吃饭的时候突然想到蒲丰投针问题。

在许多地方都遇到过此针，高中概统，大学概统，随机过程的概率部分，今天读Shannon的传记，他女儿不小心把针洒地上了。

问题描述：

在平面中有一组间隔为a的平行线，随机往平面上投掷重量均匀的长为$l(l<a)$的针，其与平行线相交的概率与$\pi$有关。

证明：

一种常见的证明方法是设运用概率统计的知识。

设针的中点到最相邻的平行线的距离为$x$，夹角为$\theta$。

考虑的区域为$x\in (0,\frac{a}{2})$，否则将针归入相邻区域。$\theta \in (0,\pi )$。

当$x<\frac{l}{2}sin\theta$时，针与线相交。

故$x$ ~ $U(0,\frac{a}{2})$，$\theta$ ~ $U(0,\pi )$，且两随机变量相互独立。

联合概率密度$f_{X\theta }(X,\theta )=f_X(x)f_{\theta }(\theta )=\frac{1}{\frac{a}{2}-0}\frac{1}{\pi -0}=\frac{2}{a\pi }$

故所求概率$P={\iint }_{x<\frac{l}{2}sin\theta }{f}_{X\theta }(X,\theta )dxd\theta =\frac{2}{a\pi }{\int }_0^{\pi }d\theta {\int }_0^{\frac{l}{2}sin\theta }dx=\frac{2}{a\pi }\frac{l}{2}2=\frac{2l}{a\pi }$
Q.E.D.

由上述结果，我们可以得到：$\pi =\frac{2l}{aP}$

明白了这个问题如何解决，编写程序简单列算式即可。

%code for testing Buffon's neddle.
%2022/12/4
clear,close all
%%%%%%%%%%%%%%%%%%%%%%%%parameters setting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a=1;%distance between parallel
l=.5;%length of Buffon's neddle
x=0;%distance of middle point of Buffon's neddle to line
th=0;%angle of neddle and line
cnt=0;%count number of crossing
test_time=10000000;
%%%%%%%%%%%%%%%%%%%%%%%%parameters setting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
tic
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%main function%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:test_time
    x=rand(1)/2;
    th=rand(1)*pi;
    if x<l/2*sin(th)
        cnt=cnt+1;
    end
end
2*l/(a*cnt/test_time)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%main function%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
toc

运行结果如下


可以看到，结果确实是随机的，而且并不是越大就一定越接近$\pi$

由此可见，L·巴杰质疑拉兹瑞尼不无道理（笑）