My opponent is dealt a real number $r\in [0, 1]$ uniformly at random and I know that with probability $1-r$ she chooses to discard that $r$ and be dealt a new number in $[0, 1]$ uniformly at random again. I am told neither $r$ nor the second outcome.
Now I am dealt a real number $s\in [0, 1]$ uniformly at random. With what probability (as a function of $s$) should I choose to discard $s$ and be dealt a new number in $[0, 1]$ uniformly at random so as to maximize chances of my number being larger than theirs?
I'm not interested in the long-run strategy, just that for a one-off.
Your win probability is your opponent's cumulative distribution function, which equals $$C(s)=\int_0^sr\,dr+\int_0^1(1-r)s\,dr=\frac12s^2+\frac12s.$$
If you redraw, your win probability will be $$E[C]=\int_0^1\left(\frac12s^2+\frac12s\right)ds=\frac16+\frac14=\frac{5}{12}.$$
So you maximize your win probability by redrawing whenever $C(s)<E[C]$, which you can solve using the quadratic formula. It turns out to be $$s<\frac{\sqrt{39}-3}{6}=0.54083\ldots$$
Note that if your first draw is $x=0.52$, you should redraw, even though your second draw will probably be less than $x$! The intuition here is that since your opponent has a better-than-average strategy, you think that you're probably already losing with $x=0.52$, so you're okay with a desperate strategy that risks losing by a greater margin.