There are 3 sections on this web page MCintro.html Study the first two sections and skip/ignore the third section. Read em: 1. Introduction to Monte Carlo Methods 2. Monte Carlo Calculation of Pi Ignore it: 3. Monte Carlo Computation of Population Distribution
"For example, consider a circle inscribed in a unit square. Given that the circle and the square have a ratio of areas that is ?/4, the value of ? can be approximated using a Monte Carlo method:[4] Draw a square on the ground, then inscribe a circle within it. Uniformly scatter some objects of uniform size (grains of rice or sand) over the square. Count the number of objects inside the circle and the total number of objects. The ratio of the two counts is an estimate of the ratio of the two areas, which is ?/4. Multiply the result by 4 to estimate ?. In this procedure the domain of inputs is the square that circumscribes our circle. We generate random inputs by scattering grains over the square then perform a computation on each input (test whether it falls within the circle). Finally, we aggregate the results to obtain our final result, the approximation of ?."
In class today, we threw bags of fruit snacks at targets down on the floor below us. Our domain of inputs was the 9 carpet square area with the colored large construction paper sheet within it and the sheet of paper taped inside of that. We generated random inputs by throwing the fruit treat bag at the target area. Too much skill at hitting the target would mean its NOT random. But we were far enough away, plus the treats bounced too. Count where each of the ten shots landed. Get the ratios.
Homecoming Eve class: Excel and Monte Carlo. The random numbers would be random x,y pairs, such as (0.5, 0.3) and (0.9, 0.8) and (0.7, 0.7) and (0.9, 0.9). How far is each of the above (x, y) points from the origin, from the point (0, 0)? Distance between point (x1, y1) and point (x2, y2) is: The square root of ( (x2 - y1) squared = (y2 - y1) squared ) Since we have point (x1, y1) here as (0, 0), the origin, its just that much simpler: Distance from (0.9, 0.8) to the origin is: square root of ( (0.9 - 0) squared + (0.8 - 0) squared ) square root of ( (0.9) squared + (0.8) squared ) square root of ( 0.81 + 0.64 ) or square root of ( 1.45 ) = 1.204 So the distance of that point (0.9, 0.8) from the origin point (0, 0 ) is 1.204 and it would be OUTSIDE of a circle of radius 1, for any circle having its center at (0, 0) and having radius = 1 and diameter = 2.