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Problem: Find the probability of 3 randomly thrown darts landing in the same half of the board. More generally, if $n$ points picked uniformly randomly on a disk, find the probability of them lying in the same half of that disk.

I tried relating the problem to 2 other problems:

  1. Picking points on the circumference, then the probability of all points lying in the same half is $\frac{n}{2^{n-1}}$. But the idea behind this answer uses a sum of disjoint probabilities. The same idea with a disk won't lead to disjoint cases.
  2. The idea in 3b1b's video on the Putnam problem. In our case we can start with a random n-gon on the disk and then try to write the probability of this $n$-gon containing the center of the disk in its interior. The answer we're looking for will be the complement of this probability. Would appreciate some help in extending this.
I ran a quick MC sim for the situation after struggling with logical approaches for a while, looks like the required probability is $\frac{1}{2}$. [Check this][2] for theoretical answers to the question, the consensus there being $\frac{3}{4}$. 
import numpy as np

def gen1(n):
    r = np.sqrt(np.random.uniform(0, 1, n))
    theta = np.random.uniform(0, 2 * np.pi, n)
    x = r * np.cos(theta)
    y = r * np.sin(theta)
    return x, y
def gen2(n):
    pointsx,pointsy = [],[]
    while(len(pointsx)<n):
        x = np.random.uniform(-0.5,0.5)
        y = np.random.uniform(-0.5,0.5)
        while(x2+y2>1):
            x = np.random.uniform(-0.5,0.5)
            y = np.random.uniform(-0.5,0.5)
        pointsx.append(x)
        pointsy.append(y)
    return np.array(pointsx),np.array(pointsy)


def check_same_half(x, y):
    angles = np.arctan2(y, x)
    min_angle = np.min(angles)
    max_angle = np.max(angles)
    return max_angle - min_angle <= np.pi


def monte_carlo_sim(trials, n, gen):
    count_same_half = 0
    for _ in range(trials):
        x, y = gen(n)
        if check_same_half(x, y):
            count_same_half += 1
    return count_same_half / trials


trials = 100000
n = 3


result_1 = monte_carlo_sim(trials, n, gen1)
print(f"Simulation 1: Probability of points lying in the same half-disk: {result_1:.4f}")
result_2 = monte_carlo_sim(trials, n, gen2)
print(f"Simulation 2: Probability of points lying in the same half-disk: {result_2:.4f}")

Would love to hear more thoughts on the problem and if I've made any errors somewhere.

Hex1729
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  • Define "the same half". For an obvious interpretation (decide what the halves are before throwing the darts), the answer is $\frac{1}{4}$. – Naïm Camille Favier Jul 18 '24 at 16:22
  • My reading of the code is we can draw the boundary in any way that fits the darts after they're thrown. Using the geometry of inscribed triangles that probability can be proved as $3/4$. – Oscar Lanzi Jul 18 '24 at 16:56
  • Is "half" a semicircle (semidisk), chosen after seeing the darts? Or could it be the inner circle of half the area (or perhaps the outer ring)? – Henry Jul 18 '24 at 17:57
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    Here is a nice proof for the complement of your question. – zetko Jul 18 '24 at 18:17
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    With probability $1$ no point is zero. Then, the directions are uniformly distributed on the unit circle. So the answer is the same as for points on a circle. – WimC Jul 18 '24 at 19:08
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    Your check_same_half function is flawed. When the difference between min_angle and max_angle is greater than pi, the third angle could also have a difference greater than pi from one of the others, thereby satisfying the condition. – Daniel Mathias Jul 18 '24 at 21:55
  • modified python code – Daniel Mathias Jul 18 '24 at 22:04
  • @DanielMathias ahh good catch – Hex1729 Jul 22 '24 at 20:06
  • @Henry In other words, the problem asks "by what probability does there exist a diameter of the disk such that all chosen points are on one side of it". And no semicircles/semidisks are fixed from the start. – Hex1729 Jul 22 '24 at 20:10

1 Answers1

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The “darts on a board” version is equivalent to the “points on a circle” version. When you throw darts on a board, you can radially project the darts onto the circumference of the dartboard; these projected points are uniformly distributed on the circumference. The darts will lie in some half of the board if and only if the projected points lie in some common semicircle.

Mike Earnest
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