If a large number of fair N-sided dice are rolled, the average of the simulated rolls is likely to be close to the mean of 1,2,...N i.e. the expected value of one die. For example, the expected value of a 6-sided die is 3.5.
Given N, simulate 1e8 N-sided dice rolls by creating a vector of 1e8 uniformly distributed random integers. Return the difference between the mean of this vector and the mean of integers from 1 to N.
function dice_diff = loln(N)
A=randi([1,N],1e8,1)
M=mean(A)
B=1:N
m=mean(B)
dice_diff =abs(M-m);
end
Here is my code, but it can't work out as it needs too long time to creat A.
7 Comments
Time Descending
后面都加分号就可以解决
It works as expected
N = 6;
A = randi([1,N],1e8,1);
M = mean(A);
B = 1:N;
m = mean(B);
dice_diff = abs(M-m);
Result:
>> dice_diff
dice_diff =
1.7416e-04
The difference is 0.000174 which is the expected result. If it take a long time, you can try to use fewer random numbers.
Thanks Ameer for the help, easy and straightforward.
Best.
I am glad to be of help.
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