Strategy for game where larger number wins. Drawn from standard uniform distribution with one redraw allowed Two players are playing a game where they each draw a secret random number uniformly between 0 and 1. If they are not satisfied with their draw they may redraw. The players do not know whether or not the other has chosen to re-draw. The players then compare their numbers and he/she who holds largest number wins $1. What is the best strategy assuming both players are perfectly rational? 
Idea: 
I believe we want to try and find the Nash Equilibrium of this game. 
A strategy that seems intuitive is to find some threshold $x$ where I redraw if my number $\leq x$. Let's imagine that our opponent has a similar strategy and his/her threshold is $y$. 
I'm not too sure how to go from here. Also, how do I show that picking a threshold to reroll is the best strategy?
The answer is not to redraw if one draws below 0.5 by the way. Further we are not necessarily trying to maximize our expectation, rather our probability of winning.
 A: Firstly, I will explain why maximizing the expected value does not work (BruceET's answer is wrong). Suppose the players play the following game with dice, whoever gets the larger number wins. Player one has a die with values $1,1,1,1,1,103$ and player 2 has a die with values $3,3,3,3,3,3$. The expected value of the first die is $18$ and of the second is $3$. Despite this, player 2 has a probability of $5/6$ to win.
Now to my approach. Assume that player 1 will redraw if his first number is less than $z$ and that player 2 will redraw if his number is less than $w$. Let us find the expected value for player 1, which in this case is 1 dollar multiplied by the probability of him winning, so we just need to find his chances of winning.

*

*Both players redraw.

*Player 1 redraws, player 2 stays.

*Player 1 stays, player 2 redraws.

*Both players stay with their first number.

We will compute the probability of player 1 to win in each scenario. Let $X$ denote the uniform random variable for player 1, and $Y$ for player 2. Assume $z \geq w$. This assumption is valid since both players will end up using the same 'threshold' to decide whether to redraw ($z=w$).

*

*$P(X < z) P(Y<w) P(X>Y) = \frac{z w}{2}.$

*$P(X < z) P(Y > w) P(X > Y \,| \, Y > w) = z P(w < Y < X) = z\int_{w}^{1}\int_{w}^{x} \, d y \, d x = \frac{z(w-1)^2}{2}$.

*$P(X > z) P(Y < w) P(X > Y \,| \, X > z) = w P(X > Y \cap X > z) = w\int_{z}^{1}\int_{0}^{x} \, d y \, d x= \frac{w(1-z^2)}{2}$.

*$P(X > z) P(Y > w) P(X > Y \,| \, X > z \cap Y > w) = \int_{z}^{1}\int_{w}^{x} \, d y \, d x= \frac{1-z^2}{2} - w(1 - z)$.

Adding everything up we get
$$\frac{1}{2} (z w + z w^2 + z - w - z^2 w + 1 -z^2).$$
Now we want to maximize player 1's expected value given that we know $w$. This is done by computing the partial derivative with respect to $z$ and equating to $0$, then solving for $z$.
$$w + w^2 + 1 - 2 z w - 2z = 0$$
Since both players will use the optimal strategy, $z$ must equal $w$. Subbing in $z$ for $w$ and solving:
\begin{align*}
    &z^2 + z - 1 = 0,\\
    &z = \frac{\sqrt{5} - 1}{2}\\
    &\boxed{z \approx 0.618034.}
\end{align*}
An interesting observation is that the optimal value for $z$ is related to the golden ratio! $\boxed{z = \phi - 1.}$
