Strategy to maximize the expected sum of $3$ numbers each drawn from $\sim U(0, 1)$ Consider a uniform random number generator $\sim U(0, 1)$.
We can generate a maximum of $10$ numbers sequentially. We are only allowed to keep $3$ numbers at a time. If we currently have $3$ numbers, then we cannot draw the next number until we discard $1$ of our current numbers.
What is the optimal strategy to maximize the expectation of the sum of the $3$ numbers we have in the end? What is the expectation of the sum of the $3$ numbers?
I think I know how to solve this problem for the case where we are allowed to keep one number. I would work backwards and start at the $10$th draw, which has expectation $0.5$. If the $9$-th draw is $\leq 0.5$, I will discard it and proceed with the $10$th draw. Then I can compute the expectation of a $9$th draw, which would come out to be $0.5 \cdot 0.75 + 0.5 \cdot 0.5 = 0.625 $. And then I can repeat this process all the way to the beginning, and I should see this number increase monotonically. We will also see our threshold of deciding to draw again or not will decrease monotonically for each subsequent draw, i.e., on the first draw, we are much more likely to be more daring than, say, on the $9$th draw. I don't know if there's an easy way to come up with a generic formula for this (feel free to let me know if there is) since it can become tedious to recurse all the way back to the first draw.
So the above is how I would approach the problem if we were allowed to keep $1$ number. I don't know how to generalize this approach to being allowed to keep more than $1$ number. It is obvious to me that we should at least draw $3$ numbers, and we should never have less than $3$ numbers on hand. It's also clear that we should never discard more than $1$ number. With this information, it is still not clear to me how to approach this problem.
 A: Your approach for keeping a single number works here as well.  In the single number case, at each round you have a threshold that says you should stand pat if your number is above that.  For the three number case, the threshold stays the same and you compare your smallest number with the threshold.
Imagine a less favorable game where my adversary can force me to pick two numbers that can not be discarded.  I pick the two highest.  Once the adversary does that, I am back to the one number game and should follow its strategy.  Your game has an advantage to me that if the new number is very high I have the option to discard one of the numbers I would have fixed.  The critical observation is that the threshold is decreasing. If I would not discard the lowest number this round, I would not discard it in any later round either.  If I would discard the lowest number this round, it is just like playing the more restrictive game.
On average, you will stand pat later in the three number game because all the numbers have to be above the threshold.  The strategy does not change.  Look at the threshold for the one number game.  If the lowest number is below that, throw it away and draw again.
The threshold with $n$ games to play is $\frac {a(n)}{2^{2^n-1}}$ where $a(n)$ is given in OEIS A167424, which starts $1, 5, 89, 24305, 1664474849, 7382162541380960705, 139566915517602820239076685726696149889$
