Probability of random geodesics on the half-sphere intersecting 4 end points (a,b,c,d say) are chosen uniformly randomly and connected a to b and c to d by two geodesics on the 2-dim half-sphere. Here, uniform means that, probability that a point lies on a surface area of size $da$ is proportional to $da$. What is the probability that the two curves cross?
 A: A very similar question is considered here. (here, the OP is asking about two segments in the unit circle). There are two possibilities: the four points form the vertices of a convex quadrilateral (in which case they intersect if they are two diagonals, the probability of which is $1/3$), or not (so that one point lies inside the triangle formed by the other three). In that case, there is probability $0$ that the segments intersect. So, this reduces the question to computing the probability that one of the points lies within the triangle formed by the other three. Now, for this last, there is an ancillary problem of finding the expected area of a triangle formed by three random points. This is addressed here, and the solution works, mutatis mutandis in the hemisphere case (you look at the triangle $OAB$ where $0$ is the north pole, the distribution of the angles is uniform, the distribution of $OA$ and $OB$ is easy to compute, the final answer will be slightly messy), and do some combinatorics to get the final answer. I am too lazy to go through with all the details, but it is clearly doable.
A: Alternatively to Igor Rivin's method, here is what I think could lead to the result.
To get four independent uniformly distributed random points on the hemisphere, we first chose two (independent, uniformly distributed) diameters of its boundary in the equatorial plane, say AOB and COD, with an angle $\theta$ between them, distributed uniformly on $[0,\pi/2]$. Then we chose two planes containing AOB and COD respectively, making with the equatorial plane an angle of $\xi$ (or $\eta$, respectively) uniformly distributed on $[0,\pi]$ ; they intersect the hemisphere on half-circles. Each point M on the half-circle is defined by the angle $\alpha$ (or $\gamma$) from OA (or OC) to OM, and the half-circle is equipped with the probability measure $\sin\alpha\ d\alpha/2$ (or $\sin\gamma\ d\gamma/2$). With this choice of conditional distribution, if M is chosen randomly it will be uniformly distributed on the half-sphere, I (strongly) believe.
Clearly then, $ab\subset$ AB and $cd\subset$ CD intersect iff both contain the point M$_0$ where the two half-circles intersect (at angles $\alpha_0$ on AB, $\gamma_0$ on CD): this (given $\theta$, $\xi$ and $\eta$) has probability $\frac12 \sin^2\alpha_0\times\frac12 \sin^2\gamma_0$.
These angles $\alpha_0$ and $\gamma_0$ of M$_0$ can be determined using spherical trigonometry (considering the spherical triangle with $\theta$ as one side and $\xi,\eta$ as adjacent angles). You get$$\tan\alpha_0=\frac{2\sin\eta}{\cot(\theta/2)\sin(\xi+\eta)+\tan(\theta/2)\sin(\xi-\eta)}$$and$$\tan\gamma_0=\frac{2\sin\xi}{\cot(\theta/2)\sin(\xi+\eta)-\tan(\theta/2)\sin(\xi-\eta)}$$
Integrating in $\theta$, $\xi$ and $\eta$ should then give the desired probability that $ab$ and $cd$ intersect.
A: Away from a zero set, $(a,b,c,d)$ gives you crossing geodesics if and only if $(a,b,c,-d)$ doesn't. It follows that the probability is exactly $0.5$, since the map $(a,b,c,d)\mapsto (a,b,c,-d)$ is measure preserving.
