[unreality]

You have an infinite plane filled with infinite parallel lines, all parallel to the same line, each line a unit distance apart... kind of like a giant sheet of paper with straight lines drawn across at regular intervals.

You drop a needle of unit length (ie, the same length as the distance between the parallel lines) onto the infinite plane.

What is the probability that it will cross one of the lines?

Specifications:

* by "crossing a line", I mean having points along the needle that fall on either side of the line. So a needle that rests exactly along a line is NOT considered to cross the line. Neither is a needle that is perfectly vertical with its bottom point on one line and its top point on another (but all other vertical needles WILL cross one of the lines obviously)

* all points and lines are geometric, ie, infinitely thin

Would you, at first glance, expect this problem to be related to the number π (pi)? It is

This problem can be solved with calculus for sure - but I'm asking a secondary question: can the problem be solved WITHOUT calculus?

Good luck

[Prof. Templeton]

If I remember correctly, the example I saw was 2 times the unit distance between the lines (X) times the number of drops of the pin (D) divided by number of times it hits the line (H).

Or Π ≅ 2XD/H

Pretty sure that was the form when the pin was the same as the unit distance.

[unreality]

Prof T: yeah I think that's a way you can use to estimate pi. I saw a java applet that let you do it, it took a lot of drops to get close to pi though lol

~~

more generally: what is the probability?

[unreality]

Here is the answer

the probability is 2/π (2 divided by pi)

But why?

[HoustonHokie]

This is going to be incomplete, because I don't have the time to do it all right now.

To me, the relation to pi seemed obvious because the probability of crossing a line is directly dependent upon the rotational angle of the needle. Angles make me think of circles and pi, so it made sense.

The question of probability boils down to the location of the center of the needle and the angle it makes with the parallel lines. If the center of the needle lies directly on one of the lines, there is only one angle (exactly parallel) where the pin doesn't cross a line. So if that situation occurs, the chance is approximately 100% of crossing the line. If, on the other hand, the center of the needle lies at the midpoint between parallel lines, there is 0% chance of the needle crossing a line. In between, the percent chance for a given location can be defined in terms of the rotation angle.

In my mind, I can see something like a sine wave with amplitude = 1 at the lines and amplitude = 0 at the midpoint defining the probability for a given location, and the area under it defining the ultimate probability independent of location. That's probably the basis for the calculus approach. Maybe these thoughts will give someone else the tools to get that explanation. If not, I'll work on it later when I have more time.

Edited January 4, 2009 by HoustonHokie

[Guest]

I'm not sure about the logical proof, but here's the calculus one:

Only the angles of 0 through pi/2 need be considered, because every other angle around the circle can be equated to an angle between 0 and pi/2 by mirroring and or turning your head.

The overall probability is therefore the average of all the probabilities when the needle lands between 0 and pi/2.

This can be computed as the integral from 0 to pi/2 of p(x)dx divided by pi/2. The integral adds up all the possible probabilities and you divided by the range of probabilities considered (pi/2) to get the average.

It's hard to demonstrate on a forum, but if you draw it out, you can see geometrically how the probability p(x) is determined for each angle the needle lands at. Essentially when the lands, create a right triangle with the needle as the hypotenuse and find the length k of the opposite side from the angle. If we consider the unit length to be 1 for ease, the probability turns out to be k, because if you start with the right bottom of the needle touching one of the infinite lines and move up a unit length of 1, it can be seen the times the needle does not cross the line above is for the length of (1-k) and therefore does cross for k. Thus, p(x)=sin(x).

Placing this back in the formula, we have the integral from 0 to pi/2 of sin(x)dx divided by pi/2 = (-cos(pi/2)+cos(0))/(pi/2) = 1/(pi/2) = 2/pi

[unreality]

great job with the calc solutions

Is it possible to solve without calculus?

The answer could be Yes/No - if Yes, show your solution, if No, show your proof ;D

[Guest]

Another follow on problem.... how does the probability change if you bend the needle 90 degrees in the centre?

You get the same answer if you bend the needle in absolutely any other way shape or form (in the plane)

[unreality]

No one has posted a non-calculus solution yet so I'll post mine It may "cheat" in a certain part, it's up to you to decide whether it could be considered wholly non-calculus

We only need to consider two horizontal parallel lines. Imagine the distance between as 1 wide unit of some measurement. We'll just say units. Now mentally draw a vertical line perpindicular to the lines that runs between them and stops when it hits either line. Put a dot where this new line touches the bottom parallel line, and another dot halfway along the perpindicular line.

Now say the center of the circle (call this point 'c') is at distance 'd' from the bottom parallel line. This distance can range from the lower dot to the upper dot (ie, d ranges from 0 to 0.5 from the bottom line).

