Buffon's Needle: a Proof

Date: 08/13/97 at 07:51:37
From: Mohammad Shakil Akhtar
Subject: Buffon's needle

I have known the Buffon's needle problem for the last 28 years and 
have yet to come across its theoretical proof. I have used this 
theorem with great enthusiasm with students (secondary) where we use 
it to calculate the value of pi using statistical methods by throwing 
some needles (toothpicks) on a ruled paper.

The formula we use is  pi = 2(total no. of throws)/no. of cuts

No. of throws means the total number of needles thrown (like 10 
needles thrown 100 times = 1000). 

Cuts = needles landing with an intersection with any line. 

Distance between parallel lines = length of the needle.

I would be very grateful if I could get its proof. I am amazed by 
how come pi and needles. What is the connection?

I am a teacher and a graduate from London. Thanks in advance. 

Date: 08/13/97 at 13:01:12
From: Doctor Anthony
Subject: Re: Buffon's needle

Let the needles have length L1 and the parallel lines be drawn a 
distance L2 (L2 > L1) apart.  A 'success' occurs when any part of a 
needle cuts a line.

We can think of the centre of the needle being uniformly distributed 
between 0 and L2/2.  Let the smaller of the angles between the 
direction of a needle and the parallel lines be theta, so that theta 
is uniformly distributed between 0 and pi/2.

If y is the distance of mid-point of the needle from the closest line, 
then we get an intersection if:

   y < (L1/2)sin(theta)

We now draw two axes with y up the vertical axis varying from 0 to 
L2/2, and theta along the horizontal axis varying from 0 to pi/2.  
The sample space is any point within this rectangular area = 
(pi/2)(L2/2).  If you draw the curve

   y = (L1/2)sin(theta)   

from 0 to pi/2, then the area under this curve divided by the total 
area of the rectangle will give the probability of an intersection.

The area under the sine curve is INT[(L1/2)sin(theta)]

                            = -(L1/2)cos(theta) from 0 to pi/2

                            =  -L1/2[0 - 1]  = L1/2

                                    L1/2
Probability of an intersection =  ---------
                                  (pi/2)(L2/2)

                                    2L1
                 Probability   = ----------
                                   pi.L2


                       No.of cuts        2L1
  Also Probability =  ------------   = --------
                       No.of throws     pi.L2  


                       2L1.Number of throws
From this       pi =  ----------------------
                        L2.Number of cuts


[I note that you have made L1 = L2, but this is not necessary 
provided; L2 is not less than L1.]    

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 08/20/97 at 08:37:29
From: Mohammad Shakil Akhtar
Subject: Buffon's needle problem

Dear Sir, 

I was delighted to see your proof.  Most importantly, it is simple and 
elegant against all my expectations. 

Thank you very much.
Sincerely yours
M.S.Akhtar