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### Buffon's Needle: a Proof

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Date: 08/13/97 at 07:51:37
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
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
```
Associated Topics:
Middle School Pi

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