Simulate the probability of winning the game commonly known as the Piso Game in a Peryahan (see image below) here in the Philippines.

You toss a 1-peso coin onto a board with various prizes written on it. For example, if your coin lands on the section labeled "PLATE" or "CUP," the corresponding item becomes your prize.

Consider a board size of 1200 x 1200 mm and assume that the likelihood of winning in all squares is the same. Using the Monte Carlo method, determine the probabilities of wining the game using a 20-mm dia coin when the grid size is 20mm, 60mm, 100mm until 300 mm with the interval of 40 mm. Neglect the region between squares (white areas in the image above). Consider 10,000 trials for every grid size.

Answer 1

[Ayush Aniket]

Hi Norberto,

You can write a MATLAB script to simulate the Piso game by following the steps below:

• Define Parameters: Start by defining the size of the board, the diameter of the coin, and the number of trials for your simulation.

    boardSize = 1200; % mm
    coinDiameter = 20; % mm
    numTrials = 10000; % number of Monte Carlo trials

• Specify Grid Sizes: Create an array of grid sizes ranging from 20 mm to 300 mm with an interval of 40 mm.

    gridSizes = 20:40:300; % mm

• Initialize Storage for Results: Prepare an array to store the probabilities for each grid size.

    probabilities = zeros(size(gridSizes));

• Loop Over Each Grid Size: Use a for loop to iterate over each grid size and perform the simulation

    for g = 1:length(gridSizes)
        gridSize = gridSizes(g);
        successfulTrials = 0;

• Simulate Coin Tosses: Within the grid size loop, use another for loop to simulate the coin tosses using rand function for randomly placing the center of the coin.

    for trial = 1:numTrials
        % Randomly place the center of the coin on the board
        xCenter = rand * boardSize;
        yCenter = rand * boardSize;

• Calculate Coin Boundaries: Determine the minimum and maximum x and y coordinates of the coin.

    xMin = xCenter - coinDiameter / 2;
    xMax = xCenter + coinDiameter / 2;
    yMin = yCenter - coinDiameter / 2;
    yMax = yCenter + coinDiameter / 2;

• Check Coin Position: Use modulus operations to check if the coin is entirely within a single square.

    if (mod(xMin, gridSize) >= 0 && mod(xMax, gridSize) <= gridSize) && ...
       (mod(yMin, gridSize) >= 0 && mod(yMax, gridSize) <= gridSize)
        successfulTrials = successfulTrials + 1;
    end

• Calculate and Store Probability: After the trials, calculate the probability for the current grid size.

    probabilities(g) = successfulTrials / numTrials;

By following these steps, you can simulate the coin tosses and calculate the probability of winning for different grid sizes using the Monte Carlo method.

[Walter Roberson]

I have to wonder what the rules are for the case that the coin lands crossing multiple squares?

[Ayush Aniket]

Based on the description of the problem, if the coin lands such that it crosses multiple squares, it typically means that the coin does not "win" any prize.

Please let me know if there is any wrong interpretation.

[Walter Roberson]

With a 20 mm diameter coin and a 20 mm grid size, if the rule is that coins crossing multiple squares do not win, then the winning probability is very close to 0 -- the coin would have to land at the exact center of a square in order to win. Considering one dimension, there would be 1200/20 = 60 "winning" spots out of... an infinite number of potential spots. Remember, with a 20 mm grid and a 20 mm coin, if the coin lands even 3 zeptometres away from the exact centre of the grid, then it is disqualified for crossing the line by 3 zeptometres.

[John D'Errico]

@Ayush Aniket

Please don't do student homework assignments for them, when it is clear the student has done nothign but post the assignment. This does not help the student, except to teach them there will always be some lonely person willing to do their work for them if they just make a sad face.

It does not help the forum, as it then teaches that same student to continue posting their assignments with no effort made. In fact, it actively hurts the forum, since it also teches other students to do the same.

If you want to offer help, then consider offering help that is proportionate to the effort they have shown. If no effort is shown, then just try to nudge them in the correct direction. If they clearly have shown some effort, but have just made a mistake along the way, then show how to fix it.

[Walter Roberson]

    if (mod(xMin, gridSize) >= 0 && mod(xMax, gridSize) <= gridSize) && ...
       (mod(yMin, gridSize) >= 0 && mod(yMax, gridSize) <= gridSize)

When the second parameter of mod() is positive, and the first parameter to mod is finite, then the output of mod() is always non-negative, including the case where the first parameter to mod() is negative. gridSize is always positive in this context, so we can immediately tell that mod(xMin, gridSize) >= 0 will be true.

Likewise, when the second parameter of mod() is positive, and the first parameter to mod is finite, then the output of mod() is always strictly less than the second parameter. gridSize is always positive in this context, so we can immediately tell that mod(xMin, gridSize) <= gridSize is true because mod(xMin,gridSize) < gridSize for finite inputs and positive gridSize

The same holds for yMin.

We thus deduce that for finite xMin and yMin and positive gridSize, that the if statement will always come out true.

[Walter Roberson]

With a 20 mm diameter coin and a 20 mm grid size, if the rule is that coins crossing multiple squares do not win, then the winning probability is (60*60) / (2^53 * 2^53) which is 225/5070602400912917605986812821504 which is about 4.43734259186819e-29

This does not take into account the possibility that accounting for floating point round-off, that rand()*1200 might potentially not be able to create some of the exact positions needed to win.

[Walter Roberson]

N = 5e8;
M = 30;
matches = 0;
tic
for K = 1 : M
    x = rand(1,N) * 1200;
    p = find(mod(x,20) == 10);
    if ~isempty(p)
        fprintf('iteration %d match at position ', K); disp(p)
    end
    matches = matches + length(p);
end
toc

Elapsed time is 232.675331 seconds.

fprintf('Overall matches: %d out of %d\n', matches, M*N );

Overall matches: 0 out of 15000000000

[Walter Roberson]

More correctly for ensuring that the coin does not overlap the grid edges:

R = coinDiameter/2;

%...
xMinmod = mod(xMin, gridSize);
yMinmod = mod(yMin, gridSize);
if xMinmod >= R & xMinmod <= gridSize-R & yMinmod >= R & yMinmod <= gridSize-R