Comparing distributions

I was thinking if you have these various distributions, like here are four different normal distributions:

I got to wondering how you could find out if you randomly pick a number from say the green distribution, how often a number picked from say the red distribution would be greater. One way of doing it is simulating say a million trials on a computer and seeing how often one was bigger than the other. So doing that first:

The answer is 94.8768%

  But I was more interested in an exact solution so I thought of this formula:

I'll explain the reasoning in a second, let's see that it works first:

It came out to 94.8753% or pretty much the same. 

Anyway Maple makes it look really ugly but the just above is evaluating the integral in the next image up with the two distributions and over a suitable interval. And you see the result is the same to within a couple millionths as the computer simulation. Probably my formula is correct so I'll explain the reasoning:

Looking at the bounds of integration the inner integral is from r to d and the outer is from a to d, that means the integral is only being evaluated over points q,r where q is greater than r. Like for a particular r in the outer integral the inner one looks at every q greater than that r up to d and so on. Q>R

So for every possible pair of q,r such that q>r the function being evaluated is X(q)*Y(r) which is to say the probability of choosing a point q,r. There's a certain probability of picking a q from X and a certain probability of picking an r from Y so X(q)*Y(r) is the probability of picking both. So when you add up all those infinitesimal probabilities you end up with the total probability of picking a pair such that q > r for those two distributions. 

So in other words you have the probability if you pick a number from each distribution that the first is greater than the second. 

*Edit* 

This formula scales to any number of variables, incidentally. Here is an example where you have 6 independent uniform variables ranging from 0 to 1. And you want to know the likelihood of them being chosen so that each one you choose is bigger than the last one. 

So it turns out 1/6! which makes sense there are 6! possible orderings and any one should be one over that probability. The 1^6 I just wrote it like that so you could see it would be six distributions multiplied together. 

*UPDATE 2

Another interesting thing to do is something like calculate the probability that 3 uniformly random numbers between 0 and 1 will be able to form a triangle. You can set up the integral like this:

C is forced to be between b and a+b and b is greater than a. But there are 3 ways to choose which variable will be the hypotenuse so all in all there is a 50% chance that the three numbers will be able to form a triangle. 

My campaign

I thought I could run for senator and have this website where people could discuss each bill and vote on it and I promised however the vote on the site came out that is how I would vote. Each bill could have a page something like:

Ping Pong waterer

I removed the other blog post about the rain barrel water distributor. Here is my new design:

Basically there is a rain bucket with a hose going to the bottom, and on the other end there is the top of a plastic soda bottle with a ping pong ball in it. Water will I hope drain from the rain barrel that's at a higher elevation but as the water gets higher in the container being filled the ping pong ball floats until it blocks off the water from filling it any more. Then as a pet or a plant drinks the water the ping pong ball sinks with the level of the water and open the flow from the barrel again.

*EDIT*
Works like a charm here is a picture. The orange thing in the neck of the bottle is the ping pong ball. The only thing different from the design above is there has to be a hole in the bottle just below where the ping pong ball seals it off so the air can get out of the bottle.

To see how this problem is normally solved look at this:
http://en.wikipedia.org/wiki/Float_valve
Mine's a lot simpler isn't it?

Sum of sine of decreasing angles

I thought of this sum to check and see if Maple knew how to calculate it:

You see you add the sum of the sin of pi/2, pi/4, pi/8... But maple does not know what it is.

But I found that it seems to have a finite limit...

It seems to be converging to something like 2.481... which doesn't get any google results for the first few digits the way most well known mathematical limits would. 

Just a reminder the sin(pi/2) = 1, the sin(pi/4)=sq(2)/2, and then:

sin(pi/8)=

So anyway when you add these all up it apparently has a finite sum but it doesn't seem anyone has mentioned this particular limit before. 

one formula for regular polyhedra

I never realized this before but one formula does the area for every regular polygon's area given a perimeter and a number of sides. And then the limit as the number of sides goes to infinity gives a formula for the area of the circle based on circumference.