Stopping threshold in continuous blackjack

A gambler and a dealer play a continuous version of blackjack using independent $\text{Unif}(0,1)$ draws.

• The gambler goes first, drawing numbers and adding them up. After each draw he may stop or draw again. If his total ever goes above $1$, he busts and the dealer wins.
• If the gambler stops with total $a\le1$, the dealer then keeps drawing uniforms and accumulating them. Should the dealer's running total land strictly between $a$ and $1$, the dealer takes the game; should it overshoot $1$ first, the gambler takes it.

The gambler uses a threshold rule: draw again while his total is below $\alpha$ and stop once it reaches $\alpha$ or more. The value of $\alpha$ that maximises his chance of winning satisfies $e^{\alpha}=\frac{e+\alpha}{x+y\alpha}$ for some integers $x$ and $y$, where $e$ is Euler's number. Compute $x^2+y^2$.

$13$. Standing on $s$ loses with probability $q(s)=(1-s)e^{s}$, since the dealer's partial sums fall below $s$ for $n$ draws with probability $s^n/n!$ and then need one draw into $(s,1)$. Drawing once more and then standing loses with probability $s+\int_s^1 q(t)\,dt$. Setting these equal gives $(3-2\alpha)e^\alpha=e+\alpha$, so $x=3$, $y=-2$ (and $\alpha\approx0.571$).