Disk and Chords
Question
Three random chords (each chord is the segment joining two independent uniform random points on the circumference). What is the expected number of regions of the disk?
Short answer
$$ E[\#\text{regions}] = 5. $$Why
With chords in general position (no shared endpoints, no three chords meeting at one interior point — both happen with probability $0$), adding the $k$-th chord increases the number of regions by
$$ 1 + (\#\text{ intersections of this chord with previous chords}). $$So for $n$ chords,
$$ R_n = 1 + n + X_n, $$where $X_n$ is the total number of pairwise chord intersections. For $n=3$:
$$ R_3 = 4 + X_3, $$and $X_3$ is the number of intersecting pairs among the three pairs of chords.
A key fact under this endpoint-uniform model is
$$ \Pr(\text{two random chords intersect}) = \frac{1}{3}. $$Hence by linearity of expectation,
$$ E[X_3] = \binom{3}{2}\cdot \frac{1}{3} = 1, $$so
$$ E[R_3] = 4 + 1 = 5. $$Bonus: full distribution for three chords
Since $R_3 = 4 + X_3$ and $X_3 \in \{0,1,2,3\}$, the probabilities are:
$$ \Pr(R_3=4)=\Pr(X_3=0)=\frac{5}{15}=\frac{1}{3} $$$$ \Pr(R_3=5)=\Pr(X_3=1)=\frac{6}{15}=\frac{2}{5} $$$$ \Pr(R_3=6)=\Pr(X_3=2)=\frac{3}{15}=\frac{1}{5} $$$$ \Pr(R_3=7)=\Pr(X_3=3)=\frac{1}{15} $$Dependence note
It is tempting to think
$$ \Pr(C_1\cap C_2\neq \varnothing,\; C_1\cap C_3\neq \varnothing,\; C_2\cap C_3\neq \varnothing) ::= \left(\frac{1}{3}\right)^3, $$but that would require independence of the three intersection events, which does not hold.
One clean way to see $\frac{1}{15}$ is the “six points + random matching” argument:
- With probability $1$, the six endpoints are distinct. Sort them around the circle and label them $1,2,3,4,5,6$ in cyclic order.
- The three chords correspond to a perfect matching of these six points. There are $(6-1)!!=15$ perfect matchings, all equally likely.
- Exactly one matching gives three pairwise crossings: $(1,4)$, $(2,5)$, $(3,6)$.
Therefore,
$$ \Pr(\text{all three pairs intersect})=\frac{1}{15}. $$As a quick check:
$$ \Pr(E_{13}\cap E_{23}\mid E_{12}) ::= \frac{1/15}{1/3} = \frac{1}{5} \neq \left(\frac{1}{3}\right)^2 = \frac{1}{9}. $$