布冯投针

找一根铁丝弯成一个圆圈，使其直径恰恰等于平行线间的距离d。可以想象得到，对于这样的圆圈来说，不管怎么扔下，都将和平行线有两个交点。因此，如果圆圈扔下的次数为n次，那么相交的交点总数必为2n。 现在设想把圆圈拉直，变成一条长为πd的铁丝。显然，这样的铁丝扔下时与平行线相交的情形要比圆圈复杂些，可能有4个交点，3个交点，2个交点，1个交点，甚至于都不相交。 由于圆圈和直线的长度同为πd，根据机会均等的原理，当它们投掷次数较多，且相等时，两者与平行线组交点的总数可望也是一样的。这就是说，当长为πd的铁丝扔下n次时，与平行线相交的交点总数应大致为2n。现在转而讨论铁丝长为l的情形。当投掷次数n增大的时候，这种铁丝跟平行线相交的交点总数m应当与长度l成正比，因而有：m=kl，式中k是比例系数。为了求出k来，只需注意到，对于l=πd的特殊情形，有m=2n。于是求得k=(2n)/(πd)。代入前式就有：m≈(2ln)/(πd)从而π≈(2ln)/(dm)

-Find a wire bent into a circle, so that the diameter is precisely equal to the distance d between the parallel lines. Imagine, for this circle, no matter how dropped and parallel lines have two points of intersection. Therefore, if the circle dropped the number of n times, then the total number of the intersection point of intersection must be 2n. Now imagine the circle straightened into a long πd the wire. Obviously, this wire dropped the case of parallel lines intersect than circle complex, there may be four intersection of the three intersection 2 intersection, an intersection, and even do not intersect. As the circle and the length of the straight line with πd, according to the principle of equality of opportunity, when they throw more frequently, and equal, the total number of the intersection of two parallel lines group is also expected the same. This means that when the wire length πd dropped n times the total number of the intersection with the parallel lines intersect should b