三棒

我们可以先用一张图来说明木棍被分成三块的地方。我们将第一点表示为 $\textit{x}$，将第二点表示为 $\textit{y}$。那么，这3个片段的给定长度为$\textit{x}$、$\textit{y-x}$和$\textit{1-y}$。我们正在计算 $P(min(x, y-x, 1-y) \leq 0.2)$。这对应于3个片段中最小的一个或$\textit{min(x, y-x, 1-y)}$的长度至多为0.2或小于或等于0.2的事件。

解决此问题的第一个关键观察结果是认识到 $P(min(x, y-x, 1-y) <= 0.2)$ 等于 $1 - P(x, y-x, 1-y \geq 0.2)$。这是根据补码规则。最小值事件上限的补集是所有事件都高于最小值的事件。这具有逻辑意义：最小值小于 0.2 的“相反”是所有值都高于 0.2。

我们现在可以这样直观地划分上述样本空间。我们从 $\textit{1 x 1}$ 网格开始，随机变量 $\textit{x}$ 和 $\textit{y}$ 在各自的轴上。然后，我们沿着线 $\textit{y=x}$ 将这个空间分成两半，并将我们的结果空间限制在正方形的上半部分三角形。这是因为，通过指示 y 是两个点中的第二个点或 $\textit{maximum}$，我们将结果空间限制为仅具有 $y \geq x$ 的点。

一旦我们定义了结果空间，我们就可以计算与事件 $P(x, y-x, 1-y >= 0.2)$ 对应的三角形的子区域。第一个条件是$x \geq 0.2$。我们可以对活动空间进行相应的分区和缩小。第二个条件是$1-y \geq 0.2$或$y \leq 0.8$。我们可以再次相应地缩小活动空间。最后一个条件是$y-x \geq 0.2$。然而，我们注意到剩余空间（白色的中心三角形）已经满足这个条件。所以我们需要计算这个三角形的面积：$0.4*0.4*0.5 = 0.08$。现在，我们将事件空间的面积除以结果空间的面积，我们在上面将结果空间定义为上半三角形，即 0.5：$0.08 / 0.5 = 0.16$。

现在请记住，需要该区域来计算所需事件的补码，因此从 1：$1 - 0.16$ 中减去它，即可得到 $P(min(x, y-x, 1-y) \leq 0.2) = 0.84$。

这是演示此解决方案的程序：

import random

def simulate_trial():
   x, y = random.uniform(0, 1), random.uniform(0, 1)
   x, y = min(x, y), max(x, y)
   sides = [x, y-x, 1-y]
   return min(sides)

res = []

for _ in range(1_000_000):
   val = simulate_trail()
   res.append(val)

print(len([i for i in res if i <= 0.2] / len(res)))

Original Explanation

We can start by illustrating a diagram indicating where the stick is broken into 3 pieces. We denote the first point as $\textit{x}$ and the second as $\textit{y}$. Then, the given length of the 3 pieces is $\textit{x}$, $\textit{y-x}$, and $\textit{1-y}$. We are looking to calculate the $P(min(x, y-x, 1-y) \leq 0.2)$. This corresponds to the event that the length of the smallest of the 3 pieces, or $\textit{min(x, y-x, 1-y)}$, is at most 0.2, or less than or equal to 0.2.

The first key observation to solving this problem is to recognize that $P(min(x, y-x, 1-y) <= 0.2)$ is equal to $1 - P(x, y-x, 1-y \geq 0.2)$. This is by the complement rule. The complement to the event upper bounding the minimum, is the event that all are above the minimum. This makes logical sense: the ‘opposite’ of the minimum being less than 0.2 is for all to be above 0.2.

We can now partition the above sample space visually as such. We start with a $\textit{1 x 1}$ grid, random variables $\textit{x}$ and $\textit{y}$ on their respective axes. We then partition this space in half along the line $\textit{y=x}$, and restrict our outcome space to the upper half triangle of the square. This is because, by indicating that y is the second or $\textit{maximum}$ of the two points, we are restricting our outcome space to only points with $y \geq x$.

Once we’ve defined the outcome space, let’s compute the sub area of this triangle that corresponds to the event $P(x, y-x, 1-y >= 0.2)$. The first condition is that $x \geq 0.2$. We can partition and reduce the event space accordingly. The second condition is that $1-y \geq 0.2$, or $y \leq 0.8$. We can reduce the event space accordingly again. The last condition is that $y-x \geq 0.2$. However, we notice that the remaining space (the white, center triangle), already satisfies this condition. So we’re left to compute the area of this triangle: $0.4*0.4*0.5 = 0.08$. We now divide the area of the event space, by the area of the outcome space, which we defined above to just be the upper half triangle, or 0.5: $0.08 / 0.5 = 0.16$.

Now remember that this area was needed to calculate the complement of the desired event, so subtract it from 1: $1 - 0.16$, to get $P(min(x, y-x, 1-y) \leq 0.2) = 0.84$.

Here is a program demonstrating this solution:

import random

def simulate_trial():
   x, y = random.uniform(0, 1), random.uniform(0, 1)
   x, y = min(x, y), max(x, y)
   sides = [x, y-x, 1-y]
   return min(sides)

res = []

for _ in range(1_000_000):
   val = simulate_trail()
   res.append(val)

print(len([i for i in res if i <= 0.2] / len(res)))