# Fun with RNGs: Calculating π

So, calculating π is a fun pastime for people it seems. There are many ways to do it, but this one is mine. It’s 12 lines of code, it wastes a lot of electricity and it takes forever to converge.

{% codeblock lang:csharp %} public double EstimatePi(int numberOfTrials) { var r = new Random();

```
return 4 * Enumerable.Range(1, numberOfTrials)
.Select(o => {
var x = r.NextDouble();
var y = r.NextDouble();
return Math.Pow(x, 2) + Math.Pow(y, 2) < 1 ? 1 : 0;
})
.Average();
```

} {% endcodeblock %}

What’s going on here? First we initialize our random number generator. Then for 1 to the number of trials we specify in the argument we do the following:

- Generate two random numbers between 0 and 1. We use one for the X coordinate and one for the Y coordinate of a point.
- We test if the point (X,Y) is inside the unit circle by using the formula for a circle (x^2 + y^2 = r^2).
- If the point (X,Y) is inside the circle we return a 1 otherwise a zero.

Then we take the average of all those zeros and ones and multiply it by a magic number, 4. We have to multiply by four because the points we generate are all in the upper right quadrant of the xy-plane.

How bad is it? Here’s some output:

```
Number Of Trials Estimate of Pi
10 3.6
100 3.24
1000 3.156
10000 3.1856
100000 3.14064
1000000 3.139544
10000000 3.1426372
100000000 3.14183268
1000000000 3.141593 (Took 2:23 to complete)
```