Monte Carlo Simulation (chutes and ladders)

I'm trying to work a monte carlo simulation for a chutes and ladders game. The board is nine squares. You can only roll a 1 or a 2. If you land on square 3, you are laddered to square 6. if you land on square 8, you chute down to square 2. If you land on square 9, you win the game. I'm trying to determine the average number of rolls it takes to win the game. We are required to run this simulation several times to see if the average changes based on how many times you run the simulation. This is what I have so far. Any help is appreciated.

n=5;

for t=1:n

    pos=0;

    move=0;

    while pos(t)<9

        roll(t)=round(a+(b-a).*rand(1));

        pos(t)=pos(t)+roll(t);

        if pos(t)=3

            then pos(t)=6

        end

        if pos(t)=8

            then pos(t)=2

        end

        move(t)=move(t)+1;

    end

end

Answer 1

[arushi]

Hi E,

Below is a corrected version of your code:

% Parameters
num_simulations = 5; % Number of simulations to run
results = zeros(1, num_simulations); % Array to store results of each simulation

% Simulation loop
for t = 1:num_simulations
    pos = 0; % Starting position
    move = 0; % Move counter
    
    % Play until the player wins
    while pos < 9
        % Roll a 1 or 2
        roll = randi([1, 2]); % Random integer between 1 and 2
        pos = pos + roll; % Update position
        
        % Check for ladder and chute
        if pos == 3
            pos = 6; % Ladder from 3 to 6
        elseif pos == 8
            pos = 2; % Chute from 8 to 2
        end
        
        % Increment move counter
        move = move + 1;
    end
    
    % Store the number of moves for this simulation
    results(t) = move;
end

% Calculate the average number of rolls
average_moves = mean(results);

% Display the results
fprintf('Average number of rolls to win: %.2f\n', average_moves);

Hope this helps.