# What is the probability of a pen touching a bar given that the length of the pen is $10$ cm and the bars are regularly spaced at $15$ cm?

Problem:

If a pen of length $10$ cm is thrown out of infinitely large window having vertical bars regularly spaced at $15$ cm, then find the probability that it will touch any of the bars. (Assume that the vertical bars are infinitely long)

I don't know how to even begin with the question, so any help will be appreciated. All I know is that integration is involved in the calculation of this probability, but I don't know what to integrate, since the pen can be thrown anywhere, and at any angle.

• This is the Buffon's needle problem: en.wikipedia.org/wiki/Buffon%27s_needle – Surb Jan 29 '16 at 20:44
• @SubhadeepDey It's not from one. It's inspired by one. I got it in a test paper from our institute, which prepares us for such competitions (PCM alike. It's mainly for JEE Advanced, but also helps for olympiads and all general PCM competitions). – Ashish Gupta Jan 29 '16 at 20:50
• @Surb Thank you for the link. I'll check it out. – Ashish Gupta Jan 29 '16 at 20:56

You're basically asking about a 3D version of Buffon's needle problem. The problem is that your question does not have enough information. If you assume that the pen is thrown so that all parts of the pen are always the same distance from the window (equivalently the pen lies on a plane parallel to the window) then you're asking about Buffon's needle problem, which has a well-known solution.

A slightly more complicated version is if we allow any orientation of the pen but don't allow it rotate on its way to the window, in which case we need to know at what angle you throw it at the window. Imagine throwing straight at the window like a dart; letting it drift left or right increases the chance it hits a bar, unless of course we assume it has infinite forward speed in which case the problem can easily be solved by tweaking Buffon's problem.

Finally, if you allow the pen rotate through space, your problem no longer gives us enough information as we need to know this speed. If the pen rotates infinitely fast in all directions we're essentially throwing a ball of diameter 10, and again we need to know at what angle to the left or right you are throwing the ball when facing the window.

To help solve these issues I propose the following solution. First assume you are in an infinitely tall (above and below) cylindrical cage, so that you don't need to worry about hitting the window at an angle. If you want general rotational speed then I fear your problem may be intractible, but if you only allow one degree of freedom, which is what you would get if you held the end of the pen before throwing, there may be hope.

I will interpret the problem in the simplest terms first, the classic problem assuming the pen is always crossing with a length of 10cm and not rotating.

A simple and elegant solution:

Consider a circle of diameter 15 cm. This circle will ALWAYS touch the bars at 2 distinct points.

Since the circle touches the bars at 2 points, and has a circumference of 15pi, we can say that a part of the circle with length x has a 2x/(15pi) chance of touching the line.

As the circle gets smaller, it is noticably less circle- like, indeed, approaching that of a line. This gets us thinking, and soon we realize that, independent of shape, any given point should have the same probability of touching the bars. Furthermore, we know that probability!

We can therefore say that, for any segment, of any shape, of length X, the expected number of times that the segment touches the bars is 2x/(15pi).

Since the pen can't touch the bars more than once, the expected number of times that the segment touches the bars is equal to the chance that the segment touches the bars!

Therefore, the solution is simply 2*10/(15*pi)= 20/(15*pi)

From here, it isn't so hard to consider the pen falling at any angle (Assuming no spin).

Since we've just proved this relationship to be linear, we realize that we can just take the average length of the pen as it spins. As it spins, the length of the pen is just sin(theta)*10. The sin function can be found to have an average of 1/sqrt(2) through calculus, so the answer to that question must be 20/(15*pi*sqrt(2))