${\cal P}_m(\phi)$ that two SAWs starting at $0$ avoid a disk with center on the imaginary axis and pass one left, one right, depends only on the angle $\phi$ of the cone drawn on the left of Fig. \[f:beau\] and is clearly a decreasing function of $\phi$, with ${\cal P}_m(0)={\cal P}_m$ and ${\cal P}_m(\pi)=0$. Hence a disk of fixed size will be pushed to infinity along the imaginary axis. Under conformal mapping of geometry $\cal A$ to $\cal C$, discs map to discs and the cone to the space between two circular arcs. Assuming conformal invariance of the probabilities, all disks shown in Fig. \[f:beau\] have the same free energy. The center of a disk will thus be pushed to the stable equilibrium point above the origin, where the largest disk is drawn. A quantitative result is possible for small radius $\rho$. E.g. in the geometry ${\cal A}$ the no-hit probability for a disk centered at $x+i y$ reads $p \approx 1 - c (\frac{\rho}{2 y})^{2/3} \sin^2(\theta)$ to lowest order in powers of $\rho/y$, as extracted from [@beffara; @bauer; @Cardy2005], with an unknown constant $c$. This gives the force $k T \vec \nabla \ln p$. In the symmetric case $\theta=\pi/2$, 