Neyman's paper states the following:
Denote by $x$ the angle between the radius perpendicular to any given chord $A$ and any fixed direction. Further, let $y$ be the distance of chord $A$ from the center of circle $C$. If $A'$ denotes a point on the plane with coordinates $x$ and $y$, then there will be a one to one correspondence between the chords $A$ of length $0\leq l < 2r$ and the points of a rectangle, say $(A')$, defined by the inequalities $0< x \leq 2 \pi$ and $0 < y \leq r$. The measure of the set of chords $A$ with lengths exceeding $B$ could be defined as being equal to the area of that part of $(A')$ where $0 < y \leq \sqrt{r^2 - \frac{1}{2} B^2}$. It follows that the probability in which we are interested is $P\{l > B\} = (r^2 - (\frac{1}{2}B^2))^{\frac{1}{2}}r^{-1}$.
Denote by $x$ and $y$ the angles between a fixed direction and the radii connecting the ends of any given chord $A$. If $A''$ denotes a point on a plane with coordinates $x$ and $y$, then there will be a one to one correspondence between the chords of the system $(A)$ and the points $A''$ within the parallelogram $(A'')$ determined by the inequalities $0 < x \leq 2\pi$, $x\leq y \leq x + \pi$. The measure of the set of chords $A$ with their lengths exceeding $B$ may be defined as being equal to the area of that part of $(A'')$ where $2r\sin \frac{1}{2} y > B$. Starting with this definition $P\{l > B\} = 1 - 2\arcsin (B/2r)\pi^{-1}$.
It is easy to see how Neyman arrived at the first solution. Inspecting the right triangle formed by the radius perpendicular to $A$, half of $A$, and the radius connecting to one end of $A$, we know that for $A$ to have length greater than $B$, we require $r^2 - y^2 > \frac{1}{2} B^2$ or $y < \sqrt{r^2 - \frac{1}{2} B^2}$. The fraction of the area of $(A')$ that this subrectangle covers is $$P\{l>B\} = \frac{2\pi \sqrt{r^2 - \frac{1}{2} B^2}}{2\pi r} = \frac{\sqrt{r^2 - \frac{1}{2} B^2}}{r}$$ just as Neyman says.
What I don't understand, however, is how Neyman arrived at the second solution. If I examine a similar triangle here, isn't the central angle $\frac{1}{2}(y-x)$, so that the length of the chord is $l = 2r\sin (\frac{1}{2} (y - x))$ and not $2r\sin (\frac{1}{2} y)$? Am I imagining the wrong angles?
EDIT: It seems I get the right answer in the end anyway? If $B < 2r\sin (\frac{1}{2}(y - x))$, then $y-x > 2\arcsin (B/2r)$ so we are considering the skinnier parallelogram $0 < x \leq 2\pi$, $x+2\arcsin(B/2r) < y \leq x+\pi$. The fraction of the area of the parallelogram covered by this skinnier one is:
$$P\{l > B\} = \frac{2\pi (\pi - 2\arcsin (B/2r))}{2\pi^2} = 1 - 2\pi^{-1}\arcsin(B/2r)$$
Is this right?
Neyman was sloppy with his variables. His answer can be obtained using basic geometry: $$B < 2r\sin\left(\frac{1}{2}(y-x)\right) \Longrightarrow y > x + 2\arcsin\left(\frac{B}{2r}\right)$$ So the probability is: $$P\{l > B\}= \frac{2\pi \left(\pi - 2\arcsin\left(\frac{B}{2r}\right)\right)}{2\pi^2} = 1 - 2\pi^{-1}\arcsin\left(\frac{B}{2r}\right)$$