# Probability of intersecting three chords

Three chords are drawn randomly and uniformly in a circle. Here each chord is established using random end point method, i.e. the endpoints of any chord are two points chosen randomly, uniformity and independently on the circumference of the circle. Now what is the probability that they intersect at a single point?

In the case of two chords, it is easy to find the probability of intersecting using symmetry of the circle. But It gets tricky when it comes to three or more chords.

Can a generalized formula be obtained for N number of chords?

*edited

• The probability is zero. – Parcly Taxel May 27 '17 at 5:57
• zero, surely??? – Angina Seng May 27 '17 at 5:58
• Firstly, "random chord" needs to be defined precisely. There are various possible meanings. But in any case, for any of those meanings, it should be obvious that the probability of a triple intersection is $0$. Intersect the first two. If that succeeds, the third one has to hit that exact point$\,-\,$without aiming (since it's a random chord). – quasi May 27 '17 at 5:58
• Maybe they wanted to know the probability that the three chords intersect in three points. – Christian Blatter May 27 '17 at 7:56

In the following I'm going to compute the probability that any two of the three cords intersect. This event only depends on the order of the six endpoints $P_i$ $(0\leq i\leq5)$ on the circle. Assume that the chords connect $P_i$ with $P_{i+3}$. The probability that $P_3$ is opposite $P_0$ in the appearing order is ${1\over5}$. Conditioned on this to be the case the probability that $P_4$ is opposite $P_1$ is ${1\over3}$, and if this is true as well $P_2$ and $P_5$ are automatically in opposition. The probability we are after therefore is ${1\over15}$.
At any rate your problem seems to be finding the set $\hat C\subset\Omega$ corresponding to segments intersecting the positive $y$-axis as well as the negative $x$-axis.
Let $e^{i\phi_1}$ and $e^{i\phi_2}$ be two points on the unit circle. Then $C$ occurs iff $$0<\phi_1<{\pi\over2}\qquad\wedge\qquad \pi<\phi_2<{3\pi\over2}\qquad\wedge\qquad \phi_2-\phi_1<\pi\ ,\tag{1}$$ or $(1)$ with $\phi_1$ and $\phi_2$ interchanged. The conditions $(1)$ together with their counterparts obtained by $\phi_1\leftrightarrow \phi_2$ define two small triangles in $\Omega$ of area ${1\over32}$ each (draw a figure!). It follows that the probability of $C$ comes to $2\cdot{1\over32}={1\over16}$.