The video I referred to in the earlier post only shows a Monte Carlo solution but we can find the exact values quite easily using the density functions.
Because the two candles (or points) are assumed to be uniformly randomly distributed in the circle (or the cake), an important parameter that keeps repeating in all the cases below is the area of a circular segment which can given in terms of the central angle $t$ as $(t-\sin t)/2$. The probability that a point randomly chosen in a unit circle lies in this segment is then $(t-\sin t)/2/\pi$.
Note that the density we found in the earlier refers are for the random variable $\theta$, the angle the chord makes with the tangent. The angle subtended by the chord is twice this value.
Let $E$ denote the event that two randomly chosen points on the circle lie on the opposite side of the chord chosen according to the following methods.
(1) Circumference method
Like before, if we let $\theta$ be the angle between the tangent and the chord, the central angle becomes $2\theta$. Therefore, the probability that a point randomly chosen in the circle lies in the segment created by this chord is the ratio of the segment's area to that of the circle. Therefore,
$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\,\frac{d\theta}{\pi/2}=\frac{1}{3}-\frac{5}{4\pi^2}\approx 0.20668$
Even though we can solve this easily, Wolfram Alpha does the job perfectly.
(2) Radial line method
We know from our earlier post that the density function of the tangent angle in this case is just $\sin\theta$. Therefore the required probaility
$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\,d\theta=\frac{128}{45\pi^2}\approx 0.28820$
This can also be simplified manually, but again Wolfram Alpha minimizes our effort and shows
Note that this is $\pi$ times the expected distance between two points chosen at random inside a circle. Coincidence?
(3) Midpoint method
Again, using the density from the previous post, the probability in this case is,
$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin2\theta\,d\theta=\frac{1}{8}+\frac{2}{3\pi^2}\approx 0.19255$
Simple even without Wolfram Alpha.
(4) Random point and line at random angle method
Like the cases above we use the density function for this case that we already found in the previous post.
$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot2\sin^2\theta\,\frac{d\theta}{\pi/2}=\frac{1}{3}\approx 0.33334$
Solved with Wolfram Alpha.
(5) Random point on a radial line and a line at random angle method
Diving right into the probability with the density function at our disposal,
$\displaystyle\mathbb{P}(E)=\int_0^{\pi/2}2\left(\frac{2\theta-\sin2\theta}{2\pi}\right)\left(1-\frac{2\theta-\sin2\theta}{2\pi}\right)\cdot\sin\theta\sinh^{-1}(\tan\theta)\,\frac{d\theta}{\pi/2}\approx 0.386408$
This time Wolfram Alpha only gives us a numerical approximation. But with Mathematica, we can confirm that the exact value is
$\displaystyle \mathbb{P}(E)=\frac{427+60\pi^2-480\log 2}{180\pi^2}$
For the sake of completion, let's finish this post with one more method that was used in the video. Here, the cut is created by choosing two random points on the circumference while the two candles were chosen randomly on a randomly chosen radius.
The first point on the circumference can be chosen to be the 'south pole' of the circle. After the second point is chosen on the circumference, let the chord subtend an angle of $2t$ at the centre. Using the symmetry of the problem, we can limit the range of $t$ to be in $(0,\pi/2)$.
We now calculate the probability that the first candle lies to the right of the chord and the second on the left. Let the radii selected for the first candle make an angle $x$ with the radius perpendicular to the chord. We get a a non-zero probability only if $-t \leq x \leq t$. When $x$ lies in this range, the probability of the first candle landing to the right of the cut is $1-\cos t/\cos x$ and $0$ otherwise.
Similarly, let $y$ be the angle between the radius chosen for the second candle and the radius perpendicular to the cut. When $-t \leq y \leq t$, the probability of the second candle ending up to the left of the cut is $\cos t/\cos y$. Else, the probability is $1$.
Therefore probability in this case depends on the following two integrals.
$\displaystyle \mathcal{I}_1=\int_0^{\pi/2} \int_{-t}^t \int_{-t}^t \frac{\cos t}{\cos y}\left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$
$\displaystyle \mathcal{I}_2=\int_0^{\pi/2} \int_{-t}^t \int_{t}^{2\pi-t} \left(1-\frac{\cos t}{\cos x}\right)\,dy\,dx\,dt$
These are very hard to evaluate even with Mathematica. Numerically evaluating these shows,
$\displaystyle \mathbb{P}(E)=2\cdot\frac{\mathcal{I}_1+\mathcal{I}_2}{\pi/2\cdot 2\pi \cdot 2\pi}\approx 0.161612$
Hope you enjoyed this.
Until then
Yours Aye
Me
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