0
$\begingroup$

I tried calculating the answer to Buffon's Needle Problem, but my answer kept coming to 4/pi. Can anyone explain why? Here is what I did:

Assume I drop the needle (length x) many times. The average horizontal length will come out to be: $[4 * \int_{0}^{\pi/2} x*cos(\theta) * d\theta] / (\int_{0}^{2\pi} d\theta) = 2 / \pi$

So the average horizontal length is the total length of the needle divided by the total number of angles.

Afterwards, I solve the variation where the distance between the cracks are equal to the length of the needle, and the length of the needle = 1.

Each crack is separated by a 0.5 border to its neighbor. The middle of the needle can land within $2/\pi$ on either side to touch the crack, so the probability should be $4/\pi$.

I saw the solution, and they just stopped at my calculation of the average length. They said it was, instead, the probability. What is wrong with my reasoning?

$\endgroup$
  • $\begingroup$ Confused. What does "the average length" mean? The length is fixed, no? What gives $2/\pi$ is the probability of the needle touching a line... $\endgroup$ – leonbloy Aug 5 '16 at 2:29
  • $\begingroup$ Sorry, I mean Average length as in, if I drop the needle an infinite amount of times on the ground with random orientations, what would the average horizontal length be. $\endgroup$ – Goldname Aug 5 '16 at 2:31
  • $\begingroup$ "the average horizontal length is the total length of the needle divided by the total number of angles." There is no "total number of angles", that makes no sense. $\endgroup$ – leonbloy Aug 5 '16 at 2:48
  • $\begingroup$ Maybe I should phrase it this way: The probability of a certain horizontal length is x*cos(theta) / (2*pi). The Integral of this with respect to theta will be the average length, correct? $\endgroup$ – Goldname Aug 5 '16 at 2:51
  • $\begingroup$ No, "the probability of a certain horizontal length" is zero. See my answer for a correct derivation of the average horizontal length. $\endgroup$ – leonbloy Aug 5 '16 at 2:55
1
$\begingroup$

Le $h$ be the "horizontal length", hence $h= x \cos(\theta)$. Assuming $\theta$ is uniform on $[0,\pi/2]$, we have

$$E[h]=x E[\cos(\theta)]=x \frac{\int_{0}^{\pi/2} \cos(\theta)d\theta}{\pi/2}=x\frac{2}{\pi}$$

Now, what seems to confuse you (I'm not sure, your question is not clear) is how this average relates with $P(E)$, the probability of the event "needle touched the lines". Let's see:

$$P(E)= \int_0^x P(E|h) P(h) dh $$

Now, $P(E|h)=1$ for $h>d$ and $P(E|h)=h/d$ otherwise. If we assume $x<d$, then

$$P(E)= \int_0^x \frac{h}{d} P(h) dh =\frac{1}{d} E[h]$$

The above shows that the average horizontal length is directly related to probability of the event of interest (or equivalently, the average number of "successes"), but only when $x<d$.

$\endgroup$
  • $\begingroup$ Yes this was my question. However, I am kind of confused about your answer. What is P(h) equal to? I am having some trouble seeing where you got your second equation from, and how the third equation is evaluated. $\endgroup$ – Goldname Aug 5 '16 at 3:12
  • $\begingroup$ $P(h)$ is the probability density of $h$. The second equation corresponds to the law of total probability for a continuous variable (see eg eq 5.16 here ) . Of course, this answer assumes you know those concepts from probability theory. $\endgroup$ – leonbloy Aug 5 '16 at 12:17
  • $\begingroup$ Ahh I see. I have never taken probability theory. Is probability theory needed to understand the evaluation of the third equation? $\endgroup$ – Goldname Aug 5 '16 at 15:49
  • $\begingroup$ The third equation uses the definition of expected value en.wikipedia.org/wiki/… $\endgroup$ – leonbloy Aug 5 '16 at 16:23
  • $\begingroup$ From what you have shown, the calculated average length is correct, but I don't understand why we can't simply use the geometric distance between cracks and the average length as a simple way to calculate the probability. $\endgroup$ – Goldname Aug 5 '16 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.