Using built-in random variable [Uni] from uniform continuous distribution over [0; 1] interval we sample two numbers (x,y) then check if point is inside the circle. Operation is repeated n-times, then proportion falling into circle is being measured. This is only 1/4 of the circle (positive x and positive y) with radius 1, so 4 times proportion should give estimation of pi.

Argument n = new Argument("n = 100000");
Expression e = new Expression("4 * sum(i, 1, n, if( [Uni]^2 + [Uni]^2 <= 1; 1; 0) ) / n", n);
mXparser.consolePrintln("Res. : " + e.getExpressionString() + " = " + e.calculate());

[mXparser-v.3.0.0] Res. : 4 * sum(i, 1, n, if( [Uni]^2 + [Uni]^2 <= 1; 1; 0) ) / n = 3.14748

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