I'm working on this problem. Could you please check if I did it correctly?
I first draw $X\sim \operatorname{unif}(0,1)$. I can accept it as my winnings or reject it. If I reject, I can draw $Y\sim \operatorname{unif}(0,1)$, assuming that $X$ and $Y$ are independent. I then can choose accept or reject it. If I reject it, I can draw another $Z\sim \operatorname{unif}(0,1)$ (which is independent from $Y$) and I'm stuck with it. Come up with a strategy to maximize my winnings.
I do it backwards. Suppose I just rejected $X$ and I'm deciding on my threshold on accepting $Y$. Denote it by $k$. That is, I will accept if $Y > k$ or reject otherwise. Then my expected winnings would be $(1-k)\frac{(1-k)}{2} + k\frac{1}{2}$, the maximum of which is achieved at $k = 1/2$. The expected gain would be $\frac{5}{8}$. Using the same logic, suppose the threshold for accepting $X$ is $m$. Then I would maximize $(1-m)\frac{1-m}{2} + m \frac{5}{8}$. I then got $m=3/8$.
Therefore, my strategy is to accept $X$ if $X > 3/8$ and accept $Y$ if $Y > 1/2$.
Update Thanks to @lulu, I realized that I made an algebraic error. This is my new solution.
Suppose I just rejected $X$ and I'm deciding on my threshold on accepting $Y$. Denote it by $k$. That is, I will accept if $Y > k$ or reject otherwise. Then my expected winnings would be $(1-k)\frac{(1-k)}{2} + k\frac{1}{2}$, the maximum of which is achieved at $k = 1/2$. The expected gain would be $\frac{1}{4}$. Using the same logic, suppose the threshold for accepting $X$ is $m$. Then I would maximize $(1-m)\frac{1-m}{2} + m \frac{1}{4}$. I then got $m=3/4$.
Did I do this correctly?