Strategy game: draw a random number uniformly between 0 and 1 and if you like redraw. You win if your number is bigger than mine. You draw a random number uniformly between 0 and 1. If you like it, you can keep it. If you don't, you can have a do-over and re-draw, but then you have to keep that final result.
I do the same. You do not know whether I've re-drawn and I do not know whether you've re-drawn. Decisions are made independently. We compare our numbers and the highest one wins $1.
What strategy do you use?
 A: It seems obvious that any optimal strategy for a player is to redraw if the first round is less than some value $t$ and not redraw if the first round is greater than $t$.
If you do this then the density and cdf of your result after making your choice,

*

*when $0 \le x \le t$, has $f_t(x) = t$ and $F_t(x)=tx$, as you must have had drawn less than $t$ in the first draw and drew again


*when $t \le x \le 1$, has $f_t(x) = t+1$ and $F_t(x)=tx+x-t$,  as you may have drawn less than $t$ in the first draw and drawn again or you may have drawn more than $t$ in the first draw and not drawn again
Suppose Player A adopts this strategy with cutoff $t$ and Player B adopts this strategy with cutoff $s$.  Then the probability of Player B winning is $$\int_0^1 F_t(x) f_s(x)\, dx$$

*

*which when $0<t<s<1$ is $\int_0^t (tx)s \, dx + \int_t^s (tx+x-t)s \, dx  +  \int_s^1 (tx+x-t)(s+1)  \, dx $ and comes out at $\frac12(1-t   +s+st +st^2 -s^2 -s^2t)$ and is maximised when $s=\frac{{{t}^{2}}+t+1}{2 t+2}$ so long as that is between $t$ and $1$


*and when $0<s<t<1$ is $\int_0^s (tx)s \, dx + \int_s^t (tx)(s+1) \, dx  +  \int_s^1 (tx+x-t)(s+1)  \, dx $ and comes out at $\frac12(1-t+t^2+s-st+st^2-s^2t)$ and is maximised when $s=\frac{{{t}^{2}}-t+1}{2 t}$ so long as that is between $0$ and $t$.
As TonyK has commented, $t=\frac12$ is not optimal in terms of probability of winning even if it maximises the expected score for Player A. Facing that, Player B setting $s=\frac58$ will beat Player A more often than not, with probability $\frac{129}{256}\approx 0.5039$.  Player B setting $s=\frac7{12}$ would be even better with probability $\frac{97}{192}\approx 0.5052$.
An unbeatable strategy comes when Player A uses a cutoff where $\frac{{{t}^{2}}+t+1}{2 t+2} = t = \frac{{{t}^{2}}-t+1}{2 t}$, i.e. when $t= \frac{\sqrt{5}-1}{2}\approx 0.618$.  In that case, optimal $s$ for Player B is the same value, giving a probability for each of winning of $0.5$ and coincidentally an expected result for each player also of $\frac{\sqrt{5}-1}{2}$.  Any other $s$ (or indeed any other strategy) would give Player B a probability of winning below $0.5$.