Imagine a third horizontal parallel line cutting across and going through point c. Now the half-needle above this line can be ignored, as it will NOT intersect any of the lines. What matters is the half-needle that could be in this area. This lower half-needle could be at any angle from 0 to pi. Now imagine it at the exact angle so that the end of the half needle (the half-needle has length of 1/2 units) touches exactly at the bottom parallel line. Mark this line. Now imagine the needle in the horizontally opposite position, going the other direction and touching the line on the other side. I'll make a drawing to illustrate:

The whole angle (not the half-angle) in the zoomed-in triangle is theta.

cos(theta/2) = d/.5 = 2d

theta/2 = arccos(2d)

theta = 2*arccos(2d)

Thus, for a given distance 'd', theta is the angle at which the needle will cross a line. So the probability of this happening, as a function of 'd', is:

p(d) = 2*arccos(2d)/pi

2/pi is a constant, so our variable here is arccos(2d).

d can range from 0 to 0.5, therefore the input of the arccosine ranges from 0 to 1.

So we need to average up all of the input values of an arccosine from 0 to 1. I'm pretty sure this is what the integral is for in calculus, but I cheated and wrote a computer program... and the more iterations I ran, the closer and closer the average got to 1. We could formally define it as the limit and stuff, but we're safe in saying that, as d ranges from 0 to 0.5, the arccosine of 2d averages a value of 1

so 2/pi * 1 = 2/pi

Therefore the probability is 2/pi

[unreality]

There is an infinite sequence that works as an algorithm for calculating arccosines, when I get the time I can find it and play around with it, seeing if I can prove with it that the average arccos from 0 to pi is 1... this would be more "pure" than my computer-program solution and would wrap up the non-calculus solution nicely

[HoustonHokie]

No one has posted a non-calculus solution yet so I'll post mine It may "cheat" in a certain part, it's up to you to decide whether it could be considered wholly non-calculus

We only need to consider two horizontal parallel lines. Imagine the distance between as 1 wide unit of some measurement. We'll just say units. Now mentally draw a vertical line perpindicular to the lines that runs between them and stops when it hits either line. Put a dot where this new line touches the bottom parallel line, and another dot halfway along the perpindicular line.

Now say the center of the circle (call this point 'c') is at distance 'd' from the bottom parallel line. This distance can range from the lower dot to the upper dot (ie, d ranges from 0 to 0.5 from the bottom line).

Imagine a third horizontal parallel line cutting across and going through point c. Now the half-needle above this line can be ignored, as it will NOT intersect any of the lines. What matters is the half-needle that could be in this area. This lower half-needle could be at any angle from 0 to pi. Now imagine it at the exact angle so that the end of the half needle (the half-needle has length of 1/2 units) touches exactly at the bottom parallel line. Mark this line. Now imagine the needle in the horizontally opposite position, going the other direction and touching the line on the other side. I'll make a drawing to illustrate:

The whole angle (not the half-angle) in the zoomed-in triangle is theta.

cos(theta/2) = d/.5 = 2d

theta/2 = arccos(2d)

theta = 2*arccos(2d)

Thus, for a given distance 'd', theta is the angle at which the needle will cross a line. So the probability of this happening, as a function of 'd', is:

p(d) = 2*arccos(2d)/pi

2/pi is a constant, so our variable here is arccos(2d).

d can range from 0 to 0.5, therefore the input of the arccosine ranges from 0 to 1.

So we need to average up all of the input values of an arccosine from 0 to 1. I'm pretty sure this is what the integral is for in calculus, but I cheated and wrote a computer program... and the more iterations I ran, the closer and closer the average got to 1. We could formally define it as the limit and stuff, but we're safe in saying that, as d ranges from 0 to 0.5, the arccosine of 2d averages a value of 1

so 2/pi * 1 = 2/pi

Therefore the probability is 2/pi

IMHO, it would be splitting hairs to say that this is non-calculus. You have done all the work that forms the very definition of an integral without actually using the established calculus "rules" for performing integration of the function. One way to define the integral of a function from f(x=a) to f(x=b) is the area between the function and the x-axis between x=a and x=b. In calculus, you prove this by taking the limit of the area under the function as the difference between a & b becomes infinitesimally small. You've done exactly that by increasing the number of iterations in your program. In order to get the "exact" answer, your program would have to have infinite iterations. Of course, you can see what the answer is tending towards with a finite number of iterations and make the logical leap to that solution, but to get the exact answer I think you need the calculus. You didn't come up with the integral function so the area could be directly evaluated, and maybe that leaves you in the non-calculus realm -- but just barely. The closer you get to the exact solution, the more you're doing calculus

[bonanova]

... seeing if I can prove with it that the average arccos from 0 to pi is 1...

Arccos[x], or cos^-1[x] is the angle theta for which cos[theta] = x.

Do you want the the average value of cos^-1[x] for x in some range?

Not for x from 0 to pi, because cos^-1[x] is not defined when |x|>1.

That is, there is no angle theta for which |cos[theta]| = pi.

If you want the average value of x over the region for which cos^-1[x] goes from 0 to pi, that would be 0.

cos^-1[1] = 0

cos^-1[0] = pi/2

cos^-1[-1] = pi

[unreality]

yeah post #10 was a typo - I meant 0 to 1, not 0 to pi Everywhere else I said 0 to 1