would then optimise the probability of winning with respect to $u^*$. This is just my initial idea but I am having trouble formalising it into precise probabilistic arguments. Any ideas would be greatly appreciated.
updated idea: let $x$ be the number observed on the paper such that $x=u$ with probability $0.5$ and $x=u^2$ with probability $0.5$. The cutoff value $u^*$ is then such that if $x>u^*$, we guess $x=u$ and if $x<u^*$, we guess $x=u^2$. Then we can write: $$\mathbb{P}(win) = \mathbb{P}(x>u^*|x=u)*0.5 + \mathbb{P}(x<u^*|x=u^2)*0.5=(1-u^*)*0.5 + \sqrt{u^*}*0.5$$ We then find $u^*=0.25$ and expected value of game is just the probability of winning which is $5/8$. Does this seem correct?
If $U \sim Unif(0,1),$ notice that $Y = U^2 \sim Beta(.5, 1),$ which concentrates probability near $0$. Because the density function of $Beta(.5, 1)$ is $f_Y(y) = 1/(2\sqrt{y})$ for $0 <y <1,$ the point at which the densities of $U$ and $Y$ cross is $1/4$.
Also, $P(U > 1/4) = 3/4$ and $P(Y < 1/4) = 1/2.$ More generally, $P(U > a) = 1 - a$ and $P(Y < a) = \sqrt{a},$ for $0 < a < 1.$
The next move is yours.