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I know that we can have an estimate of $\pi$ using Monte Carlo methods. I know that we make a huge loop and each time we take randomly two numbers in the range [0,1] and calculate the number of times this point is inside a circle of radius 1, but I don't know how to implement it using Mathematica.

So does anybody know how we implement this algorithm using Mathematica?

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  • $\begingroup$ I would wholeheartedly recommend checking out this demonstration $\endgroup$ – gpap Apr 26 '16 at 14:32
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There is an example in the documentation which may get you started:

pairs = RandomReal[{-1, 1}, {10000, 2}];
4 Count[Map[Norm, pairs], _?(# <= 1 &)]/10000.

(*3.1248*)

You can plot see this as:

Graphics[{PointSize[Small], Blue, Point@Select[pairs, Norm[#] <= 1 &],
 Gray, Point@Select[pairs, Norm[#] > 1 &], Red, Thick, Circle[]}, AspectRatio -> 1]

enter image description here

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Here is an approach using RandomPoint and graphics primitives:

pts = RandomPoint[Rectangle[], 10^6];  (* generate random points on the unit square *)
rm = RegionMember[Disk[{0.5, 0.5}, 0.5]];  (* RegionMemberFunction for an embedded Disk *)

Now we count the number of points that fall in the circle and divide that by the total number of points. That should equal the ratio of the areas of the circle to that of the square. So by multiplying by 4 we get the value of pi:

4. Count[rm[pts], True] / Length[pts]

3.1417

Visually:

Graphics[{Point[pts], Red, Point[Pick[pts, rm[pts]]]}]

Mathematica graphics

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