Expected number after n rounds of uniform~[0,1] draws If we have a series of $n$ IID random variable $X_i$ that are uniform [0,1], and at each round $i$ we decide to either keep $X_i$ or discard it for the next number. What is our strategy to maximize our final number? and what is the expected number we get under this strategy?
I figured this is pretty much the coin toss problem but with a continuous uniform [0,1] distribution. Let $f(n)$ be our expected final number if we're allowed n "re-toss". The strategy is that, if we have $n$ "tosses" left, we only keep our current number if it's greater than $f(n-1)$,otherwise we go for the next number.
With Uni~[0,1], the probability that we draw a number less than $f(n-1)$ is $f(n-1)$, so
This gives the equation
$$
\begin{aligned}
f(n) &= P(X_1 < f(n-1))f(n-1) + P(X_1 >= f(n-1))E[X_1|X_1 >= f(n-1)]\\
&= f(n-1)*f(n-1) + (1-f(n-1))*\frac{f(n-1)+1}{2}\\
&= f(n-1)^2 + \frac{1-f(n-1)^2}{2}\\
&=\frac{f(n-1)^2}{2} + \frac{1}{2}
\end{aligned}
$$
My question is, with the initial condition that $f(1) = 0.5$, is there any way to make the above into a non-recursive solution? As in, a solution where we can compute $f(n)$ without computing $f(n-1),f(n-2)...$

I think you are overcomplicating this. With $X_1, X_2, \dotsc, X_n$ iid uniform on the standard interval $(0,1)$, your construction is simply computing $\max_{i=1}^n X_i$, and to find that distribution is a standard exercise, solved at How do you calculate the probability density function of the maximum of a sample of IID uniform random variables?
