# What's wrong with my calculation of Buffon's Needle Problem?

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?

• 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... – leonbloy Aug 5 '16 at 2:29
• 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. – Goldname Aug 5 '16 at 2:31
• "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. – leonbloy Aug 5 '16 at 2:48
• 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? – Goldname Aug 5 '16 at 2:51
• No, "the probability of a certain horizontal length" is zero. See my answer for a correct derivation of the average horizontal length. – leonbloy Aug 5 '16 at 2:55

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 $x<d$.

• 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. – Goldname Aug 5 '16 at 3:12
• $P(h)$ 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. – leonbloy Aug 5 '16 at 12:17
• Ahh I see. I have never taken probability theory. Is probability theory needed to understand the evaluation of the third equation? – Goldname Aug 5 '16 at 15:49
• The third equation uses the definition of expected value en.wikipedia.org/wiki/… – leonbloy Aug 5 '16 at 16:23
• 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. – Goldname Aug 5 '16 at 19:52