riddle of the day

Two needles have the same length. One is straight; the other may be bent at will. Can a clever shape outwit chance?

The Bent Needle

Strategist
Chance and uncertainty

Riddle statement

A floor is marked by parallel lines 10 centimetres apart.

You have two needles, each 10 centimetres long:

  • the first remains straight;
  • the second may be bent at as many points as you like, so that it consists of several straight segments. Once its shape has been chosen, it remains rigid.

Each needle is dropped at random onto the floor, with no orientation or position relative to the lines being favoured.

Count one intersection whenever a segment of the needle crosses a floor line. If several segments cross the same line, count every intersection separately.

Can you bend the second needle so that, on average, it produces fewer intersections than the straight one?

Additional challenge: calculate the average number of intersections per drop.

Show solution

Solution

Answer: no. No way of bending the needle produces fewer intersections on average than the straight needle.

Consider one segment of length \(\ell\). If it makes an angle \(\theta\) with the lines, its extent perpendicular to them is

$ \ell|\sin\theta|. $

Since the lines are 10 centimetres apart, the expected number of intersections made by that segment, for this orientation, is

$ \frac{\ell|\sin\theta|}{10}. $

After averaging over all orientations, the contribution of the segment is still proportional to its length. Thus there is a single constant \(C\) such that a segment of length \(\ell\) contributes

$ C\ell. $

The straight needle has length 10, so its expectation is \(10C\).

Suppose the bent needle consists of segments of lengths

$ L_1,L_2,\ldots,L_k. $

Their expectations add. Since

$ L_1+L_2+\cdots+L_k=10, $

the total expectation is

$ CL_1+CL_2+\cdots+CL_k = C(L_1+\cdots+L_k) = 10C. $

The two needles therefore have exactly the same average. Bending the needle gives no advantage.

For the additional challenge, calculate the constant:

$ \frac1\pi \int_0^\pi |\sin\theta|\,d\theta = \frac2\pi. $

Because each needle has the same length as the distance between the lines, the average number of intersections per drop is

$ \boxed{\frac2\pi}. $

Key idea: every segment contributes an expectation proportional to its length; when the segments are added, only the total length matters.