Strategy for 2-player game, drawing uniform variables and optionally redrawing Player 1 and Player 2 secretly and separately draw uniform random variables in [0,1]. They may (secretly) elect to redraw once and replace their value. Highest value wins. What is the optimal strategy?
The answer is supposed to be redraw if less than goldenratio-1, but I haven't been able to reproduce that result. The approach I tried was to create a function representing the win probability for player 1 as a function of thresholds for player 1 and player 2 below which they redraw, and maximizing the function with respect to player 1's threshold, using the symmetry that both thresholds are equal. However, the function I obtain has no extrema. Does anyone have a detailed solution?
Also, intuitively I don't understand why the answer is not to redraw if less than 0.5 since you are more likely than not to make the value worse if the first draw was above 0.5.
Finally, if anyone could provide a source for the question it would be appreciated.
 A: Let's suppose that your opponent's strategy is to redraw on $r$ or lower.
If your first draw is $x$, then the probability you win if you keep that draw is
$$
P(\text{oppenent's final score}\le x)=\begin{cases}
rx & \text{if }x\le r \\
rx+x-r & \text{if }x\ge r
\end{cases}
$$
This is because, when $x\le r$, the only way to win is if your opponent redraws and then gets less than $x$ on their second draw, while when $x\ge r$, there is the additional possibility that the opponent's first roll is between $x$ and $r$.
On the other hand, the probability you win if you redraw is
$$
P(\text{you win if you redraw})=r\cdot\frac12+(1-r)\frac{1-r}{2}=\frac{1-r+r^2}{2}
$$
This is because if your opponent redraws, you'll win half the time, and if he doesn't, you can only possibly when if you are above $r$, and even then will only win half the time.
Now, if your optimal response were to redraw on $x$ or lower, that would mean
$$
P(\text{oppenent's final score}\le x)=P(\text{you win if you redraw})
$$
Furthermore, if redrawing on $r$ or less were the equilibirum strategy, it would by symmetry be its own optimal response, implying 
$$
x=r
$$
Putting those above four equations together implies that, if $r$ is the optimal strategy, than
$$
r^2=\frac{1-r+r^2}{2}
$$
and solving this gives $r={\phi}-1$.
