What is the probability that a random chord in a sphere touches opposite hemispheres? (edited) Consider the unit sphere $\mathbb{S}^2\subset \mathbb{R}^3$, and its upper $(z>0)$ and lower $(z<0)$ hemispheres.
Draw two independent, uniformly distributed points $X,Y$ on $\mathbb{S}^2$.
Given $\theta\in[0,\pi/2]$, what is the probability that $X$ and $Y$ belong to different hemispheres among the above two, conditioning by the event that the chord $[X,Y]$ makes an angle $\theta$ with the $z$-axis?
Numerically I find that this probability is $\cos{\theta}$.
Could anyone help me justify this $\cos{\theta}$?
 A: This is not a true "no pen or paper" solution requested by fedja, but at least it avoids integrals. :-)
Let $X$ and $Y$ be independent random vectors on the unit sphere. Write $E = (X - Y) / |X - Y|$ for the unit vector parallel to the chord $XY$, and $Z = \tfrac{1}{2} (X + Y)$.

Claim: Conditionally on $E = e$, the projection $Z = \tfrac{1}{2}(X + Y)$ of the chord $XY$ onto its perpendicular bisector plane $\pi_e = \{v : v \perp e\}$ (the entire chord projects onto a single point) is uniformly distributed over the unit disk in $\pi_e$.
Given the above claim, the proof is straightforward. Indeed: given any unit vector $e$ such that $\theta = \arcsin |e \cdot (0,0,1)|$ as in the question, and conditionally on $E = e$, $X$ and $Y$ belong to different hemispheres if and only if $Z$ belongs to an ellipse, which is the projection of the equator of the unit sphere onto $\pi_e$. This ellipse has semi-axes $1$ and $\cos \theta$ (this becomes pretty clear if one draws a picture). The area of this ellipse is equal to $\cos \theta$ times the area of the unit disk, and the desired result follows. Thus, it remains to prove the claim.

Proof of the claim: Since the random variable $X \cdot Y$ is uniformly distributed over $[-1, 1]$ (Archimedes's theorem!), the random variable
$$ \|Z\|^2 = \|\tfrac{1}{2} (X + Y)\|^2 = \tfrac{1}{2} (1 - X \cdot Y) $$
is uniformly distributed over $[0, 1]$. By rotational symmetry, $\|Z\|^2$ and $E$ are independent. It follows that conditionally on $E = e$, $\|Z\|^2$ is uniformly distributed over $[0, 1]$. Again by symmetry, the conditional distribution of $Z$ (given $E = e$) is invariant under rotations of $\pi_e$, and so it follows that this conditional distribution is uniform over the unit disk on $\pi_e$, as desired.
