monte carlo simulation in matlab two dices roll

Question

isuru senevirathne 2018-5-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/400301-monte-carlo-simulation-in-matlab-two-dices-roll

编辑： Jan 2019-7-15

采纳的回答： Jan

if 2 dices were thrown & there top value were added ,what is the probability of getting a sum of 7 ?

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Jeff Miller 2018-5-13

to count the number of 7's, use a counter variable, e.g., Ctr. Set it to zero before you start the loop, and then add one to it each time your sum equals 7. When the loop finishes, the value in Ctr will be the number of 7's that were found.

Image Analyst 2018-5-13

isuru, here's a tip to format your code correctly. In MATLAB type control-a control-i to properly indent your code. Then copy it and paste it here. Highlight the code and then click the {}Code button. See this link

And note how Jan's properly formatted answer below does not use sum as the name of a variable so as to not override the built-in sum function.

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Answer 1

Jan 2018-5-13

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https://ww2.mathworks.cn/matlabcentral/answers/400301-monte-carlo-simulation-in-matlab-two-dices-roll#answer_320002

编辑：Jan 2019-7-15

在 MATLAB Online 中打开

n        = 10;               % Number or trials
throw    = randi(6, n, 2);   % Value of throws, 2 dice
SumThrow = sum(throw, 2);    % Sum of both dice
Match    = (SumThrow == 7);  % Logical array: 1 if sum of values is 7
count    = sum(Match);       % Number of matching throws

Or with modifications of your code:

count = 0;
for i=1:10
  throws1  = randi(6, 1);
  throws2  = randi(6, 1);
  sumThrow = throws1 + throws2;   % Do not use "sum" as name of a variable
  if sumThrow == 7
    fprintf('sum is 7\n')
    count = count + 1;
  else
    fprintf('false\n')
  end
end
fprintf('Number of throws with sum=7: %d\n', count);

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Answer 2

Md Jilani 2019-7-15

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/400301-monte-carlo-simulation-in-matlab-two-dices-roll#answer_383206

Hello sir, Good Day. How can I get probabilty from here ?

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Jan 2019-7-15

编辑：Jan 2019-7-15

在 MATLAB Online 中打开

Please do not post new questions in the section for answers. Prefer to open your own thread. Thanks.

My code counts the number of 7's in the variable count for n throws. Then the probability is approximately n / count.

You can determine the probability in a constructive way also:

nAll = 0;
n7   = 0;
for throw1 = 1:6
    for throw2 = 1:6
        nAll = nAll + 1;
        if throw1 + throw2 == 7
            n7 = n7 + 1;
        end
    end
end
n7 / nAll

Of course you can do this manually also, because the calculations are very easy: There are 36 possible throws with 2 dice. You get a sum of 7 for 1+6, 2+5, 3+4, 4+3, 5+2 and 6+1: 6 possibilities.

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